2 differential equation
TRANSCRIPT
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Differential Equation INTRODUCTION Differential equations are fundamental in engineering mathematics since many of the physical laws and relationships between physical quantities appear mathematically in the form of such equations. The transition from a given physical problem to its mathematical representation is called modeling. This is of great practical interest to engineer, physicist or computer scientist. Very often, mathematical models consist of a differential equations or system of simultaneous differential equations, which needs to be solved. In this chapter we shall look at classifying differential equations and solving them by various methods. DIFFERENTIAL EQUATIONS OF FIRST ORDER Definitions A differential equation is an equation which involves derivatives or differential coefficients or differentials. Thus the following are all examples of differential equations. (a) x2dx + y2 dy = 0
(b) d2x
dt2 + a2x = 0
(c) y = xdy
dx+
x2
dy/dx
(d) [1 + (ππ¦
ππ₯)
2]
β5/3
= a d2y
dx2
(e) dx
dy wy = a cos pt,
dy
dt + wx = a sin pt
(f) x2 z
x+ y
z
y = 3z
(g)
2y
t2 = a2
2y
x2
An ordinary differential equation is an equation which all the differential coefficients al with respect to a single independent variable. Thus the equations (a) (c) are all ordinary differential variables and partial differential coefficients with respect to any of them. The equations (f) and (g) are partial differential equations. The order of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the degree of the highest derivative occurring in its, after the equation has been expressed in a form free from radicals and fractions as far as the derivatives are concerned. Thus from the examples above, (a) is of the first order and first degree (b) is of the second order and first degree
(c) written as yππ¦
ππ₯ = x(
dy
dx)
2+ x2
is of the first order but of
second degree; (d) After removing radicals is written as
[1 + (ππ¦
ππ₯)
2]
β5
= a3 (π2π¦
ππ₯2)3
and is of the second order and third degree. Solution of a differential equation A solution (or integral) of a differential equation is a relation between the variable which satisfies the given
differential equation.
For example, y = c ππ₯3
3 β¦(i)
is a solution of dy
dx = x2 y β¦(ii)
The general (or complete) solution of a differential equation is that in which the number of arbitrary constants is equal to the order of the differential equation. Thus (i) is a general solution of (i) is a general solution of (ii) as the number of arbitrary constants (one constant c) is the same as the order of the equations (ii) (first order). Similarly, in the general solution of a second order differential equation, there will be two arbitrary constants. A particular solution is that which can be obtained from the general solution by giving particular values to the arbitrary constants.
For example, y = 4ex3
3 Equations of the first order and first degree It is not possible to analytically solve such equations in general, we shall, however, discuss some special methods of solution which are applied to the following types of equations: 1. Equations where variables are separable. 2. Homogeneous equations 3. Linear equations 4. Exact equations Variables Separable If in an equation it is possible to collect all functions of x and dx on one side and all the functions of y and dy on the other side, then the variables are said to be separate. Thus the general form of such an equation is f(y) dy =
(x)dx + c as solution. Homogeneous Equations
Are of the form ππ¦
ππ₯=
f(x,y)
(x,y )
Where f(x, y) and (x) (x, y) homogeneous functions of the same degree in x and y. Homogenous function: An expression of the form a0xn +
a1xn1 y + a2xn-2y2 + β¦+ anyn in which every term is of the nth degree, is called a homogenous function of degree n. This can be rewritten as xn [a0 + a1(y/x) + a2(y/x)2 + β¦+ an(y/x)n]. Thus any functions f(x, y) which can be expressed in the form xnf(y/x), is called a homogenous function of degree n in x and y. For instance x3 cos (y/x) is a homogenous function of degree 3 in x and y. To solve a homogenous equation
1. Put y = vx, then dy
dx = v + x
dy
dx
2. Separate the variables v and x, and integrate. Linear Equations of first order A different equation is said by linear if the dependent variable and its differential coefficients occur only in the first degree and not multiplied together. Thus the following differential equations are linear
1. dy
dx + 4y = 2
2. x2 d2y
dx2 + 3x
dy
dx + 4y = 2
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equation (i) is linear first order differential equation while equation(ii) in linear second order differential equation. The following equations are not linear
1. (ππ¦
ππ₯)
2+ y = 5
2. dy
dx + y1/2 = 2
3. ydy
dx = 5
Leibnitze linear equation The standard form of a linear equation of the first order, commonly known as leibnitzβs linear equation, is dy
dx + Py = Q where P, Q are arbitrary functions of x.
