kxex2244 notes
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ordinary differential equationTRANSCRIPT
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KXEX 2244 - Ordinary differential equations
Mohamad Bakri Zubir
Monday, 17 February 2014
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Contents
1 Basic concepts 4
1.1 Differential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Ordinary and partial differential equations . . . . . . . . . . . . . . . . 5
1.4 Order and degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Linear and non-linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.7 Initial and boundary conditions . . . . . . . . . . . . . . . . . . . . . . 8
2 First order differential equations 9
2.1 Standard form and differential form . . . . . . . . . . . . . . . . . . . . 9
2.2 Separable differential equations . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Linear differential equations . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3.1 Bernoulli equations . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3.2 Riccati equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.4 Homogeneous differential equations . . . . . . . . . . . . . . . . . . . . 21
2.5 Equations reducible to homogeneous form . . . . . . . . . . . . . . . . 23
2.6 Exact differential equations . . . . . . . . . . . . . . . . . . . . . . . . 27
2.6.1 Nonexact differential equation . . . . . . . . . . . . . . . . . . . 30
1
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23 Second order differential equations 33
3.1 General theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.2 Homogeneous linear equation with constant coefficients . . . . . . . . . 38
3.3 Nonhomogeneous linear equations with constant coefficients . . . . . . 42
3.4 Method of undetermined coefficients . . . . . . . . . . . . . . . . . . . 44
3.5 Reduction of order method . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.6 Variation of parameters method . . . . . . . . . . . . . . . . . . . . . . 52
3.7 Euler-Cauchy equations . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4 Series solutions of ODEs 62
4.1 Algebraic operations on power series . . . . . . . . . . . . . . . . . . . 63
4.2 Linear differential equation of second order . . . . . . . . . . . . . . . . 64
4.3 Ordinary and singular points . . . . . . . . . . . . . . . . . . . . . . . . 65
4.3.1 Ordinary points . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.3.2 Singular points . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.4 Series solution about an ordinary point . . . . . . . . . . . . . . . . . . 68
4.5 Series solution about a regular singular point - the method of Frobenius 74
A Analytic functions 88
B Convergence of power series 89
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History of revisions
These course notes are a work in progress. It will be updated from time to time.As such, it will be perpetually in draft form. New materials, new examples, withinthe scope of the syllabus, will be added on a regular basis. If you spot any errors oromissions please email me at [email protected]. It is hoped that this draft will evolveinto a self contained notes for this course.
Please check Spectrum for any release of updated version.
Version 6.1 - this version
Version 5 - 11 November 2013
Version 4 - 7 October 2013
Version 3 - 18 September 2013
Version 2 - 11 September 2013
Version 1 - 9 September 2013
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Chapter 1
Basic concepts
We introduce the terms and terminologies found in the study of differential equations.
1.1 Differential equations
A differential equation is an equation involving an unknown function and its derivatives.
Examples of differential equations:
Example 1.
dy
dx= 2x+ 7 (1.1)
eyd2y
dx2+ 5
(dy
dx
)2= 7 (1.2)
4d3y
dx3+ (cosx)
d2y
dx2+ 3x2y = 0 (1.3)
(d2y
dx2
)2+ 4y
(dy
dx
)7+ y3
(dy
dx
)2= 5x (1.4)
2y
t2 9
2y
x2= 0 (1.5)
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51.2 Notation
The Liebnitz notationsdy
dx,d2y
dx2,d3y
dx3,d4y
dx4, . . . ,
d2n
dxnare used to represent, respectively,
the first, second, third, fourth, . . . , nth derivatives of y with respect to the independentvariable x.
If the independent variable of y is understood from the context, the expressionsy, y, y, y(4), . . . , y(n) are sometimes used.
1.3 Ordinary and partial differential equations
Ordinary differential equation Differential equations where the unknown functiondepends on only one independent variable
Example 2. Equations (1.1) to (1.4) are ordinary differential equations, since theunknown function y depends only on the variable x.
Partial differential equation Differential equations where the unknown functiondepends on two or more independent variables
Example 3. Equation (1.5) is a partial differential equation having t and x as inde-pendent variables.
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61.4 Order and degree
Order: The order of a differential equation is the highest derivative that appears inthe differential equation.
Example 4. In the equation
x2(d2y
dx2
)3+ 3
(d3y
dx3
)2+ 7
dy
dx= 0
The orders ofd3y
dx3,d2y
dx2and
dy
dxare 3, 2 and 1 respectively. The highest order is 3 and
therefore the order of the differential equation is 3.
Degree: The degree of a differential equation is the power of the highest derivativeterm.
Example 5. In the equation
x2(d2y
dx2
)3+ 3
(d3y
dx3
)2+ 7
dy
dx= 0
The order of the differential equation is 3 andd3y
dx3has power 2. Therefore the degree
of the differential equation is 2.
1.5 Linear and non-linear
Linear: A differential equation is called linear if there are no multiplications amongdependent variables and their derivatives. In other words, all coefficients are functionsof independent variables.
A linear ordinary differential equation has the form
an(x)dny
dxn+ an1(x)
dn1ydxn1
+ + a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = g(x)
where ai(x) and g(x) are functions of x only.
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7Example 6. Equations (1.1) and (1.3) are linear differential equations.
Non-linear: Differential equations that do not satisfy the definition of linear are non-linear.
Example 7. Equations (1.2) and (1.4) are non-linear differential equations.
1.6 Solutions
General solution: Solutions obtained from integrating the differential equations arecalled general solutions. The general solution of a nth order ordinary differential equa-tion contains n arbitrary constants resulting from integrating n times.
Example 8. It can be shown the general solution to the differential equation y+4y = 0is y = A sin 2x+B cos 2x where A, B are constants.
Particular solution: Particular solutions are the solutions obtained by assigning spe-cific values to the arbitrary constants in the general solutions.
Example 9. A few particular solutions to the differential equation in Example 8 are(a) y = 5 sin 2x 3 cos 2x (choose c1 = 5 and c2 = 3, (b) y = sin 2x (choose c1 = 1and c2 = 2), and (c) y = 0 (choose c1 = c2 = 0).
Singular solutions: Solutions that can not be expressed by the general solutions arecalled singular solutions.
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81.7 Initial and boundary conditions
Initial condition: Constraints that are specified at the initial point, generally timepoint, are called initial conditions. Problems with specified initial conditions are calledinitial value problems.
Example 10. The constraints in the problem y+ 2y = ex; y(pi) = 1, y(pi) = 2 givenat x = pi are initial conditions.
Boundary condition: Constraints that are specified at the boundary points, gener-ally space points, are called boundary conditions. Problems with specified boundaryconditions are called boundary value problems.
Example 11. The constraints in the problem y + 2y = ex; y(0) = 1, y(1) = 1 givenat x = pi are boundary conditions because the conditions are given at the differentvalues x = 0 and x = 1.
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Chapter 2
First order differential equations
A first order differential equation is an equation that contains a function y(x) and itsderivative y(x). We will look at the different kinds of first order equations and themethods to solve them.
2.1 Standard form and differential form
A first order differential equation in standard form is the equation
dy
dx= f(x, y).
Example 12. The equationdy
dx=x+ y
y2
is a first order differential equation in standard form.
An equation is in differential form if it is expressed as
M(x, y) dx+N(x, y) dy = 0 (2.1)
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Example 13. The equation
(2x y) dx+ (x2 + y2) dy = 0
is in differential form.
2.2 Separable differential equations
Consider a differential equation in differential form (2.1). If M(x, y) = A(x), a func-tion only of x and N(x, y) = B(y), a function only of y, the differential equation isseparable, or has its variables separated. The equation
B(y) dy = A(x) dx
The equation can be solved by integrating both sides.
Example 14. Solve the equation 9ydy
dx+ 4x = 0.
Solution. Separating the variables we get 9y dy = 4x dx. Integrating9y dy =
4x dx
9y2
2= 2x2 +A, A a constant
4x2 + 9y2 = C, where C = 2A.
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Example 15. Solve the equation y = 1 + y2.
Solution. Equation is dydx = 1 + y2. Separating the variable we get
dy
1 + y2= dx. Inte-
grating, dy
1 + y2=
1 dx
tan1 y = x+A, where A is an arbitrary constant
y = tan(x+A).
Example 16. Solve the equationdy
dx=y + 1
x 4 subject to the condition y(6) = 0.
Solution. Separating the variables we getdy
1 + y=
dx
x 4 . Integrating,dy
1 + y=
dx
x 4ln |1 + y| = ln |x 4|+ lnA|1 + y| = A|x 4|
which implies
1 + y = A(x 4),where A > 0
The initial condition x = 6, y = 0 implies A = 12 . Therefore y =12x 3.
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2.3 Linear differential equations
A first order differential equation expressed in the form
dy
dx+ p(x)y = q(x) (2.2)
is called a linear differential equation, where p(x) and q(x), are functions of x only.
Equation (2.2) is solved by multiplying it by the integrating factor I(x). The integratingfactor is
I(x) = ep(x) dx.
Equation (2.2) becomes
ep(x) dx dy
dx+ p(x)e
p(x) dx y = q(x) e
p(x) dx. (2.3)
Observe that the left hand side of this equation can be expressed as
ep(x) dx dy
dx+ p(x)e
p(x) dx y =
d
dx
(ep(x) dx y).
Equation (2.3) becomes
d
dx
(ep(x) dx y) = q(x) e p(x) dx.
Integrate both sides to obtaind
dx
(ep(x) dx y) dx = q(x) e p(x) dx dxep(x) dx y =
q(x) e
p(x) dx dx+ c.
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Example 17. Solve the equation y y = e4x.
Solution. The integrating factor is I(x) = e 1 dx = ex. Multiply the equation by the
integrating factor and integrate to obtain
ex y ex y = ex e4xd
dx
(ex y
)= e3x
d
dx
(ex y
)dx =
e3x dx
ex y =e3x
3+ c
y =1
3e4x + ce3x.
Example 18. Solve the equation xy + y + 6 = 0.
Solution. In linear form
y +1
xy = 6
x
The integrating factor is I(x) = e1/x dx = eln x = x. Multiply the equation by the
integrating factor and integrate to obtain
x y + y = 6d
dx(x y) = 6
d
dx(x y) dx =
6 dx
x y = 6x+ cy = 6 + c
x.
Example 19. Solve the initial value problem y + y tanx = sin 2x, y(0) = 1.
Solution. In linear formy + (tanx) y = sin 2x
The integrating factor is I(x) = etan x dx = e ln | cos x| = secx. Multiply the equation
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by the integrating factor and integrate to obtain
(secx) y + (secx tanx) y = secx sin 2x
(secx) y + (secx tanx) y =1
cosx2 sinx cosx
d
dx(secx y) = 2 sinx
d
dx(secx y) dx =
2 sinx dx
secx y = 2 cosx+ c.
When x = 0, y = 1, implies 1 = 2 + c so that c = 3. Thereforey
cosx= 3 2 cosx+ c
Hencey = 3 cosx 2 cos2 x.
Example 20. Solve the initial value problemdy
dx=
y
2x+ 3y2 2 , y(1) = 1.
Solution. The differential form is not linear in y. However if we let y to be theindependent variable, then
dx
dy=
2x+ 3y2 2y
=2
yx+
3y2 2y
dx
dy 2yx =
3y2 2y
which is linear in x.
The integrating factor is I(y) = e2y dy = e2 ln |y| = 1y2 .
Multiply the equation by the integrating factor and integrate
1
y2dx
dy 2y3x =
3y2 2y3
d
dy
(1
y2x
)=
3
y 2y3
d
dy
(1
y2x
)dy =
3
y 2y3dy
1
y2x = 3 ln y +
1
y2+ c.
