alparcero97oy

2022-03-24

Help to find the solution of

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

Laylah Hebert

Beginner2022-03-25Added 15 answers

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

${y}_{p}(t)=\left(\begin{array}{c}\mathrm{sin}(t)\\ \mathrm{cos}(t)\end{array}\right).$

(Remark: One way you could have known this is to guess that

${y}_{p}(t)=\left(\begin{array}{c}A\mathrm{sin}(t)+B\mathrm{cos}(t)\\ C\mathrm{sin}(t)+D\mathrm{cos}(t)\end{array}\right)$

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

Beginner2022-03-26Added 15 answers

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$y}_{1},\text{}{y}_{2$ from their solutions is then a trivial step.

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

Here consider

These you should be able to solve with your previous experiences. Reconstructing

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