To solve the equation, multiply both sides by πβ« Pdx + y
(πβ« Pdx P) = Qπβ« Pdx i.e. d
dx (π¦πβ« Pdx) = ππβ« Pdx
Integrating both sides, we et π¦πβ« Pdx = β« Qeβ« Pdx dx + c as the required solution.
Note. The factor πβ« Pdx on multiplying by which the left-hand side of (1) becomes the differential coefficient of a single function, is called the integrating factor(l.F.) of the linear equation (i) So remember the following:
l.F. = πβ« Pdx and the solution is
y(l.F.) = β« Q (l.F.)dx + c
Bernoulliβs Equation
The equation dy
dx+ Py = Qyn (i)
where P, Q are functions of x, is reducible to the Leibnitzβs linear and is usually called the Bernoulliβs equation.
To solve (i), divide both sides by yn, so that yn dy
dx + Py1 n
= Q β¦(ii)
Put y1 n = z so that (1 n) yn dy
dx = dz
dx
Eq. (ii) becomes 1
1βπ
dz
dx + Pz = Q
or dz
dx + P(1 n) z = Q(1 n),
Which is Leibnitzβs linear in z and can be solved easily. Exact differential equations Def. A differential equation of the form M(x, y) dx + N(x, y) = 0 is said to be exact if its left hand member is the exact differential of some function u(x, y) i.e. du = Mdx + Ndy = 0. Its solution, therefore, is u(x, y) c. Theorem. The necessary and sufficient condition for the differential equations Mdx + Ndy = 0 to be exact is
M
y =
N
x
Method of solution. It can be shown that, the equation Mdx + Ndy = 0 becomes
d[u + β« f(y)dy] = 0 Integrating
u + β« f(y)dy = 0. But u = β« Mdx and f(y) = terms of N not containing x.
The solution of Mdx + Ndy = 0 is β« Mdx + β«(terms of N not containing x ) dy = c (Porvides of course that the equation is exact.
i.e. M
y =
N
x )
Note: While finding β« Mdx, y is treated as constant since we are integrating with respect to x. Equations Reducible To Exact Equations Sometimes a differential equation which is not exact, can be made so on multiplication by a suitable factor called an integrating factor. The rules for finding integrating factors of the equation Mdx + Ndy= 0 are as given in theorem 1 and 2 below: In the equation Mdx + Ndy = 0
Theorem 1: if
M
yβN
x
N be a function of x only =
f(x) say, then πβ« f(y)dy is an integrating factor. Clairautβs Equation* An equation of the form = y = px + f(p), where
p = dy
dx , is known as clairautβs equation
β¦..(i) Differentiating with respect to x, we have p = p
+ xdp
dx+ f(p)
dp
dx
[x + f(p)] dp
dx = 0
dp
dx = 0 , or x + f (p) = 0
dp
dx = 0, gives p = c β¦.(ii)
Linear Differential Equations (of nTH order) Definitions Linear differential equations are those in which the dependent variable its derivatives occur only in the first degree and are not multiplied together. The general linear differential equation of the nth order is of the form
πππ¦
ππ₯π + p1
dnβ1y
dxnβ1 + p2 dnβ2y
dxnβ2 + β¦ + pny = X
where p1, p2, pn and X are functions of x only. Linear differential equations with constant coefficients are of the form
dny
dxn + k1 dnβ1y
dxnβ1 + k2dnβ2y
dxnβ2 + β¦ + kny = X
where k1, k2β¦.., kn are constants and X is a function of x only. Such equations are most important in the study of electromechanical vibrations and other engineering problems.
1. Theorem: If y1, y2 are only two solutions of the equations
πππ¦
ππ₯π +dnβ1y
dxnβ1 + k2 dnβ2y
dxnβ2 + β¦ + kny = X Then
c1y1 + c2y2 + β¦+ kny = X since it can be easily shown by differentiating is that dnu
dxn+ k1dnβ1u
dxnβ1 + β¦. + knu = 0β¦.(ii)
2. Since the general solution of a differential equation of the nth order contains n arbitrary constants, if follows, from above, that if y1, y2, y3,β¦, yn, are n independent solutions of (1), then c1y1 + c2y2 + β¦.+ cnyn(= u) is its complete solution.