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When x = 1, y = 1, which implies 1 = 0 + 1 + c so that x = 3y2 ln y + 1.
2.3.1 Bernoulli equations
A Bernoulli equation is a first order differential equation of the form
dy
dx+ p(x)y = r(x)yn (2.4)
where n 6= 0, 1.If n = 0 or n = 1 then (2.4) is a linear differential equation. For any other value of n,the equation can be reduced to the linear form by making the substitutions
v = y1n anddv
dx= (1 n)yn dy
dx.
Multiply equation (2.4) by (1 n)yn to obtain
(1 n)yn dydx
+ (1 n)yn p(x)y = (1 n)yn r(x)yn
(1 n)yn dydx
+ (1 n)p(x)y1n = (1 n)r(x)
dv
dx+ (1 n)p(x)v = (1 n)r(x)
Example 21. Solve dydx + 3x2y = xex
3
y2.
Solution. The equation is not linear but Bernoullis with n = 2. Make the substitution
v = y12 = y1. Thendv
dx= y2 dy
dx.
Multiply the equation by y2 to obtain
y2 dydx 3x2y1 = xex3
dv
dx 3x2v = xex3 .
The equation is now linear, the integrating factor is I(x) = e 3x2 dx = ex3 . Then
ex3 dv
dx 3x2ex3v = x
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d
dx
(ex
3
v)
= xd
dx
(ex
3
v)dx =
x dx
ex3
v = x2
2+ c
ex3
y1 = x2
2+ c
1
y=ex
3
2(2c x2)
y =2
ex3(A x2) where A = 2c .
Example 22. Solve the equation 3dy
dx+ y = (1 2x)y4.
Solution. It is a Bernoulli equation with n = 4 when it is expressed as
dy
dx+(13
)y =
(1 2x)3
y4.
Make the substitution v = y14 = y3. Thendv
dx= 3y4 dy
dx. Multiply the equation by
3y4,
3y4 dydx 3y4 ( 13 )y = 3y4
(1 2x)3
y4
3y4 dydx y3 = (1 2x)
dv
dx v = 2x 1.
The equation is now linear, the integrating factor is I(x) = e 1 dx = ex. Multiplying
the linear equation by the integrating factor
exdv
dx exv = (2x 1)ex
d
dx
(exv
)= (2x 1)ex
d
dx
(exv
)dx =
(2x 1)ex dx
exv = (2x 1)(ex)
2(ex) dx
= (1 2x)ex +
2ex dx
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= (1 2x)ex 2ex + c
exy3 = 2xex ex + c
1
y3= 2x 1 + cex
y3 =1
cex 2x 1 .
2.3.2 Riccati equation
A Riccati equation is an equation of the form
y + a(x)y = b(x) + c(x) y2. (2.5)
Notice that the equation is linear except for the term c(x) y2.
If a particular solution u of a Riccati is known, the general solution of the equation isfound by making the substitution
y = u+1
z(2.6)
where z is a function of x.
When (2.6) is substituted into equation (2.5), it becomes a linear equation,(u+
1
z
)+ a
(u+
1
z
)= b+ c
(u+
1
z
)2u 1
z2z + au+ a
1
z= b+ cu2 + 2cu
1
z+ c
1
z2
and since u + au = b+ cu2, equation reduces to
1z2z + a
1
z= 2cu
1
z+ c
1
z2,
then multiply by z2 and rearrange,z + (2cu a)z = c.
which is a linear equation in z. Solve to obtain a general solution in z and hence in y.
Note: There is no particular algorithm to find the particular solution u. One may usetrial and error to find u by looking the functions a, b and c.
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Example 23. Solve y = (y x)2 + 1, where u(x) = x is particular solution, subjectto the condition y(0) = 12 .
Solution. The equation is y = y2 2xy + x2 + 1 or y + 2xy = x2 + 1 + y2 which isBernoulli.
First, we verify that u(x) = x is a solution to the equation. Then u(x) = 1. Substituteinto the equation
LHS = 1, RHS = (x x)2 + 1 = 1, LHS = RHS
Therefore u(x) = x is a solution.
Second, solve by taking the substitution y = x+ 1z . Then y = 1 1z2 z. Upon substitution,
1 1z2z + 2x
(x+
1
z
)= x2 + 1 +
(x+
1
z
)21 1
z2z + 2x2 + 2x
1
z= x2 + 1 + x2 + 2x
1
z+
1
z2
1z2z =
1
z2,
and multiply by z2,
z = 1.
Therefore z = x+A for some constant A. Since y = x+ 1z , then z = 1yx , so that
1
y x = x+A or y =1
A x + x.
The initial condition says, when x = 0, y = 12 , so that
1
2=
1
A 0 + 0 which implies A = 2
Therefore
y =1
2 x + x.
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Example 24. Solve the equation y+ yx = 1x2 +y2 for x > 0 with a particular solutionu(x) = 1x subject to y(1) = 2.
Solution. We verify that u(x) = 1x is a solution.
LHS = 1x2
+1/x
x= 0, RHS = 1
x2+
(1
x
)2= 0, LHS = RHS
Therefore u(x) = 1x is a particular solution.
Make the substitution y = 1x +1z so that y
= 1x2 1z2z,
y +y
x= 1
x2+ y2(
1x2 1z2z)
+1
x
(1
x+
1
z
)= 1
x2+
(1
x+
1
z
)2
1x2 1z2z +
1
x2+
1
x 1z
= 1x2
+1
x2+
2
x 1z
+1
z2
1z2z 1
x 1z
=1
z2,
and multiply by z2,
z +1
xz = 1.
The equation is linear with integrating factor I(x) = e1/x dx = eln x = x. Multiply by
the integrating factor,
x z + z = xd
dx(xz) = x
d
dx(xz) dx =
x
xz = x2
2+A =
2A x22
z =2A x2
2x
1
z=
2x
B x2 , where B = 2A.
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Then
y =1
x+
2x
B x2 .
When x = 1, y = 2 so that
2 =1
1+
2 1B 1 which implies B = 3.
Therefore
y =1
x+
2x
3 x2 .
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2.4 Homogeneous differential equations
A function f(x, y) is said to be homogeneous of degree n if for all ,
f(x, y) = nf(x, y).
Example 25. The function f(x, y) = x4 + xy3 is homogeneous of degree 4 because
f(x, y) = (x)4 + (x)(x)3
= 4(x4 + xy3)
= 4f(x, y).
Example 26. The function f(x, y) =y xey/x
yis homogeneous of degree 0 because
f(x, y) =(y)4 + (x)(ex/y)
y
=y xey/x
y
= 0f(x, y).
Example 27. The function y2xy+1 is not homogeneous since the term 1 is of degree0 whereas the other terms are of degree 2.
A differential equation of the form
dy
dx=f(x, y)
g(x, y)
is a homogeneous equation if both the functions f(x, y) and g(x, y) are homogeneousof the same degree.
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To solve a homogeneous differential equation we make the substitution y = vx. Withthis substitution, the equation becomes separable in v and x.
Example 28. Solve 2xydy
dx= y2 x2.
Solution. In standard form the equation is
dy
dx=y2 x2
2xy,
where the numerator and denominator are homogeneous of degree 2.
Let y = vx. Thendy
dx= v + x
dv
dx. Substitute into the equation
dy
dx=y2 x2
2xy
v + xdv
dx=v2x2 x2
2vx2
xdv
dx=v2 1
2v v
= 1 + v2
2v2v
1 + v2dv = 1
xdx,
and integrating, 2v
1 + v2dv =
1xdx
ln(1 + v2) = ln |x|+ lnA
1 + v2 =A
x, x > 0.
Substitute v = yx to obtain
1 +y2
x2=A
x,
which impliesx2 + y2 = Ax.
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2.5 Equations reducible to homogeneous form
The equation of the formdy
dx=a1x+ b1y + c1a2x+ b2y + c2
(2.7)
can be reduced to a homogeneous form by making a substitution depending on thevalues of the constants a1, b1, a2, b2.
Case 1
If
a1 b1a2 b2 6= 0, choose the substitutions
{x = X + h
y = Y + kwhere the pair of constants
(h, k) is found by solving the simultaneous equations
a1x+ b1y + c1 = 0
a2x+ b2y + c2 = 0.
Then equation (2.7) reduces to the homogeneous equation
dY
dX=a1X + b1Y
a2X + b2Y.
To solve this equation, make the substitution Y = V X.
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Case 2
If
a1 b1a2 b2 = 0, make the substitution z = a1 + b1y or z = a2 + b2y.
Example 29. Solve (x 2y + 4) dx+ (2x y + 2) dy = 0.
Solution. In standard form, the equation is
dy
dx= x 2y + 4
2x y + 2 .
Here
a1 b1a2 b2 = 1 22 1
6= 0.To find the pair (h, k) solve the simultaneous equation
x 2y + 4 = 02x y + 2 = 0
to obtain h = 0, k = 2.
Make the substitution
{x = X + 0
y = Y + 2so that
{dx = dX
dy = dY.
The differential equation becomesdY
dX= X 2Y
2X Y which is a homogeneous equation.
Make the substitution Y = V X so thatdY
dX= V +X
dV
dX. Equation becomes
V +XdV
dX= X 2V X
2X V XXdV
dX=
2V 12 V V
= V2 1V 2
V 2V 2 1 dV =
1
XdX,
and integrating, V 2V 2 1 dV =
1
XdX
3/2
V + 1 1/2V 1 dV =
1
XdX
3
2ln |V + 1| 1
2ln |V 1| = lnA lnX
ln|V + 1|3|V 1| = ln
B
X2, where B = A2,
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which implies
(V + 1)3
V 1 =B
X2.
Since V = Y/X,
( YX + 1)3
YX 1
=B
X2
(Y +X)3
X3 XY X =
B
X2
(Y +X)3 = B(Y X),
and since
{X = x
Y = y 2,
(y 2 + x)3 = B(y 2 x)
or
(y + x 2)3 = B(y x 2).
Example 30. Solve (2x 4y + 5) dy + (x 2y + 3) dx = 0.
Solution. In standard form, the equation is
dy
dx= x 2y + 3
2x 4y + 5 .
Here
a1 b1a2 b2 = 1 22 4
= 0.Make the substitution z = x 2y so that dz
dx= 1 2dy
dxor dydx =
12 (1 dzdx ). Equation
becomes
1
2
(1 dz
dx
)= z + 3
2z + 5
dz
dx= 1 + 2 2z + 3
2z + 5= 1 +
4z + 6
2z + 5=
4z + 11
2z + 5
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The equation is separable, 2z + 5
4z + 11dz =
1 dx
1
2 1/2
4z + 11dz =
1 dx
1 1
4z + 11dz =
2 dx
z 14
ln |4z + 11| = 2x+A
4z ln |4z + 11| = 8x+B where B = 4A
4(x 2y) ln |4(x 2y) + 11| = B
4x+ 8y + ln |4x 8y + 11| = B.
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2.6 Exact differential equations
A first order differential equation, expressed in differential form
M(x, y) dx+N(x, y) dy = 0 (2.8)
is said to be an exact differential equation if there is a function u(x, y) such that
du(x, y) = M(x, y) dx+N(x, y) dy.
Method of solution
When equation (2.8) is determined to be exact, solve either one of the equations
u
x= M(x, y)
or
u
y= N(x, y)
for u(x, y).
If we choose ux
= M(x, y), then
u =
M(x, y) dx+ k(y). (2.9)
Here the constant of integration is some function k(y).
