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.
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.)