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Solve the initial value problem:
$$\dot{\mathbf{x}}=\begin{bmatrix} 3&-1\\ -1&3 \end{bmatrix}\mathbf{x}+\begin{bmatrix} 4e^{2t}\\ 4e^{4t} \end{bmatrix},$$ with the initial condition $$\mathbf{x}(0)=\begin{bmatrix} 1\\ 1 \end{bmatrix}.$$

I come up with finding the eigenvalues and eigenvectors, and my result is $$\bf{D}=\begin{bmatrix} 2&0\\ 0&4 \end{bmatrix},\qquad \bf{P}=\begin{bmatrix} 1&-1\\ 1&-1 \end{bmatrix}.$$ But then I found that $\mathbf{P}$ is not invertible, which means that I can't figure out $e^{\mathbf{A}t}$ by using the diagonalizability of $\mathbf{A}$, where $\mathbf{A}$ is the coefficient matrix.
I have no idea what I should do next. Anyone can help?

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  • $\begingroup$ You must have made some mistake of algebra, because eigenvectors of different eigenvalues are always linearly independent. $\endgroup$ – user228113 Dec 2 '17 at 10:00
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Hint. I think you made some mistake somewhere in your calculation. It should be $$D=\begin{bmatrix} 2&0\\ 0&4 \end{bmatrix},\qquad P=\begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}.$$ So that $$A:=\begin{bmatrix} 3&-1\\ -1&3 \end{bmatrix}=P^{-1}DP.$$ and $$\exp(At)=P^{-1}\begin{bmatrix} e^{2t}&0\\ 0&e^{4t} \end{bmatrix}P.$$ Hence $$\begin{align} \mathbf{x}(t)&=\exp(At)\begin{bmatrix} 1\\ 1 \end{bmatrix}+\exp(At)\int_0^t\exp(-As)\begin{bmatrix} 4e^{2s}\\ 4e^{4s} \end{bmatrix}ds\\ &=P^{-1}\left(\begin{bmatrix} 2e^{2t}\\ 0 \end{bmatrix}+\begin{bmatrix} e^{2t}&0\\ 0&e^{4t} \end{bmatrix}\int_0^t\begin{bmatrix} e^{-2s}&0\\ 0&e^{-4s} \end{bmatrix}P\begin{bmatrix} 4e^{2s}\\ 4e^{4s} \end{bmatrix}ds\right). \end{align}$$ Can you take it from here?

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  • $\begingroup$ Oh, right! And when I continue to solve this system, I find it really complicated in calculation, such as $e^{\bf{A}t}\int_0^{t}e^{\bf{A}-x}\begin{bmatrix}4e^{2x}\\4e^{4x}\end{bmatrix}dx$. Is there any shortcut to do this? $\endgroup$ – R.C Dec 2 '17 at 10:06
  • $\begingroup$ Yes. Just this formula. $\endgroup$ – R.C Dec 2 '17 at 10:31
  • $\begingroup$ See my answer. I added a few details. The integration is not difficult. Try it! $\endgroup$ – Robert Z Dec 2 '17 at 10:36
  • $\begingroup$ Thanks so much. I've been evaluating such kind of integrals the whole afternoon. Drizzled all the way :) $\endgroup$ – R.C Dec 2 '17 at 10:39
  • $\begingroup$ Sorry to reply to you so late, 'cause I'm busy the whole week. My answer is $$e^{2t}\begin{bmatrix}2t-1\\ 2t+1\end{bmatrix}+e^{4t}\begin{bmatrix}2-2t\\ 2t\end{bmatrix}$$ Btw, I didn't check my answer with a computer. :) $\endgroup$ – R.C Dec 7 '17 at 7:23

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