To determine k(y), partially differentiate u with respect to y. Then
u
y=
y
M(x, y) dx+ k(y) = N(x, y).
Then
k(y) = N(x, y) y
M(x, y) dx. (2.10)
Finally, integrate (2.10) with respect to y and substitute the result in (2.9).
The solution to (2.8) is given byu(x, y) = c
where c is an arbitrary constant.
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28
Likewise if you start with uy
= N(x, y), then
u =
N(x, y) dy + h(x). (2.11)
Here the constant of integration is some function h(x).
To determine h(x), partially differentiate u with respect to x. Then
u
x=
x
N(x, y) dy + h(x) = M(x, y).
Then
h(x) = M(x, y) x
N(x, y) dy. (2.12)
Finally, integrate (2.12) with respect to x and substitute the result in (2.11) and thesolution to (2.8) is given by
u(x, y) = c
for some constant d.
Example 31. Solve the equation xy + y + 4 = 0.
Solution. In differential form, the equation is
(y + 4) M(x,y)
dx+ xN(x,y)
dy = 0.
Now My = 1 andNx = 1. Therefore the equation is exact.
If we choose to solveu
x= y + 4 then
u
xdx =
y + 4 dx
u = xy + 4x+ k(y), for some function k(y),
thenu
y= x+ k(y) = x = N(x, y).
Thereforek(y) = 0 which implies k(y) = A
for some constant A. Hence solution is
xy + 4x+A = c
xy + 4x u(x,y)
= c1, where c1 = cA
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29
If we choose to solveu
y= x then
u
ydy =
x dy
u = xy + h(x), for some function h(x)
thenu
x= y + h(x) = y + 4 = M(x, y).
Thereforeh(x) = 4 which implies h(x) = 4x+B
for some constant B. Hence solution is
xy + 4x+B = c
xy + 4x u(x,y)
= c2, where c2 = cB
In either choices, we obtain the same solution.
Example 32. Solve 2x sin 3y dx+ (3x2 cos 3y + 2y) dy = 0.
Solution. Let M(x, y) = 2x sin 3y and N(x, y) = 3x2 cos 3y+ 2y. Check for exactness,
M
y= 6x cos 3y,
N
x= 6x cos 3y.
Since My =Nx , the equation is exact.
Letu
x= M(x, y) = 2x sin 3y.
Then
u =
2x sin 3y dx = x2 sin 3y + k(y).
To find k(y), differentiate u with respect to y to obtain
u
y= 3x2 cos 3y + k(y) = N(x, y) = 3x2 cos 3y + 2y.
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30
This implies k(y) = 2y so that k(y) = y2. Therefore solution is
u(x, y) = x2 sin 3y + y2 = c.
2.6.1 Nonexact differential equation
Suppose the equationP (x, y) dx+Q(x, y) dy = 0 (2.13)
is non-exact, that is Py6= Q
x. The equation can be make exact by multiplying it by
F (x), a function of x only or by G(y) a function of y only. (It can also be made exactby multiplying it with a function of both x and y, I(x, y) but this is beyond the scopeof this course)
The function that the equation is multiplied by to make it exact, be it F (x) or G(y),is called the integrating factor.
How to find F (x)
Suppose the equation (2.13) becomes exact when multiplied by F (x), that is the equa-tion
FP dx+ FQdy = 0 (2.14)
is exact. Then
(FP )
y=(FQ)
x
FP
y+ 0 P = F Q
x+dF
dxQ
dF
dxQ = F
(P
y Qx
)1
F
dF
dx=
1
Q
(P
y Qx
),
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31
integrating, 1
F
dF
dxdx =
1
Q
(P
y Qx
)dx
lnF =
1
Q
(P
y Qx
)dx
F (x) = e
1Q(
Py Qx ) dx.
How to find G(y)
Exercise 1. By similar method show that
G(y) = e
1P (
Qx Py ) dy.
To see whether the integrating factor is a function of x or y, the expression in F (x)must contain x only and the expression in G(y) must contain y only. It cannot containboth.
Example 33. Show that the equation1
ydx+ 2x dy = 0 non-exact. Find the integrat-
ing factor as a function of y only, and show that by multiplying the equation by theintegrating factor it is exact. Hence solve the equation.
Solution. Let P (x, y) = 1y and Q(x, y) = 2x. ThenPy = 1y2 and Qx = 2. Since
Py 6= Qx the equation is non-exact.
Now 1
P
(Q
x Py
)dy =
1
1/y
[2 (1/y2)] dy
=
2y +
1
ydy
= y2 + ln y.
The integrating factor is then
G(y) = ey2+ln y = yey
2
.
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32
Multiply the equation by the integrating factor to obtain the equation
ey2
M(x,y)
dx+ 2xey2
N(x,y)
dy = 0.
ThenM
y= 2xyey
2
=N
x.
Therefore the equation is exact.
Solve the exact equation to obtain
x ey2
= c.
Example 34. Solve 2x tan y dx+ sec2 y dy = 0.
Solution. Let P (x, y) = 2x tan y and Q(x, y) = sec2 y. Then Py = 2x sec2 y and
Qx = 0. Therefore the equation is non-exact. We look for an integrating factor as afunction of x.
1
Q
(P
y Qx
)dx =
1
sec2 y[2x sec2 y 0] dx
=
2x dx
= x2.
The integrating factor is then
F (x) = ex2
.
Multiply the equation by the integrating factor to obtain the equation
2xex2
tan y M(x,y)
dx+ ex2
sec2 y N(x,y)
dy = 0.
ThenM
y= 2xex
2
sec2 y =N
x.
Therefore the equation is exact.
Solve the exact equation to obtain
ex2
tan y = c.
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Chapter 3
Second order differential equations
In this chapter we will study important class of second order equations. These equationshave important applications in science and engineering.
3.1 General theory
A linear ODE of order 2 has the form
a2(x)y + a1(x)y + a0(x)y = g(x) (3.1)
where a2(x), a1(x), a0(x), g(x) are functions of x called the coefficients of the equation.
Divide equation (3.1) by a2(x) to obtain
y + p(x)y + q(x)y = r(x) (3.2)
where p(x) = a1(x)a2(x)
, q(x) = a0(x)a2(x)
and r(x) = g(x)a2(x)
.
We will use this form in subsequent discussions. This form is in standard form.
If r(x) = 0 then (3.2) has the form
y + p(x)y + q(x)y = 0 (3.3)
and is then called homogeneous to indicate that all terms are of the first degree in yand its derivatives. The original equation (3.2) is called nonhomogeneous because itcontains a term r(x) which does not depend on y. (Note, however, that this use of theterm homogeneous has a different meaning for first order differential equations)
33
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34
Example 35. The equation
(1 x)y 3xy + 6y = 0
is a homogeneous second order linear differential equation.
Example 36. The equation
y + 3y + 9y = ex cosx
is a nonhomogeneous second order linear differential equation.
This chapter will only discuss two cases:
Linear differential equation with constant coefficients
ay + by + cy = r(x)
The Euler-Cauchy equation
x2y + axy + by = r(x)
Those with other variable coefficients will be considered in the next chapter.
There are results pertaining to a given a homogeneous equation (3.3) and its solutions.
Theorem 1. If y = y1(x) is a solution of (3.3) then y = c1y1(x), where c1 is anarbitrary constant, is also a solution.
If y = y1(x) and y = y2(x) are solutions of (3.3) then the linear combination
y = c1y1(x) + c2y2(x)
where c1 and c2 are arbitrary constants, is also a solution.
Note: The theorem is not applicable to nonhomogeneous or nonlinear equation.
-
35
Example 37. Verify that y = c1 sinx + c2 cosx is a solution of the homogeneousequation
y + y = 0
where c1 and c2 are arbitrary constants.
Solution. Let y1 = sinx. Then y1 = sinx = y1. This implies y1 + y1 = 0.
Similarly let y2 = cosx. Then y2 = cosx = y2. This implies y2 + y2 = 0.
This shows that y1 = sinx and y2 = cosx are both solutions of the equation. Hence bytheorem (1), their linear combination y = c1 sinx+ c2 cosx is also a solution.
Definition 1. Two functions y1(x) and y2(x) are said to be linearly independent if theequality
k1y1(x) + k2y2(x) = 0 (3.4)
holds only when k1 = k2 = 0, i.e. when y1(x) and y2(x) are not proportional tothe other. Otherwise, y1(x) and y2(x) are linearly dependent if (3.4) holds for someconstants k1 and k2, not both zero.
Example 38. Determine whether the following functions y1(x) and y2(x) are linearlydependent or independent:
(a) y1(x) = e2x, y2(x) = x+ 1
(b) y1(x) = sin 2x, y2(x) = sinx cosx
Solution.
(a) Suppose k1e2x + k2(x+ 1) = 0.
differentiate w.r.t. x : 2k1e2x + k2 = 0, (*)
differentiate again w.r.t. x : 4k1e2x = 0
but for all real x, e2x 6= 0, which implies k1 = 0substitute k1 = 0 into (*) to obtain k2 = 0
that is k1e2x + k2(x + 1) = 0 implies k1 = 0 and k2 = 0. Hence y1(x) = e
2x andy2(x) = x+ 1 are linearly independent.
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36
(b) The functions sin 2x and sinx cosx are proportional since sin 2x = 2 sinx cosx.Choose k1 = 1 and k2 = 2.
Hence there exist a k1 and k2 not both zero such that k1 sin 2x + k2 sinx cosx = 0.Therefore y1(x) = sin 2x and y2(x) = sinx cosx are linearly dependent.
To determine the linear independence of two solutions of a differential equation we canuse the Wronskian.
Definition 2. Given two differentiable functions y1(x) and y2(x), the Wronski deter-minant or Wronskian is defined as
W (y1, y2) =
y1 y2y1 y2 = y1y2 y2y1
Theorem 2. If the Wronskian of the differentiable functions y1(x) and y2(x) is nonzeroat any point, then y1(x) and y2(x) are linearly independent. (Equivalently, if y1(x) andy2(x) are linearly dependent then W (y1, y2) = 0).
Caution: For arbitrary functions y1(x) and y2(x), the Wronskian does not provide atest for dependence. As a counterexample, let y1(x) = x|x| and y2(x) = x2. Then theWronskian of y1 and y2 is ALWAYS zero, but these functions are INDEPENDENT.
1
However: If y1(x) and y1(x) are both solutions to the equation (3.3), and if the Wron-skian is zero at any point in the domain, then it is zero everywhere and y1(x) and y2(x)are dependent.1
Example 39. Show that the functions y1(x) = cos 3x and y2(x) = sin 3x are linearlyindependent for all x.
Solution. The Wronskian is
W (y1, y2) =
y1 y2y1 y2 = cos 3x sin 3x3 sin 3x 3 cos 3x
= 3 cos2 3x+ 3 sin2 3x = 3
Hence by Theorem (2), y1(x) and y2(x) are linearly independent for all x.
1http://math.berkeley.edu/~mcivor/math54s11/wronskian.pdf
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37
If we know any two linearly independent solutions, then
Theorem 3. If y1(x) and y2(x) are any two linearly independent solutions of the ho-mogeneous equations (3.3), then the general solution of (3.3) is given by
y = c1y1(x) + c2y2(x)
where c1 and c2 are arbitrary constants.
Example 40. Show that both functions y1 =1x and y2 = x
3 are solutions to theequation
x2y xy 3y = 0.What is the general solution?
Solution. We have y1 = 1x2 and y1 = 2x3 . Substitute into the LHS of the equation:LHS = x2y xy 3y = x2( 2x3 ) x( 1x2 ) 3x = 0.
