$$\begin{align}\frac{d}{dt}x_1&=-x_1+\sin(t)x_2 \\
\frac{d}{dt}x_2&=-t^2x_2\end{align}$$
What is the state transition matrix for these differential equations? How do you get it? Please show me steps. Thank you!
Edit: Thank you kind stranger for helping me with the formatting!
Best Answer
As Moo pointed out the second equation can be solved separately by solving
$$\dfrac{dx_2}{x_2}=-t^2dt\implies \ln x_2 = \ln c_1 -\dfrac{1}{3}t^3 \implies x_2(t)=c_1\exp\left[-t^3/3 \right].$$
Use this for the first equation to obtain:
$$\dfrac{dx_1}{dt}=-x_1+c_1\sin t\exp\left[-t^3/3\right].$$
As far as I can see there is no closed form solution to this ODE. But you can express it using an integral (obtainable by integrating factor or variation of parameters)
$$x_1(t)=c_1\exp (-t)\int_{1}^{t}\exp(\tau - \tau^3/3) \sin \tau d\tau + c_2 \exp(-t).$$
Combining both results we get
$$x_1(t)=c_1\exp (-t)\int_{1}^{t}\exp(\tau - \tau^3/3) \sin \tau d\tau + c_2 \exp(-t)$$ $$x_2(t)=c_1\exp\left[-t^3/3 \right]+c_2\cdot 0.$$
We conclude that the fundamental solution $\boldsymbol{X}(t)$ is given as
$$\boldsymbol{X}(t)=\begin{bmatrix} \exp (-t)\int_{1}^{t}\exp(\tau - \tau^3/3) \sin \tau d\tau & \exp(-t)\\ \exp\left[-t^3/3 \right] & 0\\ \end{bmatrix}.$$
In order to derive the state transition matrix $\boldsymbol{\Phi}(t,t')$ we need to calculate
$$ \boldsymbol{\Phi}(t,t') = \boldsymbol{X}(t)\boldsymbol{X}^{-1}(t').$$
Remark: In order to invert an invertible $2\times 2$ matrix $\boldsymbol{A}$ you can use the following formula
$$\boldsymbol{A}=\begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{bmatrix} \implies \boldsymbol{A}^{-1}=\dfrac{1}{\det \boldsymbol{A}}\begin{bmatrix} a_{22} & -a_{12}\\ -a_{21} & a_{11}\\ \end{bmatrix}. $$
If you would like to see another example you can also have a look at this answer.