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3. If y = v be any particular solution of
dny
dxn + k1 dnβ1v
dxnβ1 + β¦ + kny = X β¦.(iii)
then dnv
dxn + k1
dnβ1v
dxnβ1 + β¦ + knv = X β¦.(iv)
Adding (ii) and (iv), we have
dn(u+v)
dxn + k1
dn(u+v)
dxn + k1
dnβ1(u + v)
dxnβ1 + β¦+ kn (u +
v) = X This shows that y = u + v is the complete solution of (iii). The part u is called the complementary function (C. F) and the part v is called the particular integral (P.l.) of (iii)
The complete solution (C.S.) of (iii) is y = C.F. + P.l. Thus in order to solve the equation (iii), we have to first the C.F. i.e., the complementary function of (i), and then the P.l., i.e. a particular solution of (iii).
Operator D denoting d
dx,
d2
dx2,
d3
dx3 = D3 etc. , the
equation (iii) above can be written in the symbolic form
(Dn + k1Dn1 + β¦.+ kn) y = X,
i.e. f(D)y = X,
where f(D) = Dn + k1Dn1+β¦ + kn, i.e. a polynomial in D. Thus the symbol D stands for the operation of differential and can be treated much the same as an algebraic quantity i.e. f(D) can be factorised by ordinary rules of algebra and the factors may be taken in any other. For instance
d2y
dx2 + 2dy
dx 3y = (D2 + 2D 3) y
= (D + 3)(D 1)y or (D 1) (D + 3) y Rules for finding the complementary function
To solve the equation dny
dxn + k1 dnβ1y
dxnβ1 + k2 dnβ2
dxnβ2 + β¦+ kny =
0 β¦.(i) where k's are constants. The equation (i) in symbolic form is
(Dn + k1Dn1 + k2 Dn2 +β¦.+ kn)y= 0 β¦.(ii) Its symbolic co-efficient equated to zero i.e.
Dn + k1Dn1 + k2Dn2 + β¦+ kn = 0 is called the auxiliary equation (A. E.). Let m1, m2,β¦., mn be its roots. Now 4 cases arise. Case I. If all the roots be real and different. then (ii) is equivalent to
(D m1) (D m2) β¦. (D mn)y = 0 β¦.(iii)
Now (iii) will be satisfied by the solution of (D mn)y = 0,
i.e. by dy
dx mny = 0.
This is a Leibnitzβs linear and l.F. = eβmnx
Its solution is yeβmnx = cn, i.e. y = cnemnx Similarly, since the factors in (iii) can be taken in any order, it will be satisfied by the solutions of
(D m1)y = 0, (D m2) = 0 etc., i.e. by y = c1emnx etc. Thus the complete solution of (i) is y = c1em1x + c2em2x+ β¦ + cnemnx β¦(iv)
Case II. If two roots are equal (i.e. m1 = m2), then (iv) becomes y = (c1 + c2) em2x + c3em1x+ β¦. + cnemnx y = Cem1x + c3em3x + β¦ + cnemnx
[ c1 + c2 = one arbitrary constant C] It has only n - 1 arbitrary constants and is, therefore, not the complete solution of (i). In this case, we proceed as follows: The part of the complete solution corresponding to the repeated root is the complete solution of
(D m1) (D m1) y = 0
Putting (D β rn1) y = z. it becomes (D m1) z = 0 or dz
dx
m1z = 0 This is Leibnitz's linear in z and I.F. = eβm1x
its solution is zeβm1x = c1 or z = c1em1x
Thus (D m1) y = z = c1em1x or dy
dx m1y = c1em1x β¦.(v)
Its l.F. being eβm1x, the solution of (v) is
yeβm1x = β« π1em1xdx + c2
y = (c1x + c2) em1x Thus the complete solution of (i) is y = (c1x + c2) em1x + c3em3x + β¦+ cnemnx If, however, the A.E. has three equal roots (i.e. m1 = m2 = m3), then the complete solution is y = (c1x2 + c2x + c3) em1x + c4em4x + β¦ + Cnemnx Case III. If one pair of roots be imaginary , i.e.
m1 = + i,
m2 = i, then the complete solution is
y = c1e( + ) + c2e( )x+ c3em3x + β¦ + cnemnx
= ex (c1eix + c2eβix
) + c3em3x + β¦ + cnemnx
= ex [c1 (cos x + i sin
x) + c2(cos x i sin x)] + c3em3x + β¦ + cnemnx
[ by Eulerβs
Theorem, ei = cos + i sin ]
= ex [c1 cos x + c2 sin x_ + c3em3x + β¦+ cnemnx where C1 = c1 + c2
and C2 = i(c1 c2) Case IV. If two pair of imaginary roots be equal i.e.
m1 = m2 = + i,
m3 = m4 = i, then by case II, the complete solution is
y = ex [(c1x + c2)cos x +
(c3x + c4) sin x] + β¦ + cnemnx. Inverse Operator
Definition, 1
f(D) X is that function of x, not containing
arbitrary constants, which when operated upon by f(D) gives X.
i.e. f(D){1
f(D)} = X
Thus 1
f(D)X satisfies the equation f(D)y = X and is,
therefore, its particular integral. Obviously f(D) and 1/f(D) are inverse operators.
