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

Question

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

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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.

Answer 1

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, $$ y(x)=\begin{pmatrix}4\exp(x) \\ 0\end{pmatrix}. $$

Answer 2

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 $y_1(t) = 4e^t,\,y_2(t)=0$.

Answer 3

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:

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