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Let the following system $X'=AX+b(t)$ be a system of differential equations, where:

$A=\begin{pmatrix}1&-4\\1&1\end{pmatrix}$ and $b=\begin{pmatrix}\cos2t\\0\end{pmatrix}$

Then When trying to solve, the roots of the characteristic polynomial of the matrix are complex, therefore I'm going to use:

$$D=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$$

since $\lambda_{1,2}=1\pm2i$ then $D=\begin{pmatrix}1&2\\-2&1\end{pmatrix}$ and $P=\begin{pmatrix}2&0\\0&-1\end{pmatrix}$, then $A=PDP^{-1}$

$X'=AX=PDP^{-1}X\implies P^{-1}X'=DP^{-1}X$ if we note $P^{-1}X=\begin{pmatrix}u\\v\end{pmatrix}$ then a matrix of fundamental solutions for the new system is: $M(t)=e^t\begin{pmatrix}cos(2t)&sin(2t)\\-sin(2t)&cos(2t)\end{pmatrix}$

Then $X=Pe^t\begin{pmatrix}cos(2t)&sin(2t)\\sin(2t)&cos(2t)\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}$

All in all we get: $$X=e^t\begin{pmatrix}2c_1cos(2t)+2c_2sin(2t)\\-c_1sin(2t)-c_2sin(2t)\end{pmatrix}$$

But that is not correct, what am I missing?

C. Cristi
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  • It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = \left( \begin{array}{cc} -2i & 2i\ 1 & 1 \end{array} \right) \left( \begin{array}{cc} 1-2i & 0\ 0 & 1+2i \end{array} \right) \left( \begin{array}{cc} -2i & 2i\ 1 & 1 \end{array}\right)^{-1}$ with a diagonal matrix in the middle. – Christoph Jan 16 '19 at 07:19
  • You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $\exp(tA)=P\exp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation. – amd Jan 17 '19 at 00:39
  • @Christoph A $2\times2$ real matrix with complex eigenvalues is similar to a matrix of the form $\small{\begin{bmatrix}a&-b\b&a\end{bmatrix}}$. The OP’s decomposition is perfectly good. – amd Jan 17 '19 at 00:40
  • @amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 \times 2$. – Christoph Jan 17 '19 at 05:16
  • @amd right, i'm sorry, And how do I search for the particular solution? – C. Cristi Jan 17 '19 at 07:01
  • @Moo not really – C. Cristi Jan 17 '19 at 14:19

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