Solution of Differential equation Question:
Find solution of differential equation
$$ 3e^{4x} \frac{dy}{dx} = -16\frac{x}{y^2} $$
which satisfies the initial condition y(0)=1
Solution:
I know that I have to bring it in the general form of :
$$ \frac{dy}{dx} + P(x) y = Q(x)$$
However in the equation there is no P(x)y component so how do i do it ? 
 A: You can solve this ODE by separation of variables:
$\frac{dy}{dx}=-16\frac{x}{y^{2}}\frac{1}{3}e^{-4x}$
$y^{2}dy=-\frac{16}{3}xe^{-4x}dx$
$\int y^{2}dy=-\frac{16}{3}\int xe^{-4x}dx$
Now you can integrate both sides using standard techniques (the integral on the right hand side using partial integration).We get
$\frac{y^{3}}{3}=\frac{4}{3}xe^{-4x}+\frac{1}{3}e^{-4x}+C$
$y(x)=\left(4xe^{-4x}+e^{-4x}+\tilde{C}\right)^{1/3}$, with $\tilde{C}:=3C$
Using your initial data we find that $\tilde{C}=0$.
A: $3e^{4x} \frac{dy}{dx} = -16\frac{x}{y^2}$ write it as 
$\int y^2 dy = -16/3\int \frac{x}{e^{4x} }dx$ now apply by parts in R.H.S
A: An interesting aspect:
The differential equation has the form
\begin{align}
3 \, e^{a x} \, y' = - 16 \cdot \frac{x}{y^{2}}
\end{align}
for which
\begin{align}
3 \, y^{2} \, y' = - 16 \, x \, e^{-a x}
\end{align}
where $a$ is a constant, here $a = 4$. This last equation can be seen to have the form
\begin{align}
\frac{d (y^{3})}{dx} &= \frac{d}{da} \left(16 \, e^{-a x} \right) = \frac{d}{dx} \left[ \frac{d}{da} \left( - \frac{16}{a} \, e^{-ax} \right) \right].
\end{align}
Integration with respect to $x$ provides:
\begin{align}
y^{3} = \frac{d}{da} \left( - \frac{16}{a} \, e^{-ax} \right) + c_{1}
\end{align}
or
\begin{align}
y(x) &= \left.\left[ \frac{d}{da} \left( - \frac{16}{a} \, e^{-ax} \right) + c_{1} \right]^{1/3}\right|_{a=4} \\
&= \left.\left[ \frac{16}{a} \, \left(1 + \frac{1}{a}\right) \, e^{-ax} + c_{1} \right]^{1/3} \right|_{a=4} \\
&= \left[ (4 x + 1) \, e^{-4x} + c_{1} \right]^{1/3}.
\end{align}