Likewise y2 = 3x2 and y2 = 6x. Substitute into the LHS of the equation:
LHS = x2y xy 3y = x2(6x) x(3x2) 3x3 = 0.
Hence y1(x) and y2(x) are solutions of the equations.
The Wronskian of y1(x) and y2(x) is
W (y1, y2) =
x1 x3x2 3x2 = 3x+ x = 4x
If x 6= 0 then y1(x) and y2(x) are linearly independent. Hence the general solution forthe homogeneous equation is
y(x) = c1x1 + c2x3
where c1 and c2.
Note: If y1(x) and y2(x) are solutions of a second order homogeneous equations thelinear combination
y(x) = c1y1(x) + c2y2(x) (3.5)
is also a solution. But it does not say (3.5) is the general solution. Only when we havedetermined y1(x) and y2(x) are linearly independent, then (3.5) is the general solution.
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38
3.2 Homogeneous linear equation with constant co-
efficients
Definition 3. The differential equation
ay + by + cy = g(x) (3.6)
where a, b and c are constants, is known as a linear equation with constant coefficients.
The complementary homogeneous linear equation of (3.6) is given by
ay + by + cy = 0. (3.7)
The solution to (3.7) is found by considering the function y(x) = emx. Then
y(x) = memx and y(x) = m2emx.
Substituting these expressions into (3.7), we get
(am2 + bm+ c) emx = 0.
But emx 6= 0, hence y(x) = emx is a solution of (3.7) provided that m satisfies thequadratic equation
am2 + bm+ c = 0. (3.8)
Equation (3.8) is known as the characteristic equation. The general solution of thehomogeneous linear equation with constant coefficients (3.7) will depend on the natureof the roots m1 and m2 of the characteristic equation (3.8).
Case I: two distinct real roots (b2 4ac > 0)The two linearly independent solutions are y1 = e
m1x and y2 = em2x. The general
solution isy(x) = c1e
m1x + c2em2x.
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39
Example 41. Solve the equation y + 2y 3y = 0.
Solution. The characteristic equation is m2 + 2m 3 = 0. Solve to obtain m = 3, 1.Hence the general solution is
y(x) = c1e3x + c2ex.
Example 42. Solve the initial value problem y + 5y + 6y = 0 subject to the initialconditions y(0) = 1.6 and y(0) = 0.
Solution. The characteristic equation is m2+5m+6 = 0. Solve to obtain m = 3,2.Hence the general solution is
y(x) = Ae3x +Be2x
When x = 0, y = 1.6 so that A+B = 1.6.
Now y(x) = 3Ae3x 2Be2x and when x = 0, y = 0 so that 3A 2B = 0.
Solve the simultaneous equations to obtain A = 3.2, B = 4.8. Hence the solution is
y(x) = 3.2e3x + 4.8e2x
Case II: complex conjugate roots (b2 4ac < 0)Suppose the roots of characteristic equation are i. Then the general solution willbe
y(x) = Ae(+i)x +Be(i)x
= ex(Ae(i)x +Be(i)x)
= ex[A(cos x+ i sin x) +B(cos x i sin x)]= ex[(A+B) cos x+ i(AB) sin x]= ex[c1 cos x+ c2 sin x]
where c1 = A+B and c2 = i(AB).
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40
Example 43. Solve the equation y 2y + 10y = 0, y(0) = 4 and y(0) = 1.
Solution. The characteristic equation is m2 2m + 10 = 0 having roots m = 1 3i.The general solution is
y(x) = ex(c1 cos 3x+ c2 sin 3x)
Given y(0) = 4 implies c1 = 4.
To find c2, differentiate the general solution to obtain
y(x) = ex(3c1 sin 3x+ 3c2 cos 3x) + ex(c1 cos 3x+ c2 sin 3x)
the initial condition implies
1 = 3c2 + 4+
c2 = 1
Therefore the solution is y(x) = ex(4 cos 3x sin 3x).
Example 44. Solve the initial value problem 4y + 16y + 17y = 0 subject to thecondition y(0) = 0.5 and y(0) = 1.
Solution. The characteristic equation is 4m2 + 16m + 17 = 0 and having roots m =2 12 i. The general solution is
y(x) = e2x(c1 cos
x
2+ c2 sin
x
2
)The initial condition y(0) = 12 implies c1 = 12 .
Differentiate the general solution
y(x) = e2x(c1
2sin
x
2+c22
cosx
2
) 2e2x
(c1 cos
x
2+ c2 sin
x
2
)= e2x
[(2c1 + c2
2) cos
x
2+ (c1
2 2c2) sin x
2
]The condition y(0) = 1 implies c2 = 0. The solution is theny(x) = 12e2x cos x2 .
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41
Case III: repeated roots (b2 4ac = 0)The only root is m = b
2aso that y1(x) = e
mx is a solution. To find the second linearlyindependent solution we can use Lagranges method of reduction of order (see page 49).
The second linearly independent solution is given by y2(x) = xemx. Therefore the
general solution isy(x) = c1e
mx + xemx.
Example 45. Solve the initial value problem y 4y + 4y = 0 subject to the initialconditions y(0) = 3 and y(0) = 1.
Solution. The characteristic equation is m2 4m+ 4 = 0 having roots m = 2, 2. Thegeneral solution is y(x) = c1e
2x + c2xe2x.
The condition y(0) = 3 implies c1 = 3.
Now y(x) = 2c1e2x + c2(x2e2x + e2x) = (2c1 + c2) e2x + 2c2 xe2x and the conditiony(0) = 1 implies c2 = 5.
Hence the solution is y(x) = 3e2x 5xe2x.
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42
3.3 Nonhomogeneous linear equations with constant
coefficients
So far the homogeneous second order linear differential equations with constant coef-ficients have been considered. Such equations will only describe physical systems withfree vibration or oscillations. Most often mechanical or electrical systems are subjectedto an applied force or emf.
The nonhomogeneous equation is
ay + by + cy = g(x)
where a, b, c are constants.
Motivating example
As a motivation for solving the general equation, let us first solve the following particularnonhomogeneous equation
y 4y + 3y = 10e2x (3.9)
The left hand side of the equation can be written as
d
dx
(dy
dx y) 3
(dy
dx y)
because
d
dx
(dy
dx y) 3
(dy
dx y)
=d2y
dx2 dydx 3dy
dx+ 3y
=d2y
dx2 4dy
dx+ 3y.
By taking the substitution z =dy
dx y, equation (3.9) can be expressed as
dz
dx 3z = 10e2x.
This is a first order linear equation with integrating factor e3x. Then
d
dx(e3xz) = 10e5x
e3xz =
10e5x dx = 2e5x + A
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43
z = 2e2x + Ae3x.
Substituting z back,dy
dx y = 2e2x + Ae3x,
which is also a linear equation with integrating factor ex. This implies
d
dx(exy) = 2e3x + Ae2x
exy =2
3e3x +
A
2e2x +B
y =2
3e2x +
A
2e3x +Bex.
That is, it has the form
y(x) = c1ex + c2e
3x yc
+2
3e2x yp
.
We observed that the sum of the first two terms, yc called the complementary function,is the general solution of the corresponding complementary homogeneous equation,whilst the third term, yc is a particular solution or integral of the nonhomogeneousequation.
The above result is generalized in the theorem:
Theorem 4. The general solution of y + p(x)y + q(x)y = r(x) is
y(x) = yc(x) + yp(x)
where the complementary function yc(x) is the general solution of the homogeneousequation y + p(x)y + q(x)y = 0 and yp(x) is a particular solution.
Example 46. Verify that yp(x) =23e2x is a particular solution of y 4y + 3y =
10e2x.
Solution. If yp(x) =23e2x then yp(x) = 43e2x and yp (x) = 83e2x.
The LHS is y 4y+ 3y = 83e2x 4 ( 43e2x) + 3( 23e2x) = 10e2x which equals theRHS.
We shall discuss two methods of finding a particular solution.
-
44
3.4 Method of undetermined coefficients
The method of undetermined coefficients is suitable for linear ODEs with constantcoefficients a and b
y + ay + by = r(x) (3.10)
where r(x) is
an exponential function, ex, a power of x, xn, a sine or cosine, sinx, cosx sum or product of such functions.
The solution of (3.10) is the sum of a general solution yh of the corresponding homo-geneous equation
y + ay + by = 0 (3.11)
and a particular solution, yp of (3.10).
The function yp is chosen so that yp is similar to r(x). The unknown coefficients in ypis determined by substituting yp and its derivative into the ODE.
The following table shows r(x) and suitable choices of yp and the rules for the correctform of yp.
Term in r(x) Choice for ypkex Cex
kxn(n = 0, 1, . . . ) Knxn +Kn1xn1 + +K1x+K0
k cosxk sinx
}K cosx+M sinx
kex cosxkex sinx
}ex(K cosx+M sinx)
Table 3.1: Choices for yp
-
45
The rules are:
(a) Basic Rule. If r(x) in (3.10) is one of the functions in the first column in Table3.1, choose yp in the same line and determine its undetermined coefficients bysubstituting yp and its derivatives into (3.10).
(b) Modification Rule. If a term in your choice for yp happens to be a solution ofthe homogeneous ODE corresponding to (3.10), multiply this term by x (or byx2 if this solution corresponds to a double root of the characteristic equation ofthe homogeneous ODE).
(c) Sum/Product Rule. If r(x) is a sum of functions in the first column of Table3.1, choose yp the sum of the functions in the corresponding lines of the secondcolumn.
Example 47. Solve the equation y + 4y = 8x2.
Solution. The characteristic equation of the homogeneous equation is m2+4 = 0 whichimplies m = 2i. Therefore the solution of the homogeneous equation is
yh = c1 cos 2x+ c2 sin 2x
where c1, c2 are arbitrary constants.
We have r(x) = 8x2. From Table 3.1, the suitable choice for yp is K2x2 + K1x + K0.
Then
yp = 2K2x+K1yp = 2K2.
Since yp is also a solution, then
yp + 4yp = 8x2
2K2 + 4(K2x2 +K1x+K0) = 8x
2.
Comparing the coefficients
x2 : 4K2 = 8
x : 4K1 = 0 K2 = 2,K1 = 0,K0 = 1x0 : 2K2 + 4K0 = 0
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46
Thus the particular solution is yp = 2x2 1. Hence the general solution is
y(x) = c1 cos 2x+ c2 sin 2x+ 2x2 1.
Example 48. (Using modification rule) Solve the equation y 3y + 2y = ex.
Solution. The characteristic equation of the homogeneous equation is m23m+2 = 0which implies m = 1, 2. Therefore the solution of the homogeneous equation is
yh = c1ex + c2e
2x
where c1, c2 are arbitrary constants.
We have r(x) = ex. From Table 3.1, the suitable choice for yp is Aex. However the
term ex coincides with a term in yh. Therefore the correct yp is yp = Axex.
yp = Axex +Aex
yp = Axex + 2Aex.
Since yp is also a solution, then
y 3y + 2y = ex(Axex + 2Aex) 3(Axex +Aex) + 2Axex = ex
Aex = exA = 1.
Hence the general solution is
y(x) = c1ex + c2e
2x xex.
Example 49. Solve the equation y 5y + 6y = 4 sin 2x.
Solution. The characteristic equation of the homogeneous equation is m25m+6 = 0which implies m = 2, 3. Therefore the solution of the homogeneous equation is
yh = c1e2x + c2e
3x
where c1, c2 are arbitrary constants.
We have r(x) = 4 sin 2x. From Table 3.1, the suitable choice for yp is A cos 2x+B sin 2x.
