Solve the differential equation $y{''}(1+y^{'2})-3y^{'}y^{''2}=0$ Solve the differential equation $y{''}(1+y^{'2})-3y^{'}y^{''2}=0$
Attempt: 
$$y{''}+y^{''}y^{'2}-3y{'}y^{''2}=0\Rightarrow \frac{1+y{'^{2}}-3y^{'}y^{''}}{y{''}}=0\Rightarrow 1+y^{'2}-3y^{'}y^{''}=0$$
Substitution $y^{'}=e^t,y^{''}=e^{t}t^{'}\Rightarrow 1+e^{2t}-3e^{2t}t^{'}=0$
Is this the right approach, and if yes, how to continue?
 A: HINT:
$$y''(x)\left(1+y'(x)^2\right)-3y'(x)y''(x)^2=0\Longleftrightarrow$$

Let $y'(x)=v(x)$ which gives $y''(x)=v'(x)$:

$$v'(x)\left(1+v(x)^2\right)-3v(x)v'(x)^2=0\Longleftrightarrow$$
$$v'(x)=0\Longleftrightarrow\space\space\vee\space\space v'(x)=\frac{v(x)^2+1}{3v(x)}\Longleftrightarrow$$
$$v(x)=\text{C}_1\Longleftrightarrow\space\space\vee\space\space \frac{3v'(x)v(x)}{v(x)^2+1}=1\Longleftrightarrow$$
$$v(x)=\text{C}_1\Longleftrightarrow\space\space\vee\space\space \int\frac{3v'(x)v(x)}{v(x)^2+1}\space\text{d}x=\int1\space\text{d}x\Longleftrightarrow$$
$$v(x)=\text{C}_1\Longleftrightarrow\space\space\vee\space\space \frac{3\ln(v(x)^2+1)}{2}=x+\text{C}_1\Longleftrightarrow$$
$$v(x)=\text{C}_1\Longleftrightarrow\space\space\vee\space\space v(x)=\pm\sqrt{e^{\frac{2(x+\text{C}_1)}{3}}-1}\Longleftrightarrow$$
$$y'(x)=\text{C}_1\Longleftrightarrow\space\space\vee\space\space y'(x)=\pm\sqrt{e^{\frac{2(x+\text{C}_1)}{3}}-1}\Longleftrightarrow$$
$$y(x)=\int\text{C}_1\space\text{d}x\space\space\vee\space\space y(x)=\int\pm\sqrt{e^{\frac{2(x+\text{C}_1)}{3}}-1}\space\text{d}x$$
A: At the beginning $y^{''} $ can be cancelled. $ y = C_1 x + C_2 $ is a solution. Next,
$$ \dfrac{1}{3 y^{'}} =  \dfrac{y^{''}}{1+y'^{2} }$$
Using substitution
$$ y^{'} = \tan \phi $$
greatly simplifies it to:
$$ \int dx = 3 \int\tan \phi \cdot d\phi = 3 \log \sec \phi $$
and you can continue with it. 
A: When you divided by $y''$, you discarded the solutions with $y''=0$. 
I don't immediately see how to go through with your substitution. 
What I would do is to let $v=y'^2$. Then $v'=2y'y''$, and the equation becomes
$$
1+v-\frac32v'=0.
$$
This is linear and separable, so it can be done in several ways. We get
$$
v(x)=-1+e^{2x/3},
$$
or
$$
y'(x)=\sqrt{-1+e^{2x/3}}.
$$
It follows that 
$$
y(x)=3\sqrt{-1+e^{2x/3}}-3\arctan\sqrt{-1+e^{2x/3}}+c, \ \ \ x\geq0.
$$
At the start we also discarded the solutions $y''=0$, so
$$
y(x)=cx+d,
$$
with no domain restriction. 
