# Inhomogeneous system of differential equations

To my grief this week I had a very incoherent class about differential equations and I find myself unable to solve

Problem: $$\begin{cases} x'(t)=2x(t)-y(t)+4t \\ y'(t)=x(t)+e^{-t} \end{cases}$$

I can rewrite the above system as: $$\begin{pmatrix}x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix}2 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}+ \begin{pmatrix} 4t \\ e^{-t} \end{pmatrix}$$

The Matrix $A \in \mathbb{R}^{(2,2)}$ happens to be non diagonalizable, which is a pity because otherwise I might have been able to compute $\exp(At)$.

I figured out that for the homogeneous system I have Eigenvalue $1$ and Eigenvector (1,1). According to my class $$z(t)=\exp(\lambda t)v$$ is a solution to the homogeneous differential equation where $\lambda =1$ and $v$ is the corresponding Eigenvector. Thus for the homogenous system: $$z(t)=\begin{pmatrix}x(t) \\ y(t) \end{pmatrix} = c_1 e^t \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ Is the solution, which even checks out.

My next attempt would have been to use variation of constants but so far I have no luck with that method.

If I let $c(t)$ be the 'constant' then $x(t)=y(t)=c(t)e^t \implies x'(t)=y'(t)= c'(t)e^t+x(t)$ Plugging that back into my original differential equation I obtain that $$c'(t)=2te^{-t} + \frac{1}{2}e^{-2t} \implies c(t) -2e^{-t}t - \frac{e^{-2t}}{4}-2e^{-t}$$ So my solution should be $$x(t)=y(t)=-2t- \frac{1}{4}e^{-t}-2$$

But this does not check out

• Have you heard of Jordan form and how to compute $\exp (A)$ with its help? – Antoine Sep 29 '15 at 10:14
• @Antoine unfortunately no. In fact I do not even know what the JNF is from a linear algebraic point of view (my Professor taught us the Smith Normal Form instead). Please feel free to write down the basic concept of doing so, maybe we will eventually cover it in class. – Spaced Sep 29 '15 at 10:21
• See this. I'm confused about something, though. In the first part of your question it is implied that you're able to compute matrix exponentials, I have no idea why Antoine acts as if you don't know it, nor why seem to confirm is suspicion. – Git Gud Sep 29 '15 at 10:56
• @GitGud, frankly said I don't know how to do that, I do know the formula and I have read (and understood) the abstract proof. However I did never compute an exponential Matrix. I would know how to compute the exponential Matrix if and only if the Matrix in discussion is diagonalizable. Following C.T. Michaels Advice in his Analysis II book, it becomes very tough to compute $\exp(At)$ if the matrix is non diagonalizable. – Spaced Sep 29 '15 at 11:25
• @Spaced In that case check example 6.5 here and also this for a couple of examples on how to get the JNF from SNF. – Git Gud Sep 29 '15 at 16:01

Variations of constants is used to find a particular solution of the complete system once you know the general solution of the homogeneous system. Let $U=(1,1)$ be an eigenvector of the eigenvalue $1$ and let $V$ be another vector. Look for a solution of the form $$t\,e^t\,U+e^t\,V.$$ This leads to the equation $(A-I)U=V$.
Another way to solve the system is to reduce it to a second order equation: $$x''=2\,x'-y'+4=2\,x'-x-e^{-t}+4.$$
Subtract the second equation from the first. Then \begin{equation*} x^{\prime }(t)-y^{\prime }(t)=x(t)-y(t)+4t-e^{t} \end{equation*} from which $u(t)=x(t)-y(t)$ can be found. Next \begin{equation*} x^{\prime }(t)=2x(t)-y(t)+4t=x(t)+u(t)+4t \end{equation*} from which $x(t)$ can be obtained.