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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..

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2 Answers 2

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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.

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  • $\begingroup$ So the first integral is $t*cost*u^3$? $\endgroup$
    – ChikChak
    Commented Nov 21, 2