Answer
$y=a \cos wx$
Work Step by Step
We are given:
$y''=w^2y$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}=u\frac{du}{dy}$
Subsitute to the equation:
$\frac{du}{dx}=w^2y$
$u\frac{du}{dy}=w^2y$
$udu=w^2ydy$
Integrating both sides:
$u^2=w^2y^2+C$
$u=\pm \sqrt w^2y^2+C$
$\frac{dy}{dx}=\pm \sqrt w^2y^2+C$
We are given:
$y(0)=a$ and $y'(0)=0$
$ \rightarrow \pm \sqrt w^2a^2+C=0$
$\rightarrow C=w^2a^2$
The equation becomes:
$\frac{dy}{dx}=\pm w \sqrt y^2-a^2$
$\pm w dx=\frac{dy}{\sqrt y^2-a^2}$
To integrate right side, we let:
$t=\sec ^{-1}(\frac{1}{a})y$
The equation becomes:
$\frac{dy}{\sqrt y^2-a^2}=\int sec t dt$
$\frac{dy}{\sqrt y^2-a^2}=\ln (tan t + \sec t)$
$\frac{dy}{\sqrt y^2-a^2}=\ln(\tan \sec^{-1}\frac{1}{a}y+\sec \sec ^{-1}\frac{1}{a}y)$
$\frac{dy}{\sqrt y^2-a^2}=\cos ^{-1}(\frac{y}{a})$
Hence,
$\cos ^{-1}(\frac{y}{a})=\pm wx +C_1$
$\frac{y}{a}=\cos (wx \pm C_1)$
$y=a \cos (wx \pm C_1) $
$\rightarrow y'=aw \sin (wx \pm C_1)$
We are given:
$y'(0)=0 \rightarrow \pm aw \sin (\pm C_1)$
Hence,
$C_1=0$
The final solution is:
$y=a \cos wx$
