alparcero97oy

2022-03-24

Help to find the solution of
${y}^{\prime }=\left(\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right)\cdot y+\left(\begin{array}{c}2\mathrm{sin}\left(t\right)\\ 2\left(\mathrm{cos}\left(t\right)-\mathrm{sin}\left(t\right)\right)\end{array}\right)$

Laylah Hebert

Step 1
The homogeneous part of y, given by ${y}_{h}\left(t\right)$, solves the following system:
${y}_{h}^{\prime }=\left(\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right){y}_{h}.$
The above system can be solved directly by looking at the eigenvalue of the matrix
$\left(\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right)$
One can check that the eigenvalues are given by -1 and -3. This yields
${y}_{h}\left(t\right)={e}^{-t}{v}_{1}+{e}^{-3t}{v}_{2}$
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.
Step 2
One can check that a particular solution ${y}_{p}\left(t\right)$ is given by
${y}_{p}\left(t\right)=\left(\begin{array}{c}\mathrm{sin}\left(t\right)\\ \mathrm{cos}\left(t\right)\end{array}\right).$
(Remark: One way you could have known this is to guess that
${y}_{p}\left(t\right)=\left(\begin{array}{c}A\mathrm{sin}\left(t\right)+B\mathrm{cos}\left(t\right)\\ C\mathrm{sin}\left(t\right)+D\mathrm{cos}\left(t\right)\end{array}\right)$
and solve for A,B,C, and D.)
This gives
$y\left(t\right)={y}_{h}\left(t\right)+{y}_{p}\left(t\right)$
${e}^{-t}{v}_{1}+{e}^{-3t}{v}_{2}+\left(\begin{array}{c}\mathrm{sin}\left(t\right)\\ \mathrm{cos}\left(t\right)\end{array}\right)$

SofZookywookeoybd

Step 1
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
${u}_{1}^{\prime }=-{u}_{1}+2\mathrm{cos}\left(t\right)$
${u}_{2}^{\prime }=3{u}_{2}+4\mathrm{sin}\left(t\right)-2\mathrm{cos}\left(t\right)$.
These you should be able to solve with your previous experiences. Reconstructing from their solutions is then a trivial step.

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