Express $\frac{d^3y}{dx^3}-\frac{dy}{dx}+y=\cos(x)$ in matrix form $x'=Ax+f$

Question

As the title says, I need to write $\frac{d^3y}{dx^3}-\frac{dy}{dx}+y=\cos(x)$ in the following matrix form $x'=Ax+f$

I've rearranged the equation to be in that form, but how do I extract my coefficient matrix, $A$, from this and how should I handle the $\frac{d^3y}{dx^3}$?

So far I just have:

$\frac{dy}{dx}=y+\frac{d^3y}{dx^3}-\cos(x)$

Answer 1

Hint: Let:

$$x_1 = y, x_2 = y', x_3 = y''$$

Now we have:

• $x_1' = y' = x_2$
• $x_2' = y'' = x_3$
• $x_3' = y''' = x_2 - x_1 + \cos x$

Can you finish it off?

$$x' = \begin{bmatrix}x_1' \\x_2'\\ x_3'\end{bmatrix} = Ax + f = \begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\-1 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 \\x_2\\ x_3\end{bmatrix} + \begin{bmatrix}0 \\0\\ \cos x\end{bmatrix}$$