Differential Equation - I'm missing something stupid 
I have the follwoing ODE to solve:
$\dfrac{dy}{dx}+3y=28e^{2x}y^{-3}$


This is a Bernoulli equation so I make the substitution $ u=y^{1-n} \; \;u=y^4 \; \;y=u^{1/4}$.
$$
\begin{split}
\frac{du^{{1/4}}}{dx} + 3u^{{1/4}} &= 28e^{2x}u^{-3/4} \\
u^{{-3/4}}\frac{du}{dx}+3u^{1/4} &= 28e^{2x}u^{-3/4} \\
\frac{du}{dx} + 3u &=28e^{2x}
\end{split}
$$
So we have a linear differential equation $\dfrac{dy}{dx}+p(x)y=f(x)$ with $p(x)=3$.
Multiplying through by ${e}^{\int{{3}dx}}\;\;$
$$
\begin{split}
e^{3x}\frac{du}{dx}+3ue^{3x} &= 28e^{5x}
 \quad \text{(expanded form from the product rule)} \\
\frac{d\left[u e^{3x}\right]}{dx} &= 28e^{5x} \\
{u}{e^{3x}} &= \int{28e^{5x}dx} \\
u &=\frac{28}{5}e^{2x} + \frac{c}{e^{3x}} \\
y &= \left(\frac{28}{5}e^{2x}+\frac{c}{e^{3x}}\right)^{1/4}
\end{split}
$$

However, the answer I am given is
$$
y = \left(8e^{2x}+\frac{c}{e^{12x}}\right)^{1/4}
$$
I've tried this about 10 times now and can't see where I am making a mistake.
Note:  I'm an amateur doing self study so there's nobody I can ask.  I'm sure there is an obvious error but I can't find it.
 A: You made the first step too complicated. Just multiply with $y^3$ to get
$$
y^3y'+3y^4=28e^{2x}
$$
Now the first term can be condensed to $\frac14(y^4)'$, or $u'=4x^3y'$. You missed the constant for the exponent in the derivative.
A: You can use the Bernoulli substitution to write $ \ u = y^4 \ \Rightarrow \ du = 4y^3 \ dy \ \ . $  You would then multiply the equation through by $ \ \frac{du}{dy} \ $ on the first term and its equivalent $ \ 4y^3 \ $ the rest of the way to obtain
$$ \frac{dy}{dx} · \frac{du}{dy} \  + \ 3y · 4y^3  \ \ = \ \ 28e^{2x} · y^{-3} · 4y^3 \ \ \Rightarrow \ \ \frac{du}{dx} \  + \ 12y^4  \ \ = \ \ 112e^{2x}  $$ $$   \Rightarrow \ \ \frac{du}{dx} \  + \ 12u  \ \ = \ \ 112e^{2x} \ \ . $$
The integrating factor is then $ \ e^{12x} \ , $ giving
$$ \frac{d\left[u · e^{12x}\right]}{dx} \ \ = \ \ 112e^{14x} \ \ \Rightarrow \ \ u · e^{12x} \ \ = \ \ 8·e^{14x} \ + \ C \ \ , $$
which will take you to the solution given.  It is a bit easier to see what needs to be done if you don't try to insert the substitution written as $ \ y \ $ in terms of $ \ u \ $ right away.  (It also helps you in seeing what the Bernoulli substitution gains us on the left side of the differential equation.)