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47
Then
yp = 2A sin 2x+ 2B cos 2xyp = 4A cos 2x 4B sin 2x.
Substitute into the equation
(4A cos 2x 4B sin 2x) 5(2A sin 2x+ 2B cos 2x)+6(A cos 2x+B sin 2x) = 4 sin 2x
(2A 10B) cos 2x+ (10A+ 2B) sin 2x = 4 sin 2x.
This implies2A 10B = 010A+ 2B = 4
} A = 5
13, B =
1
13
Therefore the general solution
y(x) = c1e2x + c2e
3x +5
13cos 2x+
1
13sin 2x.
Example 50. Solve the equation y + 4y + 5y = e2x cosx.
Solution. The characteristic equation of the homogeneous equation is m2 +4m+5 = 0which implies m = 2 i. Therefore the solution of the homogeneous equation is
yh = e2x(c1 cosx+ c2 sinx)
where c1, c2 are arbitrary constants.
We have r(x) = e2x cosx. From Table 3.1, the suitable choice for e2x is Ke2x,and the suitable choice for cosx is c1 cosx + c2 sinx. Since r(x) is the product of anexponential function and a trigonometric function, by sum/product rule, the suitablechoice for yp is Ke
2x(c1 cosx + c2 sinx), which can be simplified to e2x(A cosx +B sinx).
However, this choice is incorrect since e2x cosx coincides with a term in yh. Then bymodification rule, the correct choice for yp is xe
2x(A cosx+B sinx).
Then
yp = xe2x(A cosx+B sinx)
= x [e2x(A cosx+B sinx)]yp = x
[e2x(A sinx+B cosx) 2e2x(A cosx+B sinx)]
+ e2x(A cosx+B sinx)
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48
yp = x [e2x(A cosxB sinx) 2e2x(A sinx+B cosx) 2e2x(A sinx+B cosx) + 4e2x(A cosx+B sinx)]+ e2x(A sinx+B cosx) 2e2x(A cosx+B sinx)+ e2x(A sinx+B cosx) 2e2x(A cosx+B sinx)
that is,
yp = Ax e2x cosx+Bx e2x sinx
yp = Ax e2x sinx+Bx e2x cosx 2Ax e2x cosx 2Bx e2x sinx+Ae2x cosx+Be2x sinx
= (B 2A)x e2x cosx+ (A 2B)x e2x sinx+Ae2x cosx+Be2x sinx
yp = Ax e2x cosxBx e2x sinx+ 2Ax e2x sinx 2Bx e2x cosx+ 2Ax e2x sinx 2Bx e2x cosx+ 4Ax e2x cosx+ 4Bx e2x sinx
Ae2x sinx+Be2x cosx 2Ae2x cosx 2Be2x sinxAe2x sinx+Be2x cosx 2Ae2x cosx 2Be2x sinx
= (3A 4B)x e2x cosx+ (4A+ 3B)x e2x sinx(4A+ 2B)e2x cosx+ (2A 4B)e2x sinx.
Substitute into the equation to obtain
2Be2x cosx 2Ae2x sinx = e2x cosx.
This implies A = 0 and B = 12 . The general solution is
y(x) = e2x(c1 cosx+ c2 sinx) +1
2x e2x sinx.
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49
3.5 Reduction of order method
This method can be used to find a second linearly independent solution of a secondorder linear homogeneous equation if the first solution is found by inspection or insome other way.
Lety + p(x)y + q(x)y = 0
be a homogeneous linear ODE.
Suppose y1 be a solution. The second solution is assumed to be
y2 = u y1
for some function u(x).
Then
y2 = uy1 + u
y1
y2 = uy1 + u
y1 + uy1 + u
y1= uy1 + 2u
y1 + uy1
Substitute into the equation
y + p(x)y + q(x)y = 0
uy1 + 2uy1 + u
y1 + p(uy1 + uy1) + quy2 = 0
u(y1 + py1 + qy2) + u
y1 + u(y1 + 2y1) = 0 (3.12)
Since y1 is a solution, equation (3.12) reduces to
uy1 + u(py1 + 2y1) = 0.
This implies
u = u(p+ 2
y1y1
)u
u= p 2y
1
y1
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50
Integrating u
udu =
(p 2y
1
y1
)dx
lnu =p dx 2 ln y1 =
p dx+ ln 1
y21
u = exp[p dx+ ln 1
y21
]
=1
y21e p dx
This implies
u(x) =
1
y21e p dx dx
The general solution is then
y(x) = Ay1 +By2
= Ay1 +Buy1
= Ay1(x) +B
[1
y21e p dx dx
] y1(x)
where A,B are constants.
Example 51. Solve xy + 2y + xy = 0, given that y1 = cos xx is a solution.
Solution. In standard form the equation is y+ 2xy+y = 0. By the reduction of order
method, the second solution is y = uy1 where
u(x) =
1
y21e p(x) dx dx
=
1
(cosx/x)2e 2/x dx dx
=
x2
cos2 xe2 ln x dx
=
x2
cos2 x
1
x2dx
=
sec2 x dx
= tanx.
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51
Therefore the second solution is
y2(x) = tanx cosxx
=sinx
x,
and so the general solution is
y(x) = Acosx
x+B
sinx
x.
Example 52. Solve y 1xy 4x2y = 0 given that y1(x) = ex2 is a solution.
Solution. The equation is already in standard form. The second solution is y2(x) =
u(x)ex2
where
u(x) =
1
(ex2)2e (1/x) dx dx
=
1
e2x2eln x dx
=
xe2x
2
dx
=1
4e2x
2
.
Therefore y2(x) =14e
2x2 ex2 = 14ex2
and the general solution is
y(x) = Aex2
+B 14ex
2
= Aex2
+ Cex2
, where C = 14B.
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52
3.6 Variation of parameters method
This method can be used to solve a more general nonhomogeneous linear ODEs of theform
y + p(x)y + q(x)y = r(x). (3.13)
This method gives the particular solution yp of equation (3.13). [It does not give thehomogeneous solution, yh of the homogeneous equation.]
If yh = Ay1(x) +By2(x), we seek the solution yp of the form
yp = u(x) y1(x) + v(x) y2(x).
Then
yp = uy1 + u
y1 + vy2 + vy2
= uy1 + vy2 + u
y1 + vy2.
Impose the conditionuy1 + vy2 = 0, (3.14)
then yp reduces toyp = uy
1 + vy
2. (3.15)
Differentiating (3.15), we obtain
yp = uy1 + u
y1 + vy2 + v
y2 (3.16)
Substitute into (3.13),
(uy1 + uy1 + vy
2 + v
y2) + p(uy1 + vy
2) + q(uy1 + vy2) = r
Collecting terms in u and terms in v,
u(y1 + py1 + qy1) + v(y
2 + py
2 + qy2) + u
y1 + vy2 = r (3.17)
Since y1 and y2 are solutions of the homogeneous equation, equation (3.17) reduces to
uy1 + vy2 = r (3.18)
We now have two simultaneous equations, (3.14) and (3.18),
uy1 + vy2 = 0
uy1 + vy2 = r.
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53
We can find u and v by using Cramers rule, so that
u = y2rW
and v =y1r
W
and integrating
u = y2r
Wdx and v =
y1r
Wdx.
Example 53. Solve y + y = tanx.
Solution. The homogeneous equation y+y = 0 has characteristic equation m2+1 = 0,which implies m = i. Then the homogeneous solution, yh = A cosx+B sinx.
To find the particular solution, let yp = uy1 + vy2, where y1 = cosx and y2 = sinx.The Wronskian is
W (y1, y2) =
cosx sinx sinx cosx = 1.
Then
u =
y2r(x)
Wdx =
sinx tanx
1dx =
sin
2 x
cosxdx
=
cosx secx dx
= sinx ln | secx+ tanx|
and
v =
y1r(x)
Wdx =
cosx tanx
1dx =
sinx dx
= cosx
Then
yp = (sinx ln | secx+ tanx|) cosx+ ( cosx) sinx= cosx ln | secx+ tanx|
The general solution is
y(x) = A cosx+B sinx cosx ln | secx+ tanx|.
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54
3.7 Euler-Cauchy equations
Definition 4. Euler-Cauchy equations are ODEs of the form
x2y + axy + by = 0, (3.19)
where a, b are constants.
The Euler-Cauchy equations can be solved by making the substitution y = xm or x = et.We look at both methods.
Method 1
Make the substitution y = xm. Then y = mxm1 and y = m(m 1)xm2. Thenx2 m(m 1)xm2 + axmxm1 + bxm = 0
xm[m(m 1) + am+ b] = 0xm[m2 + (a 1)m+ b] = 0.
This impliesm2 + (a 1)m+ b = 0. (3.20)
Equation (3.20) is called the characteristic equation.
The general solution of (3.19) depends upon the type of roots of (3.20).
Case I: Real distinct roots, m1 and m2. Then the solution is
y = Axm1 +Bxm2 .
Case II: Complex roots: m = a bi.Euler formula is ei = cos + i sin .
First we note that
xbi = elnxbi
= ei(b lnx) = cos(b lnx) + i sin(b lnx)
so that
xa+bi = xaxbi = xa[cos(b lnx) + i sin(b lnx)]
xabi = xaxbi = xa[cos(b lnx) i sin(b lnx)]
Then the general solution is
y = xa[A cos(b lnx) +B sin(b lnx)].
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55
Case III: Equal roots m = 1a2
. Then the first solution is
y1 = xm.
The second solution is obtained by the reduction of order method. Let y2 = uy1 = uxm
where
u(x) =
1
y21e p dx dx =
1
(xm)2e a/x dx dx =
1
x2mea lnx dx =
1
x2m1
xadx
=
1
x2(1a)/21
xadx
=
1
xdx
= lnx.
Then the second solution is y2(x) = u(x)y1(x) = ln xxm. Therefore the general solutionis
y(x) = Axm +Bxm lnx
= Ax(1a)/2 +Bx(1a)/2 lnx.
Method 2
This method is slightly longer and requires the use of the chain rule.
The Euler-Cauchy equation can also be solved by making the substitution x = et. Thedifferential equation (3.19) can be expressed as a differential equation having t as theindependent variable. This is because y is some function of x and with the substitution,y becomes a function of t.
Now if x = et then t = lnx and
dt
dx=
1
x=
1
et= et.
Then, by using the chain rule,
dy
dx=dy
dt
dt
dx=dy
dtet = et
dy
dt
d2y
dx2=
d
dx
(dy
dx
)=dt
dx
d
dt
(et
dy
dt
)= et
d
dt
(et
d2y
dt2 etdy
dt
)= e2t
(d2y
dt2 dydt
).
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56
Substituting these derivatives in (3.19) we obtain
e2t e2t(d2y
dt2 dydt
)+ aetet
dy
dt+ by = 0
d2y
dt2 dydt
+ ady
dt+ by = 0
d2y
dt2+ (a 1)dy
dt+ by = 0
This equation is a homogeneous second order with constant coefficients. By Section3.2, it has characteristic equation m2 + (a 1)m+ b = 0.Case I. Different real roots, m1 and m2.
The general solution is
y = Aem1t +Bem2t
= Axm1 +Bxm2 .
Case II. Complex roots m = a biThe general solution is
y = eat(A cos bt+B sin bt)
= xa[A cos(b lnx) +B sin(b lnx)].
Case III. Equal roots m = 1a2
The general solution is
y = Aemt +Btemt
= Axm +Bxm lnx
= Ax(1a)/2 +Bx(1a)/2 lnx.
We see that Method 1 or Method 2 yield the same general solutions.
-
57
Example 54. Solve the equation 2x2y + 3xy y = 0.