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1
D X = β« Xdx
Let 1
Dx = y
i.e. X = dy
dx
integrating w.r..t. x, y = β« Xdx
Thus 1
DX = β« Xdx
1
DβaX = eax β« Xeβaxdx
Let 1
D aX = y.
Operating by D a,
(D a). 1
D a X = (D a) y.
or X = dy
dx ay, i.e.
dy
dx ay = X
which is a leibnitzβs linear equations.
l.F. being eax, its solution is
yeax = β« Xeβaxdx, no constant being added as (ii) doesnβt contain any constant.
Thus, 1
DβaX = y = eax β« πeβax dx.
TWO OTHER METHODS OF FINDDING P.I. Method of Variation of Parameters This method is quite general and applies to equations of the from
yn + py + qy = X β¦.(i) where p, q, and X are functions of x. It gives
P.l. = y1 β«y2X
Wdx +
y2β«y1X
Wdx β¦.(ii)
where y1 and y2 are the solutions of yβ + py + qy = 0
and W = |y1 y2
y1β² y2
β² | is called the wronskian of y1, y2.
Questions: 1. What is the derivative of π(π₯) = |π₯| at = 0 ?
(A) 1 (B) β1 (C) 0 (D) does not exist
2. The minimum point of the function π(π₯) = (π₯3
3) β π₯
is at (A) π₯ = 1 (B) π₯ = β1
(C) π₯ = 0 (D) π₯ =1
β3
3. Which of the following functions is not differentiable in the domain [β1, 1]? (A) π(π₯) = π₯2
(B) π(π₯) = π₯ β 1
(C) π(π₯) = 2
(D) π(π₯) = Maximum (π₯, βπ₯)
4. The solution of the differential equation ππ¦
ππ₯+ π¦2 =
0 is
(A) π¦ =1
π₯+π
(B) π¦ =βπ₯3
3+ π
(C) πππ₯ (D) Unsolvable as equation is nonβ linear
5. If π₯ = π(π + sin π) and π¦ = π(1 β cos π) , then ππ¦
ππ₯
will be equal to
(A) sin (π
2) (B) cos (
π
2)
(C) tan (π
2) (D) cot (
π
2)
Data question 6 and 7 The complete solution of the ordinary differential equation
π2π¦
ππ₯2+ π
ππ¦
ππ₯+ ππ¦ = 0 is π¦ = π1πβπ₯ + π2πβ3π₯
6. Then π and π are (A) π = 3, π = 3
(B)π = 3, π = 4
(C) π = 4, π = 3
(π·) π = 4, π = 4
7. Which of the following is a solution of the differential equation
π2π¦
ππ₯2+ π
ππ¦
ππ₯+ (π + 1)π¦ = 0
(A) πβ3π₯ (B) π₯πβπ₯
(C) π₯πβ2π₯ (D) π₯2πβ2π₯
8. If π₯2 ππ¦
ππ₯+ 2π₯π¦ =
2 ln (π₯)
π₯ and π¦(1) = 0, then what is
(π) ?
