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3.9 Nonhomogeneous systems

Note: 3 lectures (may have to skip a little), somewhat different from §5.6 in [EP], §7.9 in [BD]

3.9.1 First order constant coefficient

Integrating factor

Let us first focus on the nonhomogeneous first order equation

⃗x′(t) = A ⃗x(t) +f⃗(t),

where A is a constant matrix. The first method we look at is the integrating factor method. For simplicity we rewrite the equation as

⃗x′(t) + P⃗x (t) = f⃗(t),

where P = −A . We multiply both sides of the equation by etP (being mindful that we are dealing with matrices that may not commute) to obtain

 tP ′ tP tP ⃗ e ⃗x (t) + e P ⃗x(t) = e f (t).

We notice that  tP tP Pe = e P . This fact follows by writing down the series definition of  tP e :

 ( 1 ) 1 PetP = P I + tP +-(tP)2 + ⋅⋅⋅ = P + tP 2 +--t2P 3 + ⋅⋅⋅ = ( 2 ) 2 1 2 tP = I + tP +--(tP) + ⋅⋅⋅ P = e P. 2

We have already seen that  ( ) d-etP = PetP dt . Hence,

 ( ) -d etP⃗x(t) = etP⃗f(t). dt

We can now integrate. That is, we integrate each component of the vector separately

 ∫ tP tP ⃗ e ⃗x(t) = e f(t) dt + ⃗c.

Recall from Exercise 3.8.7 that  tP −1 −tP (e ) = e . Therefore, we obtain

 ∫ ⃗x(t) = e−tP etPf⃗(t) dt + e−tP⃗c.

Perhaps it is better understood as a definite integral. In this case it will be easy to also solve for the initial conditions. Consider the equation with initial conditions

⃗x′(t) + P ⃗x(t) = ⃗f(t), ⃗x(0) = ⃗b.

The solution can then be written as

|----------∫---------------------| | −tP t sP −tP | |⃗x(t) = e e f⃗(s) ds + e ⃗b. | -------------0--------------------
(3.5)

Again, the integration means that each component of the vector esPf⃗(s) is integrated separately. It is not hard to see that (3.5) really does satisfy the initial condition ⃗x(0) = ⃗b .

 ∫ 0 −0P sP ⃗ −0P⃗ ⃗ ⃗ ⃗x(0) = e 0 e f(s) ds + e b = Ib = b.

Example 3.9.1: Suppose that we have the system

x′1 + 5x1 − 3x2 = et, ′ x2 + 3x1 − x2 = 0,
with initial conditions x1(0) = 1,x2(0 ) = 0 .

Let us write the system as

 [ ] [ ] [ ] ′ 5 − 3 et 1 ⃗x + 3 − 1 ⃗x = 0 , ⃗x(0) = 0 .

We have previously computed  tP e for  [5 −3] P = 3 −1 . We find  −tP e , simply by negating t .

 [ 2t 2t ] [ −2t −2t ] etP = (1 + 3t2)et − 3te 2t , e−tP = (1 − 3t)−e2t 3te −2t . 3te (1 − 3t)e − 3te (1 + 3t)e

Instead of computing the whole formula at once, let us do it in stages. First

∫ t ∫ t[ 2s 2s ] [s] esPf⃗(s) ds = (1 + 3s)e − 3se e ds 0 0 3se2s (1 − 3s)e2s 0 ∫ t[ 3s] = (1 + 3s3)se ds 0 3se ⌊||∫ t(1 + 3s)e3s ds⌋|| = |||⌈0 ∫ t 3s |||⌉ 0 3se ds [ te3t ] = (3t−1)e3t+1 (used integration by parts). ---3----

Then

 ∫ t ⃗x(t) = e−tP esP⃗f(s) ds + e−tP⃗b 0 [(1 − 3t)e−2t 3te−2t ][ te3t ] [(1 − 3t)e−2t 3te−2t ] [1 ] = −2t −2t (3t−1)e3t+1 + −2t −2t − 3te (1 + 3t)e 3 −3te (1 + 3t)e 0 [ te−2t ] [(1 − 3t)e−2t] = et (1 ) −2t + −2t −3 + 3 + te −3te [ (1 − 2t)e−2t ] = et (1 ) −2t . −3 + 3 − 2t e

Phew!