Solution. Dividing both sides of the equation by 2, the equation x2y+ 32xy 12y = 0
is an Euler-Cauchy equation. By making the substitution y = xm, the characteristicequation is m2+( 32 1)m+ 12 = 0 with roots 1 and 0.5. Therefore the general solutionis
y(x) =A
x+Bx.
Example 55. Solve the equation x2y + 3xy + 10y = 0.
Solution. Make the substitution y = xm to obtain the characteristic equation m2+(31)m+ 10 = 0 or m2 + 2m+ 10 = 0. The roots are m = 1 3i. Therefore the generalsolution is
y(x) = x[A cos(3 lnx) +B sin(3 lnx)].
Example 56. Solve the equation x2y 3xy + 4y = 0.
Solution. By making the substitution y = xm characteristic equation is m2 + (3 1)m+ 4 = 0 or m2 4m+ 4 = 0, that is (m 2)2 = 0. This equation has repeated roots2. Then the general solution is
y(x) = Ax2 +Bx2 lnx.
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58
Example 57. Solve x2y 3xy + 4y = 0, this time by making the x = et.
Solution. By making the substitution x = et we obtain the equationd2y
dt2 dydt 3dy
dt+ 4y = 0 or
d2y
dt2 4dy
dt+ 4y = 0.
The characteristic equation is m2 4m+ 4 = 0 which implies m = 2, 2. Therefore
y(t) = Ae2t +Bte2t,
hence the general solution is
y(x) = Ax2 +B lnx x2 = Ax2 +Bx2 lnx.
Example 58. Solve 2x2y 3xy 3y = 0.
Solution. The substitution x = et transforms it into the equation
2
(d2y
dt2 dydt
) 3dy
dt 3y = 0 or 2d
2y
dt2 5dy
dt 3y = 0.
The characteristic equation is 2m2 5m 3 = 0, with roots m = 12 , 3. Therefore thegeneral solution is
y(t) = Aet/2 +Be3t,
so thaty(x) = Ax1/2 +Bx3.
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59
Example 59. Solve x2y + 7xy + 13y = 0.
Solution. By making the substitution x = et we obtaind2y
dt2 dydt
+ 7dy
dt+ 13y = 0 or
d2y
dt2+ 6
dy
dt+ 13y = 0.
The characteristic equation is m2 + 6m + 13 = 0, with roots m = 3 2i. Thereforethe general solution is
y(t) = e3t(A cos 2t+B sin 2t),
so thaty(x) = x3
[A cos(2 lnx) +B sin(2 lnx)
].
Example 60. Solve x2y 2xy + 2y = 4x3.
Solution. The substitution x = et transforms the equation into the equationd2y
dt2 dydt 2dy
dt+ 2y = 4e3t or
d2y
dt2 3dy
dt+ 2y = 4e3t.
The characteristic equation is m2 3m+ 2 = 0. Then the solution to the homogeneousequation is yh = Ae
t +Be2t.
By the method of undetermined coefficient, choose yp = Ce3t. Subtitution into the
equation will give C = 2. Then
y(t) = Aet +Be2t + 2e3t,
and hence the general solution is
y(x) = Ax+Bx2 + 2x3.
Example 61. Solve x2y 4xy + 6y = 21x4.
Solution (Method 1). The homogeneous equation x2y 4xy + 6y = 21x4 is anEuler-Cauchy equation. By making the substitution y = xm we obtain the characteristicequation m2 5m+ 6 = 0 whose roots are m = 2, 3.
Therefore the homogeneous solution is
yh = Ax2 +Bx3.
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60
Let y1 = x2 and y2 = x
3. The Wronksian of y1 and y2 is
W (x2, x3) =
x2 x32x 3x2 = x4
To find yp we must divide the equation by x2 so that the coefficients of y is 1. Equation
is now y 4xy +
6
x2y = 21x6.
Then yp = uy1 + vy2 where
u = y2 r
Wdx =
x3 21x6
x4dx = 21
x7 dx = 21x
6
6 dx
=21
6x6,
and
v =
y1 r
Wdx =
x2 21x6
x4dx = 21
x8 dx = 21
x7
7 dx
= 217x7
.
Therefore
yp = uy1 + vy2 =21
6x6 x2 + 21
7x7 x3 = 21
6x4 21
7x4=
1
2x4.
Hence the general solution is
y = yh + yp
= Ax2 +Bx3 +1
2x4.
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61
Solution (Method 2). Since the equation is an Euler-Cauchy we make the substitutiony = et.
Then equation becomesd2y
dt2 5dy
dt+ 6y = 21e4t.
The homogeneous equation d2ydt2 5dydt + 6y = 0 has solution yh = Ae2t +Be3t.
As for the non homogeneous part, by the method of undetermined coefficients let yp =Ce4t. No need for modification since yp does not coincide with the terms in yh.
We have
yp = 4Ce4t
yp = 16Ce4t.
Substitute into the equation:
16Ce4t 5(4Ce4t) + 6(Ce4t) = 21e4t
42e4t = 21e4t,
which implies C = 12 .
Therefore
y(t) = Ae2t +Be3t +1
2e4t,
so that
y(x) = Ax2 +Bx3 +1
2x4
= Ax2 +Bx3 +1
2x4.
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Chapter 4
Series solutions of ODEs
In this chapter will study a class of second order equations that cannot be solved inclosed form in terms of elementary functions. The solutions of this class of equationsinvolve power series.
Definition 5. A power series about x0 is an infinite series of the form
m=0
am(x x0)m = a0 + a1(x x0) + a2(x x0)2 + (4.1)
where x is a variable and a0, a1, a2, , are constants, called the coefficients of thepower series, and x0 is a constant, called the centre of the series.
In particular, if x0 = 0, we obtain a power series in powers of x.
m=0
amxm = a0 + a1x+ a2x
2 +
All variables and constants are assumed to be real.
A power series may converge or diverge. The radius of convergence is a number Rsuch that the series (4.1) converges absolutely for |x x0| < R, and diverges for for|x x0| > R. At |x x0| = R the series may converge or diverge.The interval of all xs, including the endpoints if need be, for which the power seriesconverges is called the interval of convergence.
62
-
63
Example 62. Some of the familiar power series
ex = 1 + x+x2
2!+x3
3!+ + x
n
n!+ < x
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64
The algebraic operations are:
1. Termwise addition/substraction
y1 + y2 = (a0 + b0) + (a1 + b1)x+ (a2 + b2)x2 + + (an + bn)xn +
=i=0
(ai + bi)xi
2. Termwise multiplication
y1y2 = c0 + c1x+ c2x2 + + cnxn +
where
c0 = a0b0
c1 = a0b1 + a1b0
c2 = a0b2 + a1b1 + a2b0...
cn = a0bn + a1bn1 + a2bn2 + + anb0 =ni=0
aibni
3. Termwise differentiation
If
y(x) = a0 + a1x+ a2x2 + + anxn + =
n=0
anxn
then
y(x) = a1 + 2a2x+ 3a3x2 + + nanxn1 + =n=1
nanxn1
y(x) = 2a2 + 6a3x+ 4 3a4x2 + + n(n 1)anxn2 + =n=2
n(n 1)anxn1
4.2 Linear differential equation of second order
The most general form of the linear differential equation of second order is
p(x)y + q(x)y + r(x)y = g(x) (4.2)
where p(x), q(x), r(x), g(x) are continuous functions in an open interval S.
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65
Equation (4.2) has a power series solution about x0. The form of the power seriessolutions is determined by whether x0 is an ordinary point or a regular singular point.
If x0 is an ordinary point of the differential equation then the series solution is
y(x) =n=0
an(x x0)n
and if x0 is a regular singular point then the series solution is
y(x) =n=0
an(x x0)n+r.
By a suitable translation we will only be looking at series solution about x0 = 0. Thenif 0 is an ordinary point the series solution takes the form
y(x) =n=0
anxn
and if 0 is a regular singular solution the series solution takes the form
y(x) =n=0
anxn+r.
4.3 Ordinary and singular points
A homogeneous linear second order ODE can be written as
a(x)y + b(x)y + c(x)y = 0. (4.3)
If a(x) 6= 0 we can divide equation (4.3) by a(x) to obtain the standard formy + p(x)y + q(x)y = 0 (4.4)
where p(x) = b(x)a(x)
and q(x) = c(x)a(x)
.
4.3.1 Ordinary points
A point x0 is an ordinary point of the equation (4.4) if when x is substituted by x0in p(x) and q(x), then p(x0) and q(x0) are finite. If p(x0) or q(x0) or both are not finitethen x0 is not an ordinary point.
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66
Example 63. The equation y+xy+ (x2 4)y = 0 is in standard form with p(x) = xand q(x) = x2 4. All points x0 R are ordinary points because p(x0) = x0 andq(x0) = x
20 4 are finite.
Example 64. Consider the equation (x 1)y + xy + 1xy = 0. In standard form it isy +
x
x 1y +
1
x(x 1)y = 0 where p(x) =x
x 1 and q(x) =1
x(x 1) .
Here all points except 0, 1 are ordinary points because p(1) and q(0) are not finite.
Example 65. The equation x2(x2)2y+2(x2)y+(x+1)y = 0 in standard form isy +
2
x2(x 2) y +
x+ 1
x2(x 2)2 y = 0 where p(x) =2
x2(x 2) and q(x) =x+ 1
x2(x 2)2 .
Here all points except 0, 2 are ordinary points because p(0), q(0) and p(2), q(2) are notfinite.
4.3.2 Singular points
A point x0 is a singular point of (4.4) if p(x0) or q(x0) or both are not finite.
Example 66. In Example 63, the equation has no singular point and in Example 64,the singular points are 0 and 1. In Example 65, the singular points are 0 and 2.
There are two kinds of singular points:
regular singular point irregular singular point
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67
A singular point x0 is a regular singular point if
limxx0
(x x0)p(x0) and limxx0
(x x0)2q(x0) (4.5)
are both finite.
A singular point x0 is an irregular singular point if any one or both of (4.5) are notfinite.
Example 67. In Example 63, all points are ordinary points, therefore there is nosingular points.
Example 68. In Example 64, the points 0 and 1 are singular points because p(1) andq(0) are not finite.
For the point x0 = 0,
limx0
(x 0)p(x) = limx0
x2
x 1 = 0
andlimx0
(x 0)2q(x) = limx0
x
x 1 = 0
Therefore 0 is a regular singular point.
For the point x0 = 1,limx1
(x 1)p(x) = limx1
x = 0
and
limx1
(x 1)2q(x) = limx1
x 1x
= 0
Therefore 0 is a regular singular point.
Example 69. In Example 65, 0 and 2 are singular points because p(0), q(0) and
p(1), q(1) are not finite. Here p(x) =2
x2(x 2) and q(x) =x+ 1
x2(x 2)2 .
For the point x0 = 0,
limx0
(x 0)p(x) = limx0
2
x(x 2) =
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68
and
limx0
(x 0)2q(x) = limx0
x+ 1
(x 2)2 =1
4,
therefore 0 is an irregular singular point.
For the point x0 = 2,
limx2
(x 2)p(x) = limx2
2
x2=
1
2
and
limx2
(x 2)2q(x) = limx2
x+ 1
x2=
3
4,
therefore 2 is a regular singular point.
Example 70. The equation x2y+xy+ (x22)y = 0 is the Bessel equation of order. The point x = 0 is a singular point and the points x 6= 0 are ordinary points.
Example 71. The equation (1x2)y2xy+n(n+1)y = 0 is the Legendre equation.The points x = 1 are singular points and the points x 6= 1 are ordinary points.
