# system of differential linear equations $y'=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}y$

find the solution to the problem $y'=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}y, y(0)=\begin{pmatrix}4\\0\end{pmatrix}$

I know i have to find the eigenvalues and eigenvectors of the matrix $A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$

there's only one eigenvalue which is $1$ and only "one" eigenvector and we can choose $\begin{pmatrix} 1\\0\end{pmatrix}$

but now I dont know what to do.

what comes next?

I should say that I know how to find the exponencial of a matrix and also I want to know how to solve this kind of problem in general, so techniques to solve this particular one that wouldnt work on a more general problem dont help me much.

• If you know how to find the exponential of a matrix, why not do it that way? $y=Ce^{\pmatrix{1&1\cr0&1\cr}x}$. – Gerry Myerson Jul 11 '13 at 9:49
• @GerryMyerson it seems like you know what the answer will look like. I'd like a constructive process to find the solution. is that possible? – complexguy Jul 11 '13 at 9:50
• Depends what you call a constructive process. There is a general form for the solution to $y'=Ay$. The form is more complicated when $A$ doesn't have enough eigenvectors. Once upon a time, someone hit upon a constructive process for finding that form. But once one person has done that, and has written it down, and it has found its way into the textbooks, the rest of us stop looking for that constructive process, and just copy what we see in the books. It's enough that we can prove the formulas work; we don't have to derive them ourselves. – Gerry Myerson Jul 11 '13 at 10:00
• @GerryMyerson i guess you're right. so I just take your solution and use the initial condition to find $C$? – complexguy Jul 11 '13 at 10:01
• Look at my answer here. – Kw08 Jan 30 '17 at 11:37

Note: this method is not a general one, but is probably easier for the particular example provided.

In this case, you do not need to diagonalize the matrix. Indeed, if we write $y=(y_1,y_2)$, then the system is simply: \left\{\begin{aligned} y_1'&=y_1+y_2\\ y_2'&=y_2 \end{aligned} \right. with initial conditions $y_1(0)=4$, and $y_2(0)=0$. Notice that the second equation is easy to solve: $$y_2(x)=C\exp(x).$$ Since $y_2(x)=0$, we get $y_2=0$, thus the first equation reduces to $y_1'=y_1$. Finally, $$y(x)=\begin{pmatrix}4\exp(x) \\ 0\end{pmatrix}.$$

• thank you for your answer. i am at fault for not being 100% clear on what I want. I want to know how to solve this kind of problem in general. I just took an example – complexguy Jul 11 '13 at 9:54
• @complexguy alright, I guess I'll delete this then. – zuggg Jul 11 '13 at 9:56
• I think adding a note that this answers the original question before the edits is better – complexguy Jul 11 '13 at 9:57

Let $$I_2 = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\quad\text{ and }\quad J_2 = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$$ Notice $I_2 J_2 = J_2 I_2$, $I_2^2 = I_2$ and $J_2^2 = 0_2$, we have:

$$e^{At} = e^{I_2t + J_2t} = e^{I_2t}e^{J_2t} = (I_2 + I_2t + I_2^2\frac{t^2}{2!} + \cdots)( I_2 + J_2t+ J_2^2\frac{t^2}{2}+\cdots)\\= I_2( 1 + t + \frac{t^2}{2} + \cdots)(I_2+J_2t) = e^t (I_2+J_2t) = e^t\begin{pmatrix}1 & t\\0 & 1\end{pmatrix}$$

This gives us:

$$y(t) = e^{At}y(0) = e^t \begin{pmatrix}1 & t\\0 & 1\end{pmatrix}\begin{pmatrix}4\\0\end{pmatrix} = \begin{pmatrix}4 e^t\\0\end{pmatrix}$$

The same approach allow you to deal with other linear ODE: $y' = A y$ where $A$ is not diagonalizable. Let's say you have use a similarity transform and bring $A$ into a Jordan normal block. For each Jordan block of size $n$:

$$B = \begin{pmatrix} \lambda & 1 & 0 &\ldots & 0\\ 0 & \lambda & 1 &\ldots & 0\\ & & \ddots & \ddots & &\\ 0 & 0 & \ldots & \lambda & 1\\ 0 & 0 & 0 &\ldots & \lambda \end{pmatrix}$$

You can rewrite $B$ as $\lambda I_n + J_n$ where $J_n$ is a $n\times n$ matrix with entries on the superdiagonal all equal to 1. Once again, $I_n J_n = J_n I_n$ and $J_n^n = 0_n$ and one has in general:

$$e^{Bt} = e^{(\lambda I_n + J_n)t} = e^{\lambda t} \left(1 + J_n t + J_n^2\frac{t^2}{2!} + \cdots J_n^{n-1}\frac{t^{n-1}}{(n-1)!}\right) = e^{\lambda t} \begin{pmatrix} 1 & t & \frac{t^2}{2!} &\ldots & \frac{t^{n-1}}{(n-1)!}\\ 0 & 1 & t &\ldots & \frac{t^{n-2}}{(n-2)!}\\ & & & \ddots & &\\ 0 & 0 & \ldots & 1 & t\\ 0 & 0 & 0 &\ldots & 1 \end{pmatrix}$$

If you know how to find an exponential of the matrix, then you no longer need to find eigenvectors and generalised eigenvectors, because your solution is precisely $y(t)=exp(At)y(0)$; in your case it is $y_1(t) = 4e^t,\,y_2(t)=0$.