Let us check that this really works.

x′ + 5x − 3x = (4te−2t − 4e −2t) + 5(1 − 2t)e−2t + et − (1 − 6t)e−2t = et. 1 1 2

Similarly (exercise)  ′ x2 + 3x 1 − x2 = 0 . The initial conditions are also satisfied (exercise).

For systems, the integrating factor method only works if P does not depend on t , that is, P is constant. The problem is that in general

d [∫ ] ∫ -- e P(t)dt ⇔ P(t)e P(t)dt, dt

because matrix multiplication is not commutative.

Eigenvector decomposition

For the next method, note that eigenvectors of a matrix give the directions in which the matrix acts like a scalar. If we solve the system along these directions the computations are simpler as we treat the matrix as a scalar. We then put those solutions together to get the general solution for the system.

Take the equation

 ′ ⃗x(t) = A ⃗x(t) +f⃗(t).
(3.6)

Assume A has n linearly independent eigenvectors ⃗v ,⃗v ,...,⃗v 1 2 n . Write

⃗x (t) = ⃗v1ξ1(t) + ⃗v2ξ2(t) + ⋅⋅⋅ + ⃗vnξn(t).
(3.7)

That is, we wish to write our solution as a linear combination of eigenvectors of A . If we solve for the scalar functions ξ1 through ξn we have our solution ⃗x . Let us decompose f⃗ in terms of the eigenvectors as well. We wish to write

f⃗(t) = ⃗v g (t) + ⃗v g (t) + ⋅⋅⋅ + ⃗v g (t). 1 1 2 2 n n
(3.8)

That is, we wish to find g1 through gn that satisfy (3.8). Since all the eigenvectors are independent, the matrix E = [⃗v1 ⃗v2 ⋅⋅⋅ ⃗vn] is invertible. Write the equation (3.8) as ⃗f = E⃗g , where the components of ⃗g are the functions g1 through gn . Then ⃗g = E −1⃗f . Hence it is always possible to find ⃗g when there are n linearly independent eigenvectors.

We plug (3.7) into (3.6), and note that A⃗vk = λk⃗vk .

 ⃗x′ A⃗x ⃗f ◜-------′-----------------◞′ ◟------------------------◝′ ◜-----(------------------------◞◟-----------------------------◝) ◜------------------------◞◟ ------------------------◝ ⃗v1ξ1 + ⃗v2ξ2 + ⋅⋅⋅ + ⃗vnξn = A ⃗v1ξ1 + ⃗v2ξ2 + ⋅⋅⋅ + ⃗vnξn +⃗v1g 1 + ⃗v2g2 + ⋅⋅⋅ + ⃗vngn = A⃗v1ξ1 + A ⃗v2ξ2 + ⋅⋅⋅ + A⃗vnξn + ⃗v1g1 + ⃗v2g2 + ⋅⋅⋅ + ⃗vngn = ⃗v1λ1ξ1 + ⃗v2λ2ξ2 + ⋅⋅⋅ + ⃗vnλnξn +⃗v1g 1 + ⃗v2g2 + ⋅⋅⋅ + ⃗vngn = ⃗v1(λ 1ξ1 + g1) + ⃗v2(λ2ξ2 + g2) + ⋅⋅⋅ + ⃗vn(λnξn + gn).

If we identify the coefficients of the vectors ⃗v1 through ⃗vn we get the equations

 ′ ξ1 = λ 1ξ1 + g1, ξ′2 = λ 2ξ2 + g2, . .. ′ ξn = λnξn + gn.
Each one of these equations is independent of the others. They are all linear first order equations and can easily be solved by the standard integrating factor method for single equations. That is, for the kth equation we write
 ′ ξk(t) − λkξk(t) = gk(t).

We use the integrating factor  −λ t e k to find that

 [ ] d- ξ (t)e−λkt = e−λktg (t). dt k k

We integrate and solve for ξk to get

 ∫ λkt −λkt λkt ξk(t) = e e gk(t) dt + Cke .