Example 72. The equation y = xy is the Airy equation. All points are ordinarypoints.
4.4 Series solution about an ordinary point
Theorem 5. If x0 = 0 is an ordinary point, equation (4.2) can be solved by powerseries in the region x0 = 0 having as solution
y = A(a0 + a1x+ a2x2 + ) +B(b0 + b1x+ b2x2 + )
The two series are linearly independent and converge in the region x0 = 0.
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69
Example 73. Solve y y = 0 by using(i) constant coefficients method
(ii) power series method where series in the form
m=0
amxm.
Show that the same solution is obtained.
Solution.
Constant coefficients method
The characteristic equation is m2 1 = 0 so that m = 1. Therefore the generalsolution is
y = Aex +Bex
where A,B are arbitrary constants.
Power series method
The point x = 0 is an ordinary point. We seek a solution of the form
y = a0 + a1x+ a2x2 + =
m=0
amxm. Then
y(x) =m=1
mamxm1
y(x) =m=2
m(m 1)amxm2.
Substitute into the given equation
m=2
m(m 1)amxm2 m=0
amxm = 0,
change the dummy variables such that both series match the power of x.
n+2=2
(n+ 2)[(n+ 2) 1]an+2 x(n+2)2 n=0
anxn = 0
n=0
(n+ 2)(n+ 1)an+2xn
n=0
anxn = 0,
and put the terms in a single summation sign
n=0
[(n+ 2)(n+ 1)an+2 an
]xn = 0.
This implies(n+ 2)(n+ 1)an+2 an = 0, for n = 0, 1, 2, . . .
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70
and we obtain the recurrence relation
an+2 =an
(n+ 2)(n+ 1), for n = 0, 1, 2, . . .
Substitute the values for n
n = 0, a2 =a0
2 1
n = 1, a3 =a1
3 2 =a13!
n = 2, a4 =a2
4 3 =a04!
n = 3, a5 =a3
5 4 =a15!
...
In general, if n is even, an =an2
n(n 1) =a0n!
,
and if n is odd, an =an2
n(n 1) =a1n!
.
Then
y0 = a0 + a1x+ a2x2 + a3x
3 + a4x4 + + anxn +
= a0 + a2x2 + a4x
4 + even power of x
+ a1x+ a3x3 + a5x
5 + odd power of x
= a0 +a02!x2 +
a04!x4 + + a1x+ a1
3!x3 +
a15!x5
= a0
(1 +
x2
2!+x4
4!+
)+ a1
(x+
x3
3!+x5
5!+
)= a0 coshx+ a1 sinhx.
Therefore the general solution is
y(x) = C coshx+D sinhx
where C and D are arbitrary constants.
-
71
To show that the same solution is obtained,
y(x) = C coshx+D sinhx
= C
(ex + ex
2
)+D
(ex ex
2
)
=
(C
2 D
2
)ex +
(C
2+D
2
)ex
= Aex +Bex,
where A = C2 D2 and B = C2 + D2 .
Example 74. Solve (1 + x2)y + xy y = 0.
Solution. The point x = 0 is an ordinary point. We seek a solution of the form
y = a0 + a1x+ a2x2 + =
m=0
amxm. Then
y(x) =m=1
mamxm1
y(x) =m=2
m(m 1)amxm2.
Substitute into the given equation
(1 + x2)
m=2
m(m 1)amxm2 + xm=1
mamxm1
m=0
amxm = 0
m=2
m(m 1)amxm2 +m=2
m(m 1)amxm
+
m=1
mamxm
m=0
amxm = 0.
Make xm in the first summation,
n+2=2
(n+ 2)[(n+ 2) 1]an+2x(n+2)2 +n=2
n(n 1)anxn
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72
+
n=1
nanxn
n=0
anxn = 0.
Simplify
n=0
(n+ 2)(n+ 1)an+2xn +
n=2
n(n 1)anxn
+
n=1
nanxn
n=0
anxn = 0.
Make the index of summation start at the same number n = 2 by expanding some of thesummation,
2a2 + 6a3x+
n=2
(n+ 2)(n+ 1)an+2xn
+
n=2
n(n 1)anxn
+ a1x+
n=2
nanxn
a0 a1xn=2
anxn = 0.
Collect the terms,
(2a2 a0) + 6a3x+n=2
[(n+ 2)(n+ 1)an+2 + (n
2 1)an]
= 0.
This implies
2a2 a0 = 0 a2 = a02
6a3 = 0 a3 = 0,
and
(n+ 2)(n+ 1)an+2 + (n2 1)an = 0 an+2 = n
2 1(n+ 2)(n+ 1)
an
= n 1n+ 2
an.
-
73
Write the first few terms of the series,
n = 2 a4 = a24
=a0
4 2
n = 3 a5 = 2a35
= 0
n = 4 a6 = 3a46
=3a0
6 4 2
n = 5 a7 = 4a57
= 0
n = 6 a8 = 5a68
= 5 3a08 6 4 2
In general, if n is odd (except a1), an = 0,
and if n is even, a2k = (1)k+1 1 3 5 (2k 3)2kk!
a0.
Then
y = a0 + a1x+ a2x2 + a3x
3 + + anxn + = a0 + a2x
2 + a4x4 + a6x
6 + anxn + + a1x= a0 +
a02x2 a0
22 2!x4 +
3a023 3!x
6 + a1x
= a0
(1 +
1
2x2 1
22 2!x4 +
3
23 3!x6
)+ a1x
= a0(1 + x2)1/2 + a1x.
The general solution isy(x) = Ay1(x) +By2(x),
where the linearly independent solutions are
y1(x) =
1 + x2 and y2(x) = x.
-
74
4.5 Series solution about a regular singular point -
the method of Frobenius
Letx2y + xP (x)y +Q(x)y = 0, (4.6)
and in standard form
y +P (x)
xy +
Q(x)
x2y = 0, (4.7)
with
limx0
xP (x)
xand lim
x0x2Q(x)
x2
be finite, so that 0 is a regular singular point of the equation (4.6).
To find a solution about a regular singular point, we can use the method by the Germanmathematician, Ferdinand Georg Frobenius.
Theorem 6 (The method of Frobenius). If 0 is a regular singular point of equation(4.6) then it has at least a series solution of the form
y(x) = xrn=0
anxn =
n=0
anxn+r
where r satisfies a quadratic equation called the indicial equation.
Since r is a root of a quadratic equation, there are three cases to consider and thesolutions of (4.6) depend on these roots.
Case 1: Distinct roots not differing by an integer
Suppose the roots are r1, r2 with r1 > r2 and r1 r2 / Z. Then the two linearlyindependent solutions are
y1(x) = xr1(a0 + a1x+ a2x
2 + )and
y2(x) = xr2(A0 + A1x+ A2x
2 + )
Case 2: Double root, r1 = r2 = r
The linearly independent solutions are
y1(x) = xr(a0 + a1x+ a2x
2 + )
-
75
andy2(x) = y1(x) lnx+ x
r(A1x+ A2x2 + )
Note: The second solution y2(x) may also by obtained by using the reduction of ordermethod.
Case 3: Roots differing by an integer
Suppose the roots are r1, r2 with r1 > r2 and r1 r2 Z. Then the two linearlyindependent solutions are
y1(x) = xr1(a0 + a1x+ a2x
2 + )
andy2(x) = k y1(x) lnx+ x
r2(A0 + A1x+ A2x2 + )
where k is a constant which may turn out to be zero.
This example illustrates Case 1 where the roots do not differ by an integer.
Example 75. Solve the equation 4xy + 2y + y = 0.
Solution. In the form (4.6) the equation is
x2y + x1
2y +
x
4y = 0
with P (x) = 12 and Q(x) =x4 , and in standard form (4.7),
y +1/2
xy +
1/4
xy = 0
with p(x) = 1/2x and q(x) =1/4x as in the form (4.4).
The point 0 is a singular point because 1/20 and1/40 are not finite. It is a regular singular
point because
limx0
x 1/2x
= limx0
1
2=
1
2
and
limx0
x2 1/4x
= limx0
1
4x = 0
are both finite.
-
76
By the method of Frobenius it has a series solution of the form y =
n=0
anxn+r. Then
y =n=0
(n+ r) anxn+r1
y =n=0
(n+ r)(n+ r 1) anxn+r2
Substitute into the equation
4x
n=0
(n+ r)(n+ r 1) an xn+r2 + 2n=0
(n+ r) an xn+r1
+
n=0
an xn+r = 0
n=0
4(n+ r)(n+ r 1) an xn+r1 + 2n=0
(n+ r) an xn+r1
+
n=0
an xn+r = 0.
Make the power of x in each summation to be the same. The substitution n = m 1 inthe third summation changes xn+r to xm+r1:
m=0
4(m+ r)(m+ r 1) am xm+r1 +m=0
2(m+ r) am xm+r1
+
m1=0
am1 xm1+r = 0.
Simplify,
m=0
4(m+ r)(m+ r 1) am xm+r1 +m=0
2(m+ r) am xm+r1
+
m=1
am1 xm+r1 = 0.
-
77
Make the index of each summation starts at the same number by expanding some of thesummation,
4r(r 1) a0 xr1 +m=1
4(m+ r)(m+ r 1) am xm+r1
+ 2r a0 xr1 +
m=1
2(m+ r) am xm+r1
+
m=1
am1 xm+r1 = 0.
Simplify
[4r(r 1) + 2r] a0 xr1 +m=1
4(m+ r)(m+ r 1) am xm+r1
+
m=1
2(m+ r) am xm+r1
+
m=1
am1 xm+r1 = 0,
and express it in a single summation sign
[4r(r 1) + 2r] a0 xr1
+
m=1
[2(m+ r)(2m+ 2r 1) am + am1
]xm+r1 = 0.
This implies[4r(r 1) + 2r] a0 = 0,
and2(m+ r)(2m+ 2r 1) am + am1 = 0.
Now [4r(r 1) + 2r] a0 = 0
(4r2 2r) a0 = 0
Since, by definition a0 6= 0,
4r(r 1) = 0 this is the indicial equationr = 0,
1
2
If we let r1 =12 and r2 = 0, then r1 r2 = 12 which is not an integer. This is Case 1.
-
78
The recurrence relation is
2(m+ r)(2m+ 2r 1) am + am1 = 0
am =am1
2(m+ r)(2m+ 2r 1) , m = 1, 2, 3, . . .
If r = 12 , am =am1
2m(2m+ 1), then
m = 1, a1 =a03 2 =
a03!
m = 2, a2 =a15 4 =
1
5 4 a03!
=a05!
m = 3, a3 =a27 6 =
1
7 6 a05!
= a07!
...
In general,
am =(1)m
(2k + 1)!a0.
Therefore a linearly independent solution is
y1(x) = x1/2(a0 + a1x
2 + a2x2 + + anxn + )
= x1/2(
1 x3!
+x2
5! x
3
7!+ + anxn +
)choose a0 = 1. Then
y1(x) =
k=0
(1)k(2k + 1)!
xk+1/2.
If r = 0, am =am1
2m(2m 1) , then
m = 1, a1 =a02 1 =
a02!
m = 2, a2 =a14 3 =
1
4 3 a02!
=a04!
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79
m = 3, a3 =a26 5 =
1
6 5 a04!
= a06!
...
In general,
am =(1)m(2k)!
a0.
A linearly independent solution is
y2(x) = x0(a0 + a1x
2 + a2x2 + + anxn + )
= a0 a02!x2 +
a24!x2 + + anxn +
choose a0 = 1. Then
y2(x) =
k=0
(1)k(2k)!
xk.
