I'm stuck with this problem:
Given $$\dot{y}=A(t)y, \qquad A(t)=\begin{pmatrix} 1+\frac{\cos(t)}{2+\sin(t)}&0\\ 1&-1 \end{pmatrix}.$$ Find $\phi^{t,0}$.
So I know I have to find the fundamental matrix because $\phi^{t,0}=\psi(t)[\psi(0)]^{-1}$, where $\psi(x)$ is the fundamental matrix. But I don't know how to find it. Can anyone help?
This system splits into two subsystems. The first equation, $$\dot y_1=\left(1+\frac{\cos t}{2+\sin t}\right)y_1$$ is separable: $$ \frac{dy_1}{y_1}= \left(1+\frac{\cos t}{2+\sin t}\right)dt $$ and has a solution $$ y_1= C_1 e^{t}(\sin t +2). $$ Substituting this to the second equation, one obtains $$ \dot y_2= C_1 e^{t}(\sin t +2) - y_2 $$ or $$ \dot y_2+y_2= C_1 e^{t}(\sin t +2). $$ This is a linear first order ode, which has a solution $$ y_2(t)=C_2 e^{-t}+C_1 e^{t}\left(1+\frac25 \sin t-\frac15\cos t\right) $$ Thus, the fundamental matrix is $$ \Psi(t)= \left(\begin{array}{cc} e^{t}(\sin t +2)& 0\\ e^{t}\left(1+\frac25 \sin t-\frac15\cos t\right) & e^{-t} \end{array}\right). $$