Question

Solve
x dy/dx + y = x^2y^2 ln x
This equation is reducible to Bernoulli's equation but they want it solved by a suitable substitution.

Answer 1

Please try this substitution:
\[y\left( x \right) = \frac{1}{{z\left( x \right)}}\]

Answer 2

you should get this:
\[z' - \frac{1}{x}z = - x\ln x\]

Answer 3

excuse me - i must leave right now - I' will come back later to look at this

Answer 4

ok!

Answer 5

\[xy'+y=x^2y^2\ln{x}\]\[(xy)'=(xy)^2\ln{x}\quad,\quad z=xy\]\[\frac{z'}{z^2}=\ln{x}\]\[\left(-\frac{1}{z}\right)'=\ln{x}\]\[\frac{1}{z}=-\int\ln{x}dx\quad,\quad z=-\left(\int\ln{x}dx\right)^{-1}\]\[y=\frac{z}{x}=-\left(x\int\ln{x}dx\right)^{-1}=-\frac{1}{x(x(\ln{x}-1)+C)}\]

Answer 6

ah that makes sense thank you @nikvist