TITLE: Solving $t^2=x(t) x'(t) t^3+x'(t)$
QUESTION [5 upvotes]: Solve the following differential equation
$$t^2=x(t) x'(t) t^3+x'(t)$$
I would appreciate some help with this problem. 
Thank you very much.

REPLY [5 votes]: write the differential equation as $$3(t^3x + 1)dx= 3t^2dt = dt^3$$ using a change of variable $t^3 = s$ gives you $$3(sx + 1) dx = ds$$ which is a linear differential equation $$\frac{ds}{dx} = 3sx + 1$$
we can turn this into $d(se^{-3x^2/2}) = e^{-3x^2/2}dx$ whose solution is $$ s = t^3 =  Ce^{3x^2/2} + \int_0^x e^{3(x^2-\xi^2)/2} d\xi$$  we have found an implicit relation between $t$ and $x.$ i don't know how you will solve for $x$ if that is what you are after.