Question

Solve the system\[x'=\left[\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right]x+\left[\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right]\] Anyone walk me through, please

Answer 1

any way of U.C, Diagonalize, Variation parameter or Laplace, please

Answer 2

@oldrin.bataku @SithsAndGiggles @ybarrap @amistre64

Answer 3

I would use UC, but only because I've forgotten how to apply the other methods.

Homogeneous solution:
\[x'=\left[\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right]x\]
\[\begin{vmatrix}1-\lambda&1&0\\1&1-\lambda&0\\0&0&3-\lambda\end{vmatrix}=0~\Rightarrow~\lambda_1=0,\lambda_2=2,\lambda_3=3\]
Which gives you the following corresponding eigenvectors:
\[\eta_1=\begin{pmatrix}1\\-1\\0\end{pmatrix},\eta_2=\begin{pmatrix}1\\1\\0\end{pmatrix},\eta_3=\begin{pmatrix}0\\0\\1\end{pmatrix}\]
So you have the homogeneous solution of
\[y_c=C_1\eta_1e^{\lambda_1 t}+C_2\eta_2e^{\lambda_2 t}+C_3\eta_3e^{\lambda_3 t}\\
y_c=C_1\eta_1+C_2\eta_2e^{2 t}+C_3\eta_3e^{3 t}\]

Answer 4

Sorry, meant \(x_c\) instead of \(y_c\), but you get the idea. As a guess for the nonhomogeneous solution, I'd try
\[x_p=e^t\vec{a}+te^{2t}\vec{b}+te^{3t}\vec{c}\]
I'm not entirely sure if that would work, but I haven't tried it yet.

Answer 5

I use that guess because you can write the last vector as three vectors:
\[e^t\begin{pmatrix}1\\0\\0\end{pmatrix}+e^{2t}\begin{pmatrix}0\\1\\0\end{pmatrix}+e^{3t}\begin{pmatrix}0\\0\\1\end{pmatrix}\]