# Ordinary differential equation and first integral - help!

Ok, so I started learning ODE, and got my first H.W., but I have no idea even how to begin!

The question is to find the "first integral" of the following ODE:

$3t·(\cos t)u^2u'+(\cos t -t\sin t)u^3=0$

Later, I need to find "the first integral" of the equation that I get from comparing the previous "first integral" to a const. And if possible, to solve this ODE.

Of course, this is only the first ODE for this question, but I have no idea how to start solving it..

As a first strategy in such situations, try if you can identify functions $f$ and $g$ such that your ODE reads as $$f(t)g'(u)u'+f'(t)g(u)=0$$ since then you can employ the product rule to find that your equation is the same as $$(f(t)g(u))'=0$$ which has an easy first integral.
• So the first integral is $t*cost*u^3$? Commented Nov 21, 2016 at 18:22
• $u(t)=u(t_0)\sqrt[3]{\dfrac{t_0\cos t_0}{t\cos t}}$ is the obvious solution. It is of course only valid between two zeros of $t\cos t$. Commented Nov 22, 2016 at 21:06
$$\frac{du}{u}=\frac{t\sin t-\cos t}{3t\cos t}dt$$ $$\frac{du}{u}=(\frac{1}{3}\tan t-\frac{1}{3t})dt$$