If we are looking for just any particular solution, we could set Ck to be zero. If we leave these constants in, we get the general solution. Write ⃗x(t) = ⃗v1ξ1(t) + ⃗v2ξ2(t) + ⋅⋅⋅ +⃗vnξn(t) , and we are done.

Again, as always, it is perhaps better to write these integrals as definite integrals. Suppose that we have an initial condition ⃗x(0) = ⃗b . Take ⃗a = E−1⃗b to find ⃗b = ⃗v1a1 +⃗v2a 2 + ⋅⋅⋅ + ⃗vnan , just like before. Then if we write

|-----------∫------------------------| | λkt t −λks λkt | | ξk(t) = e e gk(s) ds + ake , | -------------0-----------------------

we actually get the particular solution ⃗x(t) = ⃗v1ξ1(t) +⃗v2ξ2(t) + ⋅⋅⋅ + ⃗vnξn(t) satisfying ⃗x(0) = ⃗b , because ξk(0) = ak .

Example 3.9.2: Let  [ ] A = 13 31 . Solve  ′ ⃗x = A⃗x + ⃗f where  [ t] f⃗(t) = 2e2t for  [ 3∕16] ⃗x(0) = −5∕16 .

The eigenvalues of A are − 2 and 4 and corresponding eigenvectors are [1] −1 and [1] 1 respectively. This calculation is left as an exercise. We write down the matrix E of the eigenvectors and compute its inverse (using the inverse formula for 2 × 2 matrices)

 [ 1 1] 1 [1 − 1] E = , E −1 =-- . −1 1 2 1 1

We are looking for a solution of the form ⃗x = [ 1]ξ1 + [1]ξ2 −1 1 . We first need to write f⃗ in terms of the eigenvectors. That is we wish to write  ⃗ [2et] [1] [1] f = 2t = −1 g1 + 1 g 2 . Thus

[ ] [ ] [ ][ ] [ ] g1 −1 2et 1 1 −1 2et et − t g = E 2t = -- 1 1 2t = et + t . 2 2

So g1 = et − t and g2 = et + t .

We further need to write ⃗x(0) in terms of the eigenvectors. That is, we wish to write  [ ] ⃗x(0) = 3∕16 = [1] a + [1]a −5∕16 −1 1 1 2 . Hence

[ ] [ ] [ ] a1 −1 3∕16 1∕4 a2 = E −5∕16 = −1∕16 .

So  1 a1 = ∕4 and  −1 a2 = ∕16 . We plug our ⃗x into the equation and get that

[ ] [ ] [ ] [ ] [ ] [ ] 1 ξ′ + 1 ξ′= A 1 ξ + A 1 ξ + 1 g + 1 g −1 1 1 2 − 1 1 1 2 − 1 1 1 2 [ ] [ ] [ ] [ ] = 1 (−2ξ1) + 1 4ξ2 + 1 (et − t) + 1 (et + t). − 1 1 −1 1

We get the two equations

 ′ t 1- ξ1 = − 2ξ1 + e − t, where ξ1(0) = a1 = 4, −1 ξ′2 = 4 ξ2 + et + t, where ξ2(0) = a2 = ---. 16
We solve with integrating factor. Computation of the integral is left as an exercise to the student. Note that you will need integration by parts.
 ∫ −2t 2t t −2t et -t 1- −2t ξ1 = e e (e − t) dt + C 1e = 3 − 2 + 4 + C 1e .

C 1 is the constant of integration. As  1 ξ1(0) = ∕4 , then 1 1 1 ∕4 = ∕3 + ∕4 + C1 and hence  −1 C1 = ∕3 . Similarly

 ∫ et t 1 ξ2 = e4t e−4t(et + t) dt + C2e4t = −-−--− ---+ C2e4t. 3 4 16

As ξ2(0) = −1∕16 we have −1∕16 = −1∕3 − 1∕16 + C2 and hence C2 = 1∕3 . The solution is

 [ ]( ) [ ]( ) [ 4t −2t ] 1 et −-e−2t 1-−-2t 1 e4t −-et 4t +-1 e-−3e--+ 3−1162t ⃗x(t) = − 1 3 + 4 + 1 3 − 1 6 = e−2t+e4t+2et+ 4t−5 . 3 16

That is,  e4t−e−2t- 3−12t x1 = 3 + 16 and  e−2t+e4t+2et 4t−5 x2 = 3 + 16 .

