Show that the vector x is a solution of the system and satisfying the initial condition.

Question

Show that vector $x = [\sin(t) - t\cos(t), t\sin(t)]$ is a solution of the system $$x'= [0 1, -1 0]x + [0, 2\sin(t)]$$ satisfying the initial condition $x(0) = [0, 0].$ My question is what do I do after I have taken the derivative of vector x. That is $x'_1= t\sin(t)$ and $x'_2 = \sin(t) + t\cos(t).$

Answer 1

Verify that $x$ is indeed a solution.

\begin{align} \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix} \sin(t)-t\cos(t) \\ t\sin(t) \end{bmatrix}+ \begin{bmatrix} 0 \\ 2\sin(t) \end{bmatrix}&=\begin{bmatrix} t\sin(t) \\ -\sin(t)+t\cos(t)\end{bmatrix}+\begin{bmatrix} 0 \\ 2\sin(t) \end{bmatrix}\\&=\begin{bmatrix} t\sin(t) \\ \sin(t)+t\cos(t) \end{bmatrix}\end{align}

which is equal to your derivative.

Also, you have to check that the initial condition holds.

Answer 2

I assume that "[01,-10]" is the matrix $\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$. Do you understand what "[01,-10]x" means? It is the matrix product, $\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix}x_2 \\ -x_1\end{bmatrix}$

You have calculated the derivative of x, so the left side of the given equation. Now find the value of the right side and see if they are the same.