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