Exercise 3.9.1: Check that x1 and x2 solve the problem. Check both that they satisfy the differential equation and that they satisfy the initial conditions.

Undetermined coefficients

We also have the method of undetermined coefficients for systems. The only difference here is that we have to use unknown vectors rather than just numbers. Same caveats apply to undetermined coefficients for systems as for single equations. This method does not always work. Furthermore if the right hand side is complicated, we have to solve for lots of variables. Each element of an unknown vector is an unknown number. So in system of 3 equations if we have say 4 unknown vectors (this would not be uncommon), then we already have 12 unknown numbers that we need to solve for. The method can turn into a lot of tedious work. As this method is essentially the same as it is for single equations, let us just do an example.

Example 3.9.3: Let  [−1 0] A = −2 1 . Find a particular solution of  ′ ⃗ ⃗x = A ⃗x + f where  ⃗ [et] f (t) = t .

Note that we can solve this system in an easier way (can you see how?), but for the purposes of the example, let us use the eigenvalue method plus undetermined coefficients.

The eigenvalues of A are − 1 and 1 and corresponding eigenvectors are [1] 1 and [0] 1 respectively. Hence our complementary solution is

 [ ] [ ] ⃗xc = α1 1 e−t + α 2 0 et, 1 1

for some arbitrary constants α 1 and α 2 .

We would want to guess a particular solution of

⃗x = ⃗aet + ⃗bt + ⃗c.

However, something of the form ⃗aet appears in the complementary solution. Because we do not yet know if the vector ⃗a is a multiple of [0] 1 , we do not know if a conflict arises. It is possible that there is no conflict, but to be safe we should also try  t ⃗bte . Here we find the crux of the difference for systems. We try both terms ⃗aet and ⃗btet in the solution, not just the term ⃗btet . Therefore, we try

 t t ⃗x = ⃗ae + ⃗bte + ⃗ct + ⃗d.

Thus we have 8 unknowns. We write  [ ] ⃗a = aa12 ,  [ ] ⃗b = bb12 ,  [ ] ⃗c = cc12 , and  [ ] d⃗= dd12 . We plug ⃗x into the equation. First let us compute ⃗x′ .

 ( ) [ ] [ ] [ ] ⃗x′ = ⃗a + ⃗b et +⃗btet + ⃗c = a1 + b 1 et + b 1 tet + c1 . a2 + b 2 b 2 c2

Now ⃗x′ must equal A ⃗x + f⃗ , which is

A⃗x + ⃗f = A⃗aet + A ⃗btet + A⃗ct + A ⃗d +f⃗= [ ] [ ] [ ] [ ] [ ] [ ] = − a1 et + − b1 tet + − c1 t + − d1 + 1 et + 0 t. − 2a1 + a2 −2b1 + b2 − 2c1 + c2 − 2d1 + d 2 0 1
We identify the coefficients of et , tet , t and any constant vectors.
a 1 + b1 = −a1 + 1, a 2 + b2 = −2a1 + a2, b1 = −b1, b2 = −2b1 + b2, 0 = −c1, 0 = −2c1 + c2 + 1, c1 = −d1, c = −2d + d . 2 1 2
We could write the 8 × 9 augmented matrix and start row reduction, but it is easier to just solve the equations in an ad hoc manner. Immediately we see that b1 = 0 , c1 = 0 , d1 = 0 . Plugging these back in, we get that c2 = − 1 and d2 = − 1 . The remaining equations that tell us something are
 a1 = − a1 + 1, a2 + b2 = − 2a1 + a 2.
So a = 1∕2 1 and b = − 1 2 . Finally, a 2 can be arbitrary and still satisfy the equations. We are looking for just a single solution so presumably the simplest one is when a2 = 0 . Therefore,
 [1∕2] [ 0 ] [ 0 ] [ 0 ] [ 1et ] ⃗x = ⃗aet + ⃗btet + ⃗ct + d⃗= et + tet + t + = t2 . 0 − 1 −1 −1 −te − t − 1

That is, x1 = 12et , x 2 = − tet − t − 1 . We would add this to the complementary solution to get the general solution of the problem. Notice also that both  t ⃗ae and ⃗ t bte were really needed.