The general solution is then
y(x) = Ay1(x) +By2(x)
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80
This next example illustrates Case 2 where the indicial equation has double root.
Example 76. Solve xy + (x+ 1)y + y = 0.
Solution. In the form (4.6) the equation is
x2y + x(x+ 1)y + xy = 0.
In standard form (4.7), the equation is
y +x+ 1
xy +
1
xy = 0
with p(x) = x+1x and q(x) =1x .
The point 0 is a singular point because 0+10 and10 are not finite. It is a regular singular
point because
limx0
x x+ 1x
= limx0
(x+ 1) = 1
and
limx0
x2 1x
= limx0
x = 0
are both finite.
By the method of Frobenius it has a series solution of the form y =
n=0
anxn+r. Then
Substitute into the equation
x
n=0
(n+ r)(n+ r 1) an xn+r2 + (x+ 1)n=0
(n+ r) an xn+r1
+n=0
an xn+r = 0
n=0
(n+ r)(n+ r 1) an xn+r1 +n=0
(n+ r) an xn+r
+
n=0
(n+ r) an xn+r1 +
n=0
an xn+r = 0
-
81
Make the power of x to be xn+r1,
m=0
(m+ r)(m+ r 1) am xm+r1 +
m1=0(m 1 + r) am1 xm1+r
+
m=0
(m+ r) am xm+r1 +
m1=0
am1 xm1+r = 0,
and simplify
m=0
(m+ r)(m+ r 1) am xm+r1 +m=1
(m+ r 1) am1 xm+r1
+m=0
(m+ r) am xm+r1 +
m=1
am1 xm+r1 = 0.
Make the index of summation in each starts at the same number,
r(r 1) a0 xr1 +m=1
(m+ r)(m+ r 1) am xm+r1
+
m=1
(m+ r 1) am1 xm+r1
+ r a0 xr1 +
m=1
(m+ r) am xm+r1
+
m=1
am1 xm+r1 = 0,
and simplify
[r(r 1) + r] a0 xr1
+
m=1
[(m+ r)(m+ r) am + (m+ r) am1
]= 0.
This implies [r(r 1) + r] a0 = 0 and since a0 6= 0, the indicial equation is
r(r 1) + r = 0 so that r = 0, 0.
This also implies(m+ r)(m+ r) am + (m+ r) am1 = 0
so that the recurrence relation is
am =am1m+ r
,
-
82
and when r = 0,
am =am1m
.
Substitute for the first few values of m,
m = 1, a1 = a0
m = 2, a2 =a1
2=a02
m = 3, a3 =a2
3=
1
3 a0
2=a03!
m = 4, a4 =a3
4=14(a0
3!
)=a04!
...
am = (1)m a0m!.
Therefore the first solution is
y1(x) = a0 + a1x+ a2x2 + a3x
3 + + anxn
= a0 a0x+ a02x2 a0
3x3 + + (1)m a0
n!xn
= a0
[1 x+ x
2
2 x
3
3!+ + (1)mx
n
n!xn +
]= a0e
x.
Taking a0 = 1 we have y1(x) = ex.
The second solution can be obtained by the reduction of order method. The secondsolution is y2(x) = u(x)y1(x) where
u(x) =
1
(ex)2e x+1x dx dx
=
e2xe
1 1x dx
=
e2xexln x dx
=
ex
1
xdx
=
1
x
[1 + x+
x2
2!+x3
3!+ + x
n
n!+
]dx
-
83
=
1
x+ 1 +
x
2!+x2
3!+ + x
n1
n!+ dx
= lnx+ x+x2
2 2! +x3
3 3! + +xn
n n! +
Then the second solution is
y(x) = lnx+ x+x2
2 2! +x3
3 3! + +xn
n n! +
and therefore the general solution is
y(x) = Aex +Bex[lnx+ x+
x2
2 2! +x3
3 3! + +xn
n n! + ]
This example illustrates Case 3, where the roots r1 r2 differs by an integer.
Example 77. Solve the Bessels equation x2y + xy +(x2 1
4
)y = 0.
Solution. The method of Frobenious says the equation has a series solution of the form
y =
n=0
anxn+r.
Substitute y, y, y into the equation
x2n=0
(n+ r)(n+ r 1) an xn+r2 + xn=0
(n+ r) an xn+r1
+
(x2 1
4
) n=0
an xn+r = 0
n=0
(n+ r)(n+ r 1) an xn+r +n=0
(n+ r) an xn+r
+
n=0
an xn+r+2 = 0
n=0
an4xn+r = 0.
-
84
Make the power of x in the summation to be xm+r,
m=0
(m+ r)(m+ r 1) am xm+r +m=0
(m+ r) am xm+r
+
m2=0
am2 xm2+r+2 = 0
m=0
am4
xm+r = 0
m=0
(m+ r)(m+ r 1) am xm+r +m=0
(m+ r) am xm+r
+m=2
am2 xm+r = 0
m=0
am4
xm+r = 0.
Make the index of summation in each start at 2,
r(r 1) a0 xr + (1 + r)r a1 x1+r +m=2
(m+ r)(m+ r 1) am xm+r
+ r a0 xr + (1 + r) a1x
1+r +
m=2
(m+ r) am xm+r
+
m=2
am2 xm+r = 0
a04xr a1
4x1+r
m=2
am4
xm+r = 0,
and simplify [r2 1
4
]a0 x
r +
[(1 + r)2 1
4
]a1 x
1+r
+
m=2
([(m+ r)2 1
4
]am + am2
)xm+r.
This implies[r2 14
]a0 and since a0 6= 0, the indicial equation is
r2 14
= 0, which implies r1 =1
2, r2 = 1
2.
-
85
This also implies[(1 + r)2 14
]a1 = 0 and since for r1 =
12 , r2 = 12 , (1 + r)2 14 6= 0,
then a1 = 0.
Also [(m+ r)2 1
4
]am + am2 = 0
(m+ r 1
2
)(m+ r +
1
2
)am = am2
am =am2
(m+ r 12 )(m+ r + 12 ), m = 2, 3, 4, . . .
When r1 =12 , the recurrence relation is
am =am2m(m+ 1)
, m = 2, 3, 4, . . .
m = 2, a2 =a02 3
m = 3, a3 =a13 4 = 0
m = 4, a4 =a24 5 =
14 5
a02 3 =
a02 3 4 5 =
a05!
m = 5, a5 = 0
m = 6, a6 =a46 7 =
16 7
a02 3 4 5 =
a07!
...
The first linearly independent solution is
y1(x) = x1/2(a0 + a1x+ a2x
2 + a3x3 + a4x
4 + a5x5 + a6x
6 + )
= x1/2(a0 a0
3!x2 +
a05!x4 a0
7!x6 +
)=
a0x
(x x
3
3!+x5
5! x
7
7!+
)
= a0
sinx
x.
-
86
The second solution is
y2(x) = ky1(x) lnx
+ x1/2(A0 +A1x+A2x
2 +A3x3 +A4x
4 +A5x5 +A6x
6 + )
and the constants K,A0, A2, A3, . . . may be determined by substituting y2(x) into thedifferential equation.
It can be shown the second solution is y2(x) = A0cos xx .
Discussion
What happens if we try to find y2(x) by using r2 = 12?
The coefficient of x1+r is[(1 + r)2 14
]a1, for r2 = 12 , (1 + r2)2 14 = 0. Here A1
may be 0 and may not be 0. We take it to be non zero.
The recurrence relation is
Am =Am2
(m 1)m, m = 2, 3, 4, . . .
m = 2, A2 =A01 2
m = 3, A3 =A12 3
m = 4, A4 =A23 4 =
13 4
A01 2 =
A01 2 3 4 =
A04!
m = 5, A5 =A34 5 =
14 5
A12 3 =
A05!
m = 6, A6 =A45 6 =
15 6
A01 2 3 4 =
A06!
m = 7, A7 =A56 7 =
16 7
A05!
=A0
7!...
The series solution is then
y2(x) = x1/2(A0 +A1x+A2x2 +A3x3 +A4x4 +A5x5
+A6x6 +A7x
7 )
-
87
= x1/2(A0 +A1x A0
2!x2 A1
3!x3 +
A04!x4 +
A15!x5
A06!x6 A0
7!x7 +
)
= x1/2(A0 A0
2!x2 +
A04!x4 A0
6!x4 +
)
+ x1/2(A1x A1
3!x3 +
A15!x5 A1
7!x7 +
)
=A0x
(1 x
2
2!+x4
4! x
6
6!+
)
+A1x
(x x
3
3!+x5
5! x
7
7!+
)
= A0cosxx
+A1sinxx.
It seems that by taking the smaller value r2 we obtain the general solution. However thesmaller value does not give the general solution all the time. Sometimes you will notend up with any solution if start with the smaller r2.
In general, if is not an integer, the solution to the Bessels equation
x2 y + xy + (x2 2) y = 0
isy = AJ(x) +BJ(x)
where the Bessel function J is given by
J(x) =(x
2
) k=0
(1)kk! (+ k + 1)
(x2
)2k
J(x) =(x
2
) k=0
(1)kk! (+ k + 1)
(x2
)2k
where (. . . ) is a gamma function. You will learn about Bessel and gamma functionsin KXEX3244.
-
Appendix A
Analytic functions
Definition 7. A function f(x) is analytic at x0 if its Taylor series about x0 exists, thatis f(x) can be represented by the series
f(x) =n=0
an(x x0)n, where an = f(n)(x0)
n!
88
-
Appendix B
Convergence of power series
Theorem 7 (DAlembert Ratio test). For a positive series
n=0
Un = U0 + U1 + + Un + , Un > 0
if
limn
Un+1Un
= m (finite value)
then
(i) the series converges for m < 1,
(ii) the series diverges for m > 1,
(iii) the series fails for m = 1.
Example 78. For what values of x does the series
n=2
xn
n(n 1) converges?
Solution. A few terms expansion of the series is
x2
2 1 +x3
3 2 + +xn
n(n 1) + (B.1)
Taking the ratio of the n-th term to the n 1-th term,Un+1Un
=xn+1
n(n+1)xn
n(n1)
= n 1n+ 1 |x|
89
-
90
Then
limn
Un+1Un = |x| limn n 1n+ 1 = |x|
Case 1: If |x| < 1 then the series (B.1) converges.
If |x| = then
Case 2: If x = 1 series (B.1) is
1
2 1 +1
3 2 +1
4 3 + +1
n(n 1) + (B.2)
By the technique of partial fraction
1
n(n 1) =1
n 1 1
n
Then (B.2) can be expressed as
S2 =
(1 1
2
)+
(1
2 1
3
)+
(1
3 1
4
)+ +
(1
n 1 1
n
)+
As n, S2 1. Therefore S2 converges.
Case 3: If x = 1 series (B.1) is
1
2 1 1
3 2 +1
4 3 + +(1)nn(n 1) + < S2
that is, the series converges as well.
Therefore the radius of convergence is 1 x 1.
For the endpoints the following special might be helpful.
Geometric series
The seriesa+ ar + ar2 + + arn1 +
where a and b are constants, converges to S =a
1 r if |r| < 1 and diverges if |r| > 1.
The p series
The series1
1p+
1
2p+
1
3p+ + 1
np+
-
91
where p is a constant, converges for p > 1 and diverges for p 1.The series with p = 1
1 +1
2+
1
3+ + 1
n+
is called the harmonic series. This series diverges. However it can be shown that
limn
(1 +
1
2+
1
3+ + 1
n lnx
)=
where the Eulers constant = 0.577215 . . .
Theorem 8 (Cauchys nth root). For a posit