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

Let us ﬁrst focus on the nonhomogeneous ﬁrst order equation

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

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

We notice that . This fact follows by writing down the series deﬁnition of ,

We have already seen that . Hence,

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

Recall from Exercise 3.8.7 that . Therefore, we obtain

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

The solution can then be written as

(3.5) |

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

Example 3.9.1: Suppose that we have the system

with initial conditions .Let us write the system as

We have previously computed for . We immediately have , simply by negating .

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

Then

Phew!

Let us check that this really works.

Similarly (exercise) . The initial conditions are also satisﬁed as well (exercise).

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

because matrix multiplication is not commutative.

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

Take the equation

(3.6) |

Assume that has linearly independent eigenvectors . Let us write

(3.7) |

That is, we wish to write our solution as a linear combination of eigenvectors of . If we can solve for the scalar functions through we have our solution . Let us decompose in terms of the eigenvectors as well. Write

(3.8) |

That is, we wish to ﬁnd through that satisfy (3.8). We note that since all the eigenvectors are independent, the matrix is invertible. We see that (3.8) can be written as , where the components of are the functions through . Then . Hence it is always possible to ﬁnd when there are linearly independent eigenvectors.

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

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

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 example for the equation we writeWe use the integrating factor to ﬁnd that

We integrate and solve for to get

If we are looking for just any particular solution, we could set to be zero. If we leave these constants in, we will get the general solution. Write , 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 . We take and note , just like before. Then if we write

we actually get the particular solution satisfying , because .

Example 3.9.2: Let . Solve where for .

The eigenvalues of are and 4 and corresponding eigenvectors are and respectively. This calculation is left as an exercise. We write down the matrix of the eigenvectors and compute its inverse (using the inverse formula for matrices)

We are looking for a solution of the form . We also wish to write in terms of the eigenvectors. That is we wish to write . Thus

So and .

We further want to write in terms of the eigenvectors. That is, we wish to write . Hence

So and . We plug our into the equation and get that

We get the two equations

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.is the constant of integration. As , then and hence . Similarly

As we have and hence . The solution is

That is, and .

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

We also have the method of undetermined coeﬃcients for systems. The only diﬀerence here is that we will have to take 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 will 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 . Find a particular solution of where .

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 are and 1 and corresponding eigenvectors are and respectively. Hence our complementary solution is

for some arbitrary constants and .

We would want to guess a particular solution of

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

Thus we have 8 unknowns. We write , , , and . We plug into the equation. First let us compute .

Now must equal , which is

That is, , . We would add this to the complementary solution to get the general solution of the problem. Notice also that both and were really needed.

Exercise 3.9.2: Check that and solve the problem. Also try setting 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 had done was very simple.

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

Suppose we have the equation

(3.9) |

Further, suppose we have solved the associated homogeneous equation and found the fundamental matrix solution . The general solution to the associated homogeneous equation is for a constant vector . Just like for variation of parameters for single equation we try the solution to the nonhomogeneous equation of the form

where is a vector valued function instead of a constant. Now we substitute into (3.9) to obtain

But is the fundamental matrix solution to the homogeneous problem, so , and

Hence . If we compute , then . We integrate to obtain and we have the particular solution . Let us write this as a formula

Note that if is constant and we let , then and hence we get a solution , which is precisely what we got using the integrating factor method.

Example 3.9.4: Find a particular solution to

(3.10) |

Here is most deﬁnitely not constant. Perhaps by a lucky guess, we ﬁnd that solves . Once we know the complementary solution we can easily ﬁnd a solution to (3.10). First we ﬁnd

Next we know a particular solution to (3.10) is

Adding the complementary solution we have that the general solution to (3.10).

Exercise 3.9.3: Check that and 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 for a vector of arbitrary constants. But that is precisely the complementary solution.

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

where is a constant matrix. If is of the form , then as two derivatives of cosine is again cosine we can try a solution of the form

and we do not need to introduce sines.

If the is a sum of cosines, note that we still have the superposition principle. If , then we would try for the problem , and we would try for the problem . Then we sum the solutions.

However, if there is duplication with the complementary solution, or the equation is of the form , 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.

If we have the system

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

Let , …, be the eigenvalues and , …, be eigenvectors. Again form the matrix . We write

We decompose in terms of the eigenvectors

And again .

Now we plug in and doing the same thing as before we obtain

We identify the coeﬃcients of the eigenvectors to get the equations

Each one of these equations is independent of the others. We solve each equation using the methods of chapter 2. We write , and we are done; we have a particular solution. If we have found the general solution for through , then again 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

The eigenvalues were and , with eigenvectors and . Therefore and . Therefore,

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

For each equation we use the method of undetermined coeﬃcients. We try for the ﬁrst equation and for the second equation. We plug in to get We solve each of these equations separately. We get and . And hence and . So our particular solution isThis solution matches what we got previously in § 3.6.

Exercise 3.9.4: Find a particular solution to , , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coeﬃcients.

Exercise 3.9.5: Find the general solution to , , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coeﬃcients.

Exercise 3.9.6: Find the general solution to , , a) using eigenvector decomposition, b) using undetermined coeﬃcients.

Exercise 3.9.7: Find the general solution to , , a) using eigenvector decomposition, b) using undetermined coeﬃcients.

Exercise 3.9.8: Take the equation

a) Check that

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

Exercise 3.9.101: Find a particular solution to , , a) using integrating factor method, b) using eigenvector decomposition, c) using undetermined coeﬃcients.