Exercise 3.9.2: Check that x1 and x2 solve the problem. Also try setting a2 = 1 and again check these solutions. What is the difference between the two solutions we can obtain in this way?

As you can see, other than the handling of conflicts, undetermined coefficients works exactly the same as it did for single equations. However, the computations can get out of hand pretty quickly for systems. The equation we considered was pretty simple.

3.9.2 First order variable coefficient

Just as for a single equation, there is the method of variation of parameters. For constant coefficient systems, it is essentially the same thing as the integrating factor method we discussed earlier. However, this method works for any linear system, even if it is not constant coefficient, provided we somehow solve the associated homogeneous problem.

Suppose we have the equation

⃗x′ = A (t)⃗x + f⃗(t).
(3.9)

Further, suppose we have solved the associated homogeneous equation  ′ ⃗x = A (t)⃗x and found a fundamental matrix solution X(t) . The general solution to the associated homogeneous equation is X (t)⃗c for a constant vector ⃗c . Just like for variation of parameters for single equation we try the solution to the nonhomogeneous equation of the form

⃗xp = X (t)⃗u(t),

where ⃗u (t) is a vector valued function instead of a constant. We substitute x⃗p into (3.9) to obtain

X′(t)⃗u(t) + X (t)⃗u′(t)= A(t)X (t)⃗u(t)+f⃗(t). ◟-------------------◝ ◜′-------------------◞ ◟--------◝ ◜--------◞ ⃗xp(t) A(t)⃗xp(t)

But X (t) is a fundamental matrix solution to the homogeneous problem. So X′(t) = A(t)X (t) , and

 ′ // ′ ′ // /X /(t/)⃗u (t) + X (t)⃗u (t) = /X/(/t)⃗u(t) + ⃗f(t).

Hence X(t)⃗u ′(t) = ⃗f(t) . If we compute [X(t)]−1 , then ⃗u′(t) = [X (t)]−1 ⃗f(t) . We integrate to obtain ⃗u and we have the particular solution ⃗x = X(t)⃗u (t) p . Let us write this as a formula

|----------∫-----------------| | −1 ⃗ | | ⃗xp = X (t) [X (t)] f(t) dt.| ------------------------------

If A is constant and  tA X(t) = e , then  −1 −tA [X(t)] = e . We get a solution  tA ∫ −tA ⃗ ⃗xp = e e f(t) dt , which is precisely what we got using the integrating factor method.

Example 3.9.4: Find a particular solution to

 [ ] [ ] ⃗x ′ =--1--- t −1 ⃗x + t (t2 + 1). t2 + 1 1 t 1
(3.10)

Here  [ ] A = t21+1 t1 −t1 is most definitely not constant. Perhaps by a lucky guess, we find that  [ ] X = 1t − 1t solves X ′(t) = A(t)X (t) . Once we know the complementary solution we can easily find a solution to (3.10). First we find

 [ ] [X(t)]−1 = --1--- 1 t . t2 + 1 − t 1

Next we know a particular solution to (3.10) is

 ∫ ⃗x = X (t) [X(t)]−1 ⃗f(t) dt p [1 − t]∫ 1 [ 1 t][ t] = ------ (t2 + 1 ) dt [t 1] t[2 + 1 −]t 1 1 1 − t∫ 2t = t 1 − t2 + 1 dt [ ][ ] 1 − t t2 = t 1 − 1t3 + t [ 14 ] 3 = 3t . 23t3 + t

Adding the complementary solution we find the general solution to (3.10):

 [1 −t][c1] [ 1t4 ] [ c1 − c2t + 1t4 ] ⃗x = + 2 33 = 3 2 3 . t 1 c2 3t + t c2 + (c1 + 1)t + 3t

Exercise 3.9.3: Check that  1 x1 = 3t4 and  2 x2 = 3t3 + t really solve (3.10).

