Question

Find the general solution of the following differential equation

Answer 1

\[y \prime - \frac{ y }{ 3x } = y^{4} * \ln x\]

Answer 2

what techniques are you learning ?

Answer 3

\[y'-\frac{y}{3x}=y^4\ln x\\
y^{-4}y'-y^{-3}\frac{1}{x}=\ln x\]
Bernoulli equation. Do you know how to proceed from here?

Answer 4

Yes, I thought it could be Bernoulli, but I was not sure

Answer 5

But seems difficult. How should I proceed?

Answer 6

Substitute \(u=y^{-3}~\Rightarrow~u'=-3y^{-4}~y'\).

So your equation becomes
\[-\frac{1}{3}u'-u\frac{1}{3x}=\ln x\]
Now you have a pretty straightforward linear equation.

Answer 7

Thanks

Answer 8

You're welcome!