<s>[INST] Differential equation exercise. I am tasked with solving 
\begin{cases} 
y''(t) &=& \frac{(y(t)')^2}{y} - 2\frac{y'(t)}{y^4(t)} \\
y(0)   &=& -1 \\ 
y'(0)  &=& -2
\end{cases} 
I proceed by setting $v(s) = y' (y^{-1}(s))$ reducing the problem to
\begin{cases} 
v'(s) &=& \frac{v(s)}{s} - \frac{2}{s^4} \\ 
v(-1) &=& -2 
\end{cases} 
This is a first order linear ODE, after multiplying by $\frac{1}{|s|}$ (I drop the absolute value and change the sign since $s$ will be in a negative interval). I obtain $$-\frac{v'(s)}{s} + \frac{v}{s^{2}} = \frac{2}{s^5}$$ 
This gives me $$\frac{v(s)}{s} = \frac{1}{(-2s^4)} + \frac{5}{2} \implies y'(t) =  \frac{-1 + 5y(t)^4}{2y(t)^3}$$ 
And I can't quite manage to integrate the reciprocal of this to get my $y(t)$.
Are my calculations correct? How should I do this?  I am utilizing this method because it's the one I am expected to use at the exam (I have already been told that it can make things more difficult).
 [/INST] \begin{cases} 
y''(t) &=& \frac{(y(t)')^2}{y} - 2\frac{y'(t)}{y^4(t)} \\
y(0)   &=& -1 \\ 
y'(0)  &=& -2
\end{cases} 
This is an ODE of the autonomeous kind. The usual way to reduce the order is the change of function :
$$\frac{dy}{dt}=F(y) \quad\to\quad \frac{d^2y}{dt^2}=\frac{dF}{dy}\frac{dy}{dt}=F\frac{dF}{dy}$$
$$F\frac{dF}{dy}=\frac{F^2}{y} -2\frac{F}{y^4}$$
$$y\frac{dF}{dy}-F =-\frac{2}{y^3}$$
There is no difficulty to solve this linear ODE. You will obtain :
$$F(y)=c_1y+\frac{1}{2y^3}\quad\to\quad \frac{dy}{dt}=c_1y+\frac{1}{2y^3}$$
With conditions $y(0)=-1$ and $y'(0)=-2\,$ then $\,-2=c_1(-1)+\frac{1}{2(-1)^3} \quad\to\quad c_1=\frac{3}{2}$
$$F(y)=\frac{3}{2}y+\frac{1}{2y^3}\quad\to\quad \frac{dy}{dt}=\frac{3}{2}y+\frac{1}{2y^3} = \frac{3y^4+1}{2y^3}$$
$$t=\int \frac{2y^3dy}{3y^4+1}=\frac{1}{6}\ln(3y^4+1)+c_2$$
with conditions : $0=\frac{1}{6}\ln(3(-1)^4+1)+c_2 \quad\to\quad c_2=-\frac{1}{6}\ln(4)$
$$t(y)=\frac{1}{6}\ln(3y^4+1)-\frac{1}{6}\ln(4)$$
$$t(y) =\frac{1}{6}\ln(\frac{3y^4+1}{4})$$
With sign according to condition $y(0)=-1$ , the inversion of $t(y)$ leads to :
$$y(t)=-\left(\frac{4e^{6t}-1}{3} \right)^{1/4}$$
</s>