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\textbf{Ordinary Differential Equations}

\textit{\textbf{Classification}}
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\textbf{Question}

Find a solution of $y''+y=0$ given that $y(\pi/2)=2y(0)$ and
$y(\pi/4)=3.$


\textbf{Answer}

$y= A\cos x + B\sin x$, for any $A$ or $B$, gives a solution to
$y''+y=0$.

To satisfy the given conditions:
\begin{eqnarray*}
0 &amp; = &amp; y(\pi/2)-2y(0)+B-2A\\
3 &amp; = &amp; y(\pi/r) = \frac{A}{\sqrt{2}}+\frac{B}{\sqrt{2}}
\end{eqnarray*}
provided that
\begin{eqnarray*}
A &amp; = &amp; \sqrt{2}\\
B &amp; = &amp; 2\sqrt{2}
\end{eqnarray*}
So the solution is
$$y = \sqrt{2} \cos x + 2 \sqrt{2} \sin x$$

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