# Solve the initial value problem $\dot{\bf x}=\begin{bmatrix}1 & -1\\ 0 & 1\end{bmatrix} \bf x+\begin{bmatrix} t\\ 0 \end{bmatrix}$

Solve the following initial value problem: $$\dot{\bf x}=\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix} \bf x+\begin{bmatrix} t\\ 0 \end{bmatrix},\qquad \mathbf{x}(0)=\begin{bmatrix} 0\\ 1 \end{bmatrix}.$$

This system can be solved by using the elimination method, but here I want to use another approach as following:

Let $\mathbf{A}=\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix},$ and $\mathbf{\beta}=\begin{bmatrix} t\\ 0 \end{bmatrix}$.
Note that $$\mathbf{A}-\mathbf{I}=\begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix},\qquad \text{and}\qquad (\mathbf{A}-\mathbf{I})^2=\begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix}.$$ Therefore, $e^{(\mathbf{A}-\mathbf{I})t}=\mathbf{I}t+(\mathbf{A}-\mathbf{I})t$, and \begin{aligned} e^{\mathbf{A}t}&=e^{(\mathbf{A}-\mathbf{I})t}\cdot e^{\mathbf{I}t}\\ &=\mathbf{I}te^{\mathbf{I}t}+(\mathbf{A}-\mathbf{I})te^{\mathbf{I}t}\\ &=\begin{bmatrix} te^t & -te^t\\ 0 & te^t \end{bmatrix}. \end{aligned} Thus, the complementary solution is $$\mathbf{x_c}=e^{\mathbf{A}t}\cdot \mathbf{x_0}=\begin{bmatrix} -te^t\\ te^t \end{bmatrix},$$ and the particular solution is \begin{aligned} \mathbf{x_p}&=e^{\mathbf{A}t}\int_0^te^{-\mathbf{A}\tau}\beta(\tau)d\tau\\ &=\begin{bmatrix} t+(t-2)e^t+2\\ 0 \end{bmatrix}. \end{aligned} Hence, the solution is given by $$\mathbf{x}=\mathbf{x_c}+\mathbf{x_p}=\begin{bmatrix} t-2e^t+2\\ te^t \end{bmatrix}.$$

But unfortunately, I find this answer is incorrect. I don't know where I'm wrong.
Any help is appreciated!

You should have

$$e^{(\mathbf{A}-\mathbf{I})t}=\color{red}{\mathbf{I}}+(\mathbf{A}-\mathbf{I})t$$

• Ooooops, thank you for your hint!
– R.C
Dec 14, 2017 at 1:05

The exponential matrix of $\mathbf{A}=\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}$ is not correct: $e^{\mathbf{A}t}\not=\begin{bmatrix} te^t & -te^t\\ 0 & te^t \end{bmatrix}.$ It should be $$e^{\mathbf{A}t}=\sum_{n\geq 0}\mathbf{A}^n\frac{t^n}{n!}= \sum_{n\geq 0}\begin{bmatrix} 1 & -n\\ 0 & 1 \end{bmatrix}\frac{t^n}{n!}= \begin{bmatrix} e^t & -te^t\\ 0 & e^t \end{bmatrix}.$$

• See you again! :P
– R.C
Dec 14, 2017 at 1:10