Question

Solve the given differential equation by using the substitution \(u=y'\)
\[y''+(y')^2+1=0, u\frac{du}{dy}=y''\]
\[udu+u^2dy+dy=0\]
\[udu=-u^2dy-dy=-dy(u^2+1)\]
\[\int \frac{u}{u^2+1}du=-\int dy\]
\[v=u^2+1,\frac{dv}{2}=udu\]
\[\int \frac{dv}{v}=-\int dy\]
\[\ln(u^2+1)=y+\ln(c_1)\]
\[u^2=c_1e^y-1\]
\[u=\pm\sqrt{c_1e^y-1}=\frac{dy}{dx}\]
That's as far as I can get.