Help to find the solution of $y' = \begin{pmatrix}-2 & 1 \\1 &-2\end{pmatrix} \cdot y + \begin{pmatrix}2\sin(t) \\ 2(\cos(t) - \sin(t))\end{pmatrix}$

Solution 1:

Homogeneous Portion

The homogeneous part of $y$, given by $y_h(t)$, solves the following system:

$$y_h' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}y_h.$$

The above system can be solved directly by looking at the eigenvalue of the matrix $\begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}$. One can check that the eigenvalues are given by $-1$ and $-3$. This yields

$$y_h(t) = e^{-t}v_1 + e^{-3t}v_2$$

where $v_1$ and $v_2$ are arbitrary vectors in $\mathbb{R}^2$ (am assuming that is what you are working with) to be determined by providing the relevant initial data.

Inhomogeneous Portion

One can check that a particular solution $y_p(t)$ is given by

$$y_p(t) = \begin{pmatrix} \sin(t) \\ \cos(t) \end{pmatrix}.$$

(Remark: One way you could have known this is to guess that $y_p(t) = \begin{pmatrix} A \sin(t) + B \cos(t) \\ C \sin(t) + D\cos (t)\end{pmatrix}$ and solve for $A,B,C,$ and $D$.)

This gives

$$\begin{aligned} y(t) &= y_h(t) + y_p(t) \\ &= \boxed{e^{-t}v_1 + e^{-3t}v_2 + \begin{pmatrix} \sin(t) \\ \cos(t) \end{pmatrix}}. \end{aligned}$$

Solution 2:

This kind of matrix structure has some very simple eigenvectors. In general for circulant matrices one can use a Fourier basis.

Here consider $u_1=y_1+y_2$ and $u_2=y_1-y_2$ to get the decoupled system of scalar equations \begin{align} u_1'&=-u_1+2\cos(t)\\ u_2'&=3u_2+4\sin(t)-2\cos(t). \end{align} These you should be able to solve with your previous experiences. Reconstructing $y_1,y_2$ from their solutions is then a trivial step.