In the variation of parameters, just like in the integrating factor method we can obtain the general solution by adding in constants of integration. That is, we will add X (t)⃗c for a vector of arbitrary constants. But that is precisely the complementary solution.

3.9.3 Second order constant coefficients

Undetermined coefficients

We have already seen a simple example of the method of undetermined coefficients for second order systems in § 3.6. This method is essentially the same as undetermined coefficients for first order systems. There are some simplifications that we can make, as we did in § 3.6. Let the equation be

⃗x′′ = A ⃗x + F⃗(t),

where A is a constant matrix. If ⃗ F(t) is of the form ⃗ F 0cos(ωt) , then as two derivatives of cosine is again cosine we can try a solution of the form

⃗xp = ⃗ccos(ωt),

and we do not need to introduce sines.

If the F⃗ is a sum of cosines, note that we still have the superposition principle. If F⃗(t) = ⃗F0 cos(ω 0t) + ⃗F1co s(ω 1t) , then we would try ⃗a cos(ω0t) for the problem ⃗x′′ = A⃗x + ⃗F 0cos(ω0t) , and we would try ⃗bcos(ω 1t) for the problem ⃗x ′′ = A ⃗x + ⃗F1cos(ω 1t) . Then we sum the solutions.

However, if there is duplication with the complementary solution, or the equation is of the form  ′′ ′ ⃗x = A⃗x + B ⃗x +F⃗(t) , then we need to do the same thing as we do for first order systems.

You will never go wrong with putting in more terms than needed into your guess. You will find that the extra coefficients will turn out to be zero. But it is useful to save some time and effort.

Eigenvector decomposition

If we have the system

x⃗′′ = A ⃗x + ⃗f(t),

we can do eigenvector decomposition, just like for first order systems.

Let λ1,λ2,...,λn be the eigenvalues and ⃗v1,⃗v2,...,⃗vn be eigenvectors. Again form the matrix E = [⃗v1 ⃗v2 ⋅⋅⋅ ⃗vn] . Write

⃗x (t) = ⃗v1ξ1(t) + ⃗v2ξ2(t) + ⋅⋅⋅ + ⃗vnξn(t).

Decompose ⃗f in terms of the eigenvectors

f⃗(t) = ⃗v1g1(t) + ⃗v2g2(t) + ⋅⋅⋅ + ⃗vngn(t),

where, again, ⃗g = E−1f⃗ .

We plug in, and as before we obtain

 ⃗x′′ A⃗x f⃗ ◜ ------′---′----------------◞′′◟-------------------------◝′′ ◜ ---(--------------------------◞ ◟-----------------------------◝) ◜------------------------◞ ◟------------------------◝ ⃗v 1ξ1 + ⃗v2ξ2 + ⋅⋅⋅ +⃗vnξn = A ⃗v1ξ1 + ⃗v2ξ2 + ⋅⋅⋅ + ⃗vnξn + ⃗v1g1 + ⃗v2g2 + ⋅⋅⋅ + ⃗vngn = A ⃗v1ξ1 + A⃗v2ξ2 + ⋅⋅⋅ + A⃗vnξn + ⃗v1g1 +⃗v2g 2 + ⋅⋅⋅ + ⃗vngn = ⃗v1λ1ξ1 +⃗v2λ2ξ2 + ⋅⋅⋅ + ⃗vnλnξn + ⃗v1g1 +⃗v2g2 + ⋅⋅⋅ + ⃗vngn = ⃗v1(λ1ξ1 + g 1) +⃗v2(λ2ξ2 + g2) + ⋅⋅⋅ + ⃗vn(λn ξn + gn).

