Answer
$y=\sec x$
Work Step by Step
We are given:
$yy''=2(y')^2+y^2$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}=u\frac{du}{dy}$
Subsitute to the equation:
$y\frac{du}{dx}-2u^2-y^2=0$
$u\frac{du}{dy}-\frac{2}{y}u^2=y$
Let $t=u^2 \rightarrow \frac{dt}{dx}=2u\frac{du}{dx}$
The equation becomes:
$\frac{1}{2}\frac{dt}{dx}-\frac{2}{y}t=y$
$\frac{dt}{dx} -\frac{4}{y}t=-2y$
Integrating factor:
$I=e^{-\int \frac{4}{y}dy}=e^{-4\ln y}=y^{-4}$
Hence,
$\frac{d}{dy}(y^{-4}t)=2y^{-3}$
Integrating both sides:
$y^{-4}t=-y^{-2}+C$
$t=-y^{2}+Cy^4$
$u^2=Cy^4-y^2$
$u=\pm \sqrt Cy^4-y^2$
$\frac{dy}{dx}=\pm \sqrt Cy^4-y^2$
$\pm dx=\frac{dy}{y\sqrt Cy^2-1}$
$\rightarrow \pm dx=\frac{dt}{t\sqrt t^2-1}$
Integrating both sides:
$\sec^{-1}t=\pm x+ C$
$t=\pm \sec x+C_1$
since $\sec (-x)=\sec x$
$\rightarrow \sqrt Cy=\pm \sec (x+C_1)$
We are given:
$y(0)=1 \rightarrow \sqrt C=\pm sec C_1$
and $y'(0)=0 \rightarrow \sqrt Cy'=\pm \sec x+C_1\tan x+C_1$
$\rightarrow \pm \sec C_1 \tan C_1=0$
We get:
$C_1=0 \wedge C=1$
Since $\sec x $ can not be zero:
$\tan C_1 =0 \rightarrow C_1=0$
The final solution is:
$y=\sec x$
