Review
I For the linear equation with constant coefficients,
ay ′′ + by ′ + cy = 0,
when the roots of the characteristic equation are equal, thegeneral solution is
y = Aeλt + Bteλt .
I For the linear homogeneous equation withnot-necessarily-constant coefficients, if we have one solutionu, we can find the general solution Au + Bvu using themethod of reduction of order.
I We look for a solution y = vu, and when we plug this into theequation, the coefficients of v will cancel out and leave afirst-order equation for v ′.
I Solve that equation, and then integrate to find v .
The non-homogeneous equation
Consider the non-homogeneous second-order equation withconstant coefficients:
ay ′′ + by ′ + cy = F (t).
I The difference of any two solutions is a solution of thehomogeneous equation.
I Suppose we have one solution u. Then the general solution isu plus the general solution of the homogeneous equation.
I Proof, let y be any solution. Then y − u solves thehomogeneous equation, so y = u+ a solution of thehomogeneous equation.
The non-homogeneous equation
I Suppose we have one solution u. Then the general solution isu plus the general solution of the homogeneous equation.
I So, solving the equation boils down to finding just onesolution.
I But there is no foolproof method for doing that (for anyarbitrary right-hand side F (t)).
I We can do it in some useful common cases.
Constant coefficients or not?
I What we did up to now works for any second-order linearequation.
I What we will do next assumes constant coefficients.
I Textbook does not emphasize the transition
The method of undetermined coefficients
This method applies to a second-order linear equation withconstant coefficients if the right-hand side F (t) has one of a fewparticularly simple forms:
I A polynomial
I An exponential times a polynomial: eαtP(t). The exponent isa constant times t.
I Complex exponentials are allowed, so we also can handleP(t) cos t,
I P(t) sin t
I eαt cos t, etc.
Also the right-hand side can be a linear combination of expressionsof those forms, since if
au′′ + bu′ + cu = F (t)
andav ′′ + bv ′ + cv = G (t)
then y = au + bv satisfies
ay ′′ + by ′ + cy = aF (t) + bG (t)
The method of undetermined coefficients
I Assume that the sought-after solution y has the same form asthe right-hand side P(t).
I That assumption mentions some unknown coefficients.
I Plug in that solution to the differential equation and simplify
I You get equations for the coefficients.
I If they are solvable you are done.
Example: y ′′ − 3y ′ − 4y = 3e2t
I 3e2t is (a constant times) an exponential.
I Assume y = Ae2t .
I Plug and grind:
y ′ = 2Ae2t
y ′′ = 4Ae2t
y ′′ − 3y ′ − 4y = 4Ae2t − 3(2Ae2t − 4Ae2t
= −6Ae2t
= 3e2t if this is to be a solution
A = −1
2the equation is solvable, so it works.
Example: y ′′ − 3y ′ − 4y = 3e2t
I We found the particular solution
y = −1
2e2t
I To find the general solution we need to solve thehomogeneous equation too.
I The characteristic equation is
λ2 − 3λ− 4 = 0
which has roots λ = 4 and λ = −1.
I So the general solution of the homogeneous equation is
ae4t + be−t
I and the general solution of the non-homogeneous equation is
ae4t + be−t − 1
2e2t .
Special cases can cause trouble
I If the proposed solution of the non-homogeneous equation isactually already a solution of the homogeneous equation, thenthe equations for the coefficients cannot be solved.
I For example, in the preceding problem, the homogeneousequation had solutions e−t and e4t . What if the right-handside had been e4t?
y ′′ − 3y ′ − 4y = e4t
I Then we would have assumed y = Ae4t , but when we plug itin, we get 0 on the left, which can never be equal to e4t onthe right.
I So the method needs modification in such cases.
What to do in a special caseI If the proposed solution of the non-homogeneous equation is
already a solution of the homogeneous equation, then theassumed form should be multiplied by a factor of t.
I For example:y ′′ − 3y ′ − 4y = e4t
Since e4t is a solution of the homogeneous equation, weinstead assume
y = Ate4t .
I Now we plug and grind:
y ′ = Ae4t(4t + 1)
y ′′ = Ae4t(4 + 4(4t + 1)) = Ae4t(16t + 8)
y ′′ − 3y ′ − 4y = Ae4t(16t + 8)− 3Ae4t(4t + 1)− 4Ate4t
= −2Ae4t
I So if we take A = −1/2 we have a solution.I Note that the te4t terms canceled out. That’s because e4t
solves the homogeneous equation
Another special case
I If the right-hand side is already a solution of the homogeneousequation, and
I if in addition the characteristic equation has double roots, then
I multiply by t2 instead of only t.
I For exampley ′′ + 2y ′ + 1 = e−t
I The characteristic equation is (λ+ 1)2 = 0, so thehomogeneous equation has solutions e−t and te−t . So theright side is a solution of the homogeneous equation, but so iste−t (which we would otherwise try as a solution). So insteadwe try
y = At2e−t
and you can check that it works.
Summary of the Method of Undetermined Coefficients
I It’s for linear non-homogeneous second-order equations withconstant coefficients.
I Assume a solution that has the same form as the right handside.
I That is, a polynomial, or an exponential or trig function timesa polynomial.
I Use letters for the polynomial coefficients and solve for them.