(A) π (B) 1 (C) 1/π (D) 1/ π2
9. By a change of variable π₯(π’, π£) = π’π£, π¦(π’, π£) = π£ππ’
is double integral, the integrand π(π₯, π¦) changes to (π’π£, π£ππ’)π(π’, π£) . Then, π(π’, π£) is (A) 2π£/π’ (B) 2uv (C) π£2 (D) 1
10. For π2π¦
ππ₯2 + 4ππ¦
ππ₯+ 3π¦ = 3π2π₯, the particular integral
is
(A) 1
15π2π₯ (B)
1
5π2π₯
(C) 3π2π₯ (D) πΆ1πβπ₯ + πΆ2πβ3π₯
11. The solution of the differential equation ππ¦
ππ₯+ 2π₯π¦ =
πβπ₯2 with π¦(π) = 1 is
(A) (1 + π₯)π+π₯2
(B) (1 + π₯)πβπ₯2
(C) (1 β π₯)π+π₯2
(D) (1 β π₯)πβπ₯2
12. The solution of ππ¦
ππ₯= π¦2 with initial value π¦(π) = 1
bounded in the interval (A) ββ β€ π₯ β€ β
(B) ββ β€ π₯ β€ 1
(C) π₯ < 1, π₯ > 1
(D) β2 β€ π₯ β€ 2
13. The partial differential equation π2π
ππ₯2 +π2π
ππ¦2 +ππ
ππ₯+
ππ
ππ¦= 0 has
(A) Degree 1 order 2
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(B) degree 1 order 1 (C) degree 2 order 1 (D) degree 2 order 2
14. If π(π₯, π¦) and π(π₯, π¦) are functions with continuous
second derivatives, then π(π₯, π¦) + ππ(π₯, π¦) can be expressed as an analytic ffinction πππ₯ + ππ(π =
ββ1) , when
(A) ππ
ππ₯= β
ππ
ππ₯,
ππ
ππ¦=
ππ
ππ¦
(B) ππ
ππ¦= β
ππ
ππ₯,
ππ
ππ₯=
ππ
ππ¦
(C) π2π
ππ₯2+
π2π
ππ¦2=
π2π
ππ₯2+
π2π
ππ¦2= 1
(D) ππ
ππ₯+
ππ
ππ¦=
ππ
ππ₯+
ππ
ππ¦= 0
15. It is given that π¦β²β² + 2π¦β² + π¦ = 0, π¦(π) = 0, π¦(1) =
0. What is (π. 5) ?
(A) 0 (B) 0.37 (C) 0.62 (D) 1.13
16. Let π = π¦π₯ What is π2π
ππ₯ππ¦ at π₯ = 2, π¦ = 1 ?
(A) 0 (B) ln2
(C) 1 (D) 1
ln 2
17. Given that οΏ½ΜοΏ½ + 3π₯ = 0, and π₯(π) = 1, οΏ½ΜοΏ½(0) = 0,
what is π₯(1) ?
(A) β0.99 (B) β0.16 (C) 0.16 (D) 0.99
18. The solution of π₯ππ¦
ππ₯+ π¦ = π₯4 with the condition
π¦(1) =6
5 is
(A) π¦ =π₯4
5+
1
π₯
(B) π¦ =π₯4
5+ 1
(C) π¦ =4π₯4
5+
4
5π₯
(D) π¦ =π₯5
5+ 1
19. An analytic function of a complex variable π§ = π₯ +
ππ¦ is expressed as π(π§) = π’(π₯, π¦) + ππ£(π₯, π¦) where
π = ββ1. If π’ = π₯π¦, the expression for π£ should be
(A) (π₯+π¦)2
2+ π (B)
π₯2βπ¦2
2+ π
(C) π¦2βπ₯2
2+ π (π·)
(π₯βπ¦)2
2+ π
20. The Blasius equation, π3π
ππ3 +π
2
π2π
ππ2 = 0, is a
(A) second order nonlinear ordinary differential equation (B) third order nonlinear ordinary differential equation (C) third order linear ordinary differential equation (D) mixed order nonlinear ordinary differential equation
21. Consider the differential equation ππ¦
ππ₯= (1 + π¦2)π₯.
The general solution with constant π is
(A) π¦ = tan π₯2
2+ tan π
(B) π¦ = tan2(π₯
2+ π)
(C) π¦ = tan2(π₯
2) + π
(D) π¦ = tan (π₯2
2+ π)
22. A series expansion for the function sin π is
(A) 1 βπ2
2!+
π4
4!β β―
(B) 1 + π +π2
2!+
π3
3!+ β―
(C) π βπ3
3!+
π5
5!β β―
(D) π +π3
3!+
π5
5!+ β―
23. Consider the differential equation π₯2(π2π¦/ππ₯2) +
π₯(ππ¦/ππ₯) β 4π¦ = 0 with the boundary conditions of π¦(π) = 0 and π¦(1) = 1. The complete solution of the differential equation is (A) π₯2
(B) sin (ππ₯
2)
(C) ππ₯ sin (ππ₯
2)
(D) πβπ₯ sin (ππ₯
2)
Answers
1 D
2 A
3 D
4 A
5 C
6 C
7 C
8 D
9 A
10 B
11 B
12 C
13 A
14 B
15 A
16 C
17 D
18 A
19 C
20 B
21 D
22 C
23 A