Solving the differential equation for general solution.

Question

Find general solution of x(1-xlny)y' +y=0 my attempt: I tried substituting 1-xlny=t and rephrased the given equation in terms of t and x but couldn't solve thereafter. Help appreciated and other approaches are welcome.

Answer 1

Write the given differential equation as $$xdy+ydx-x^2\ln ydy=0 \implies d(xy)-x^2\ln ydy=0$$ Dividing by $x^2y^2$, $$\frac{d(xy)}{(xy)^2}=\frac{\ln ydy}{y^2}$$ Integrating both sides, $$-\frac{1}{xy}+C=\int\frac{\ln ydy}{y^2}$$ Applying Integration by Parts to the second integral gives, $$-\frac{1}{xy}+C=-\frac{\ln y}{y}+\int{\frac{dy}{y^2}}=-\frac{\ln y}{y}-\frac{1}{y}$$ $$\therefore \frac{1}{x}=1+\ln y+Cy$$

Answer 2

$$x(1-x\ln y)y' +y=0 $$ $$x(1-x\ln y)=-yx' $$ It's Bernoulli's equation: $$x'y+x=x^2 \ln y$$ And it's also separable: $$\dfrac {d(xy)}{dy}=x^2 { \ln y}$$ $$\dfrac {d(xy)}{(xy)^2}=\dfrac { \ln y}{y^2}dy$$ Integrate.