# Non linear first order ODE, not exact and not separable

I am trying to solve the following differential equation $$tx'(t)=x(t)\left(\ln(x(t))-\ln(t)\right)$$ and I have hit a brick wall. I tried separating but no luck. The equation is not exact $$\int t dx=tx+C$$ and $$\int -(\ln x-\ln t) dt = -t\ln x-t+t\ln t+C$$ The only way I know to solve non linear first order ODEs is Bernoulli but this clearly isn't. What other methods are there for solving non linear first order Equations? P.S. this is not homework, I am self learning for exams. Second p.s. the solution is $$x(t)=te^{ct+1}$$

You may assume that: $$\ln(x(t))=u(t)$$

you obtain:

$$tu'-u+\ln(t)=0$$

An even easier approach than is to substitute $u = \ln x - \ln t$, then the equation becomes

$$tu' + 1 = u$$

which is separable

set $$x(t)=tv(t)$$ then we have $$t\left(t\frac{dv(t)}{dt}+v(t)\right)=t\log(v(t))\cdot v(t)$$ then we obtain $$\frac{dv(t)}{dt}=\frac{-v(t)+\log(v(t)\cdot v(t)}{t}$$ $$\int\frac{\frac{dv(t)}{dt}}{-v(t)+\log(v(t))\cdot v(t)}dt=\int\frac{1}{t}dt$$ this gives $$\log\left(-1+\log(v(t)\right)=\log(t)+C_1$$ and we obtain $$v(t)=e^{e^{C_1t}+1}$$