MAT244-2013S > Term Test 1

TT1--Problem 4

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Victor Ivrii:
$\renewcommand{\Re}{\operatorname{Re}}$

Changyu -- wrong signs in the last 2 terms. Then your solution would collapse to $\cos(2x)+x\sin(2x)$ rather than to $\cos(2x)-x\sin(2x)$ as your solution. Definitely if you going through complex form you need to work only with half of roots so you look for
\begin{equation*}
\Re \bigl(C_1 e^{2ix} + C_2 x e^{2ix}\bigr).
\end{equation*}

Because of initial data, the real form is preferable but there is more. Note $y'(0)=y'''(0)=0$ and equation contains only even derivatives. Therefore if $y(x)$ is a solution, $y(-x)$ is also a solution and since solution is unique we conclude that $y(x)=y(-x)$ so it is an even function i.e.
\begin{equation*}
y(x) = C_1 \cos(2x) +C_3 x\sin(2x)
\end{equation*}
and now everything goes very fast: $y(0)=C_1$, $y''(0)=-4C_1+4 C_3$.

Devin, each approach has its own advantages and disadvantages and you need to be comfortable with both. Also note that $\cos(x)$ and $\sin(x)$ are not only real but also even and odd respectively which often makes life easier; $\cosh(x)$ and $\sinh(x)$ are used by the same reason and sometimes they give you an edge over pair $e^x$ and $e^{-x}$.