Originally Posted by

**HallsofIvy** That's very clever, guys, but it doesn't actually answer the question asked, does it?

Royal Flush, yes, if you let $\displaystyle y_1= y$ and $\displaystyle y_2= \frac{dy}{dt}$, then we have immediately that $\displaystyle \frac{dy_1}{dt}= y_2$ and then $\displaystyle \frac{d^2y}{dt^2}= \frac{d}{dt}\frac{dy_1}{dt}= \frac{dy_2}{dt}$ so that $\displaystyle \frac{d^2y}{dt^2}+ y= \frac{dy_2}{dt}+ y_1= 0$ so that $\displaystyle \frac{dy_2}{dt}= -y_1$

And your initial conditions are $\displaystyle y_1(0)= 0$ and $\displaystyle y_2(0)=1$.

This can also be written as the "matrix" equation:

$\displaystyle \frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dt}= \begin{pmatrix}y_2 \\ -y_1\end{pmatrix}= \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}$ subject to the initial condition $\displaystyle \begin{pmatrix}y_1(0) \\ y_2(0)\end{pmatrix}= \begin{pmatrix}0 \\ 1\end{pmatrix}$.

In the same way, the equation y"+ y= sin(2t) becomes the pair of equations $\displaystyle \frac{dy_1}{dt}= y_2$ and $\displaystyle \frac{dy_2}{dt}= -y_1+ sin(2t)$ with the same intial conditions.

That also can be written as a "matrix" equation:

$\displaystyle \frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dt}=$$\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}+ \begin{pmatrix}0 \\ sin(2t)\end{pmatrix}$

with initial condition $\displaystyle \begin{pmatrix}y_1(0) \\ y_2(0)\end{pmatrix}= \begin{pmatrix}0 \\ 1\end{pmatrix}$.

Of course, the eigenvalues of the matrix $\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ are i and -i which lead to the complex exponentials (sine and cosine) solutions Prove It and AllenCuz show.