We identify the coefficients of the eigenvectors to get the equations

 ′′ ξ1 = λ1ξ1 + g1, ξ′′2 = λ2ξ2 + g2, ... ′′ ξn = λnξn + gn.
Each one of these equations is independent of the others. We solve each equation using the methods of chapter 2. We write ⃗x(t) = ⃗v1ξ1(t) + ⃗v2ξ2(t) + ⋅⋅⋅ + ⃗vnξn(t) , and we are done; we have a particular solution. We find the general solutions for ξ1 through ξn , and again ⃗x(t) = ⃗v ξ (t) + ⃗v ξ(t) + ⋅⋅⋅ + ⃗v ξ (t) 1 1 2 2 n n is the general solution (and not just a particular solution).

Example 3.9.5: Let us do the example from § 3.6 using this method. The equation is

 [ ] [ ] ⃗x′′ = − 3 1 ⃗x + 0 cos(3t). 2 −2 2

The eigenvalues are − 1 and − 4 , with eigenvectors [12] and [−11] . Therefore E = [12 1 −1] and E −1 = 1[1 1] 3 2 −1 . Therefore,

[g1] 1 [1 1 ][ 0 ] [ 2 cos(3t) ] = E −1⃗f(t) = -- = 3−2 . g2 3 2 −1 2 cos(3t) 3 cos(3t)

So after the whole song and dance of plugging in, the equations we get are

ξ′′= − ξ + 2-cos(3t), 1 1 3 ′′ 2 ξ2 = − 4ξ2 −--cos(3t). 3
For each equation we use the method of undetermined coefficients. We try C1 cos(3t) for the first equation and C 2cos(3t) for the second equation. We plug in to get
−9C cos(3t) = − C co s(3t) + 2cos(3t), 1 1 3 2 −9C 2cos(3t) = − 4C2 cos(3t) −-co s(3t). 3
We solve each of these equations separately. We get − 9C 1 = − C1 + 2∕3 and − 9C2 = − 4C2 − 2∕3 . And hence C 1 = −1∕12 and C2 = 2∕15 . So our particular solution is
 [ ] ( ) [ ] ( ) [ ] 1 −-1 1 -2- 1∕20 ⃗x = 2 1 2 co s(3t) + −1 15 cos(3t) = −3∕10 co s(3t).

This solution matches what we got previously in § 3.6.

3.9.4 Exercises

Exercise 3.9.4: Find a particular solution to x′ = x + 2y + 2t , y′ = 3x + 2y − 4 , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coefficients.

Exercise 3.9.5: Find the general solution to  ′ x = 4x + y − 1 ,  ′ t y = x + 4y − e , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coefficients.

Exercise 3.9.6: Find the general solution to x′′= − 6x + 3x + cos(t) 1 1 2 , x′′ = 2x − 7x + 3co s(t) 2 1 2 , a) using eigenvector decomposition, b) using undetermined coefficients.

Exercise 3.9.7: Find the general solution to  ′′ x1 = −6x 1 + 3x2 + cos(2t) ,  ′′ x2 = 2x 1 − 7x2 + 3cos(2t) , a) using eigenvector decomposition, b) using undetermined coefficients.

Exercise 3.9.8: Take the equation

 [1 −1] [t2] ⃗x′ = t 1 ⃗x + . 1 t −t

a) Check that

 [ ] [ ] ⃗xc = c1 tsin t + c2 tcost −tco st tsint

is the complementary solution. b) Use variation of parameters to find a particular solution.

Exercise 3.9.101: Find a particular solution to x′ = 5x + 4y + t , y′ = x + 8y − t , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coefficients.

Exercise 3.9.102: Find a particular solution to  ′ t x = y + e ,  ′ t y = x + e , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coefficients.

Exercise 3.9.103: Solve  ′ x 1 = x2 + t ,  ′ x2 = x 1 + t with initial conditions x1(0) = 1 , x 2(0) = 2 , using eigenvector decomposition.

Exercise 3.9.104: Solve  ′′ x1 = −3x 1 + x2 + t ,  ′′ x2 = 9x1 + 5x2 + co s(t) with initial conditions x1(0) = 0 , x2(0) = 0 ,  ′ x1(0) = 0 ,  ′ x2(0 ) = 0 , using eigenvector decomposition.