I Use an extra factor of t if the right side already solves thehomogeneous equation, or an extra factor of t2 if in additionthe characteristic equation has multiple roots.
I For proof that it works, see pages 181-182 of the text. Youjust use letters for the coefficients of the right-hand side andplug and grind as you do when solving a particular example.
Variation of Parameters
I This method “works” on any second-order non-homogeneousequation, constant coefficients or not.
I But the “solution” involves an integral, so it may be harder towork with.
I Also it requires have a fundamental set of solutions of thehomogeneous equation, which may not be easy if the equationdoesn’t have constant coefficients.
I Therefore use the method of undetermined coefficients if it isapplicable.
Variation of Parameters
We consider the equation
y ′′ + p(t)y ′ + q(t)y = g(t)
and suppose we have somehow found a fundamental set of twosolutions y1 and y2 of the homogeneous equation
y ′′ + p(t)y ′ + q(t)y = 0.
The basic idea is to look for a solution in the form
y = uy1 + vy2
where u and v are not constants, but functions of t.
Variation of Parameters
Our plan is to plugy = uy1 + vy2
into the equation
y ′′ + p(t)y ′ + q(t)y = g(t)
So we start by differentiating y :
y ′ = u′y1 + uy ′1 + v ′y2 + vy ′2
Now we assume u′y1 + v ′y2 = 0. Then
y ′ = uy ′1 + vy ′2
y ′′ = u′y1 + uy ′′1 + v ′y ′2 + v ′′2
So now plug the expressions for y ′ and y ′′ into the originalequation. Specifically, plug
y ′ = uy ′1 + vy ′2
y ′′ = u′y1 + uy ′′1 + v ′y ′2 + v ′′2
intoy ′′ + p(t)y ′ + q(t)y = g(t)
The coefficient of u is y ′′1 + py ′1 + qy1, which is zero since y1 is asolution of the homogeneous equation. The coefficient of v issimilarly zero since y2 is a solution. We are left with
u′y ′1 + v ′y ′2 = g(t).
I We have proved that if we solve the equations
u′y1 + v ′y2 = 0 which we assumed above
u′y ′1 + v ′y ′2 = g(t)
then y = uy1 + vy2 will solve the non-homogeneous equation.
I But these are algebraic equations for u′ and v ′.
I The solution has the Wronskian W (y1, y2) in thedenominator, which is nonzero.
I So it’s always possible to solve for u′ and v ′.
I Then if we can integrate the answers, we have our solution.
Solving
u′y1 + v ′y2 = 0
u′y ′1 + v ′y ′2 = g(t)
we get
u′ = −y2g
W
v ′ =y1g
W
where W is the Wronskian
W = y1y′2 − y2y
′1
Therefore a solution is
u = −∫
y2(t)g(t)
W (t)dt + c1
v =
∫y1(t)g(t)
W (t)dt + c2
y = uy1 + vy2
An example: y ′′ + 4y = 3 csc t
I Although the coefficients are constant, the right side is not apolynomial times an exponential.
I So we can’t use the method of undetermined coefficients.
I We can solve the homogeneous equation, since thecoefficients are constant.
I The details of this example are on pages 185-187, presentedas a motivation for the method of variation of parameters.
An example: y ′′ − 3y ′ − 4y = t2
I What method should we use?
I Undetermined coefficients, since we have a polynomial on theright.
I What form should we assume for the solution?I y = a + bt + ct2. No exponential, since the right side is a
polynomial.I Plugging this into the equation we find it will work if
2c − 3b − 4a = 0
−6c − 4b = 0
−4c = 1
I It is possible to solve these equations:c = −1/4, b = 3/8, a = −13/32
I What is the general solution?
y = −13
32+
3
8t − t2
4+ Ae−t + Be4t .
An example: y ′′ − 3y ′ − 4y = t2
I What method should we use?I Undetermined coefficients, since we have a polynomial on the
right.I What form should we assume for the solution?
I y = a + bt + ct2. No exponential, since the right side is apolynomial.
I Plugging this into the equation we find it will work if
2c − 3b − 4a = 0
−6c − 4b = 0
−4c = 1
I It is possible to solve these equations:c = −1/4, b = 3/8, a = −13/32
I What is the general solution?
y = −13
32+
3
8t − t2
4+ Ae−t + Be4t .
An example: y ′′ − 3y ′ − 4y = t2
I What method should we use?I Undetermined coefficients, since we have a polynomial on the
right.I What form should we assume for the solution?I y = a + bt + ct2. No exponential, since the right side is a
polynomial.I Plugging this into the equation we find it will work if
2c − 3b − 4a = 0
−6c − 4b = 0
−4c = 1
I It is possible to solve these equations:c = −1/4, b = 3/8, a = −13/32
I What is the general solution?
y = −13
32+
3
8t − t2
4+ Ae−t + Be4t .
An example: y ′′ − 3y ′ − 4y = t2
I What method should we use?I Undetermined coefficients, since we have a polynomial on the
right.I What form should we assume for the solution?I y = a + bt + ct2. No exponential, since the right side is a
polynomial.I Plugging this into the equation we find it will work if
2c − 3b − 4a = 0
−6c − 4b = 0
−4c = 1
I It is possible to solve these equations:c = −1/4, b = 3/8, a = −13/32
I What is the general solution?
y = −13
32+
3
8t − t2
4+ Ae−t + Be4t .