How to solve this matrix equation

Question

Consider the system of ODE in $\Bbb R^2 $

$\dfrac{dY}{dt}=AY$ where $Y(0)=$ \begin{bmatrix} 0 \\ 1\end{bmatrix} $t>0$

where $ A=$ \begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix}

and $Y(t)=$\begin{bmatrix} y_1(t) \\ y_2(t)\end{bmatrix}

My try: $dy_1(t)=-y_1(t)+y_2(t)$ and $dy_2(t)=-y_2(t)$

On solving the second equation I got $y_2(t)=e^{-t}$

Putting this in the first one I got : $dy_1(t)+y_1(t)=e^{-t}$

On solving the homogeneous and complementary function I got

$y_1(t)=Ae^{-t}+te^{-t}$

Putting $t=0$ we get $A=0$ so $y_1(t)=te^{-t}$.

Answer 1

The solution of the second differential equation, i.e., $dy_1(t)+y_1(t)=e^{-t}$ is $y_1(t)=ce^{-t}+te^{-t}$ and according to the boundary condition, it would be $y_1(t)=te^{-t}$.

Answer 2

My comment above explains your error.

For another perspective, you can use the matrix exponential. The solution to the IVP is given by

$$\mathbf{y}(t)=e^{At}\mathbf{y}(0)$$

You just need to compute $e^{At}$. But note that $A=D+N$ where $D$ is $\text{diag}(-1,-1)$, $N$ is nilpotent with $N^2=0$, and $D$ and $N$ commute. Therefore

$$e^{At}=e^{(D+N)t}=e^{Dt}e^{Nt}=\begin{pmatrix}e^{-t}&0\\0&e^{-t}\end{pmatrix}\left(I+Nt+\frac{1}{2}N^2t^2+\cdots\right)=\begin{pmatrix}e^{-t}&0\\0&e^{-t}\end{pmatrix}\left(I+Nt\right)$$

But $$I+Nt=\begin{pmatrix}1&t\\0&1\end{pmatrix}$$

So the solution is

$$\mathbf{y}(t)=\begin{pmatrix}e^{-t}&0\\0&e^{-t}\end{pmatrix}\begin{pmatrix}1&t\\0&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}te^{-t}\\e^{-t}\end{pmatrix}$$