Help to find the solution ofy′=(−211−2)⋅y+(2sin⁡(t)2(cos⁡(t)−sin⁡(t)))

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Laylah Hebert · Accepted Answer

Step 1The homogeneous part of y, given by&amp;nbsp;yh(t), solves the following system:yh′=(−211−2)yh.The above system can be solved directly by looking at the eigenvalue of the matrix(−211−2)One can check that the eigenvalues are given by -1 and -3. This yieldsyh(t)=e−tv1+e−3tv2where&amp;nbsp;v1&amp;nbsp;and&amp;nbsp;v2&amp;nbsp;are arbitrary vectors in&amp;nbsp;R2&amp;nbsp;(am assuming that is what you are working with) to be determined by providing the relevant initial data.Step 2One can check that a particular solution&amp;nbsp;yp(t)&amp;nbsp;is given byyp(t)=(sin⁡(t)cos⁡(t)).(Remark: One way you could have known this is to guess thatyp(t)=(Asin⁡(t)+Bcos⁡(t)Csin⁡(t)+Dcos⁡(t))and solve for A,B,C, and D.)This givesy(t)=yh(t)+yp(t)e-tv1+e-3tv2+sin(t)cos(t)

SofZookywookeoybd · Accepted Answer

Step 1This kind of matrix structure has some very simple eigenvectors. In general for circulant matrices one can use a Fourier basis.Here consider u1=y1+y2 and u2=y1−y2 to get the decoupled system of scalar equationsu1′=−u1+2cos⁡(t)u2′=3u2+4sin⁡(t)−2cos⁡(t).These you should be able to solve with your previous experiences. Reconstructing y1, y2 from their solutions is then a trivial step.

Help to find the solution of \(y' = \begin{pmatrix}-2

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