<s>[INST] I need some help with this bernoulli's equation that is given in my assignment Solve:
$$y′−\frac{6y}{x}=\frac{y^5}{x^{13}}$$
I have tried again and again but my answer is still wrong. Can someone show me the steps to solving this question?
 [/INST] For Bernoulli equation of the form
$$y^\prime+p(x)y=q(x)y^\beta$$
we use substituation $u=y^{1-\beta}$. So with $\beta=5$ we have $u=y^{1-5}=\dfrac{1}{y^4}$ then $u'=-\dfrac{4y'}{y^5}$ then
\begin{eqnarray*}
-\frac{y^5u^\prime}{4}-6\frac{y}{x}  &=& \frac{1}{x^{13}}y^5\\
-\frac{u^\prime}{4}-6\frac{1}{x}\frac{1}{y^4}  &=& \frac{1}{x^{13}}\\
u^\prime+\dfrac{24}{x}u                &=&  -4\frac{1}{x^{13}}\\
x^{24}\left(u^\prime+\dfrac{24}{x}u \right) &=& -4 x^{24}\frac{1}{x^{13}}\\
x^{24}u'+24ux^{23}&=& -4 x^{11}\\
(x^{24}u)'&=& -4 x^{11}\\
x^{24}u&=& -\dfrac13 x^{12}+C\\
\frac{1}{y^4}   &=& -\frac{1}{3x^{12}}+\frac{C}{x^{24}}                               
\end{eqnarray*}
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