How can I estimate this second order differential equation?. $ \ ((y'(x))^2+1)^{3/2}\cos(x) = y''(x) \ \ ,y(0)=0, y'(0)=1 \ $

Question

$$ \ ((y'(x))^2+1)^{3/2}\cos(x) = y''(x) \ \ ,y(0)=0, y'(0)=1 \ $$ where do I start?. I know using the Maclaurin series. what should I use an infinite product or infinite sum. it doesn't work well.

Answer 1

$$((y'(x))^2+1)^{3/2}\cos(x) = y''(x) $$ $$\implies \frac{d}{dx} y'(x)=((y'(x))^2+1)^{3/2}\cos(x)$$ $$\implies \frac{d~(y'(x))}{(y'(x))^2+1)^{3/2}} =\cos(x)~ dx$$ Putting $~y'=\tan u~$ and then integrating we have $$\frac{y'}{\sqrt{y'~^2+1}}= \sin x ~+c\qquad \text{where $~c~$ is a constant}$$ $$\implies y'~^2=(\sin x +c)^2~(y'~^2+1)$$ $$\implies y'~=\frac{\sin x +c}{\sqrt{1-(\sin x +c)^2}}$$ $$\implies y=k+\int \frac{\sin x +c}{\sqrt{1-(\sin x +c)^2}}~dx \qquad \text{where $~c~$and$~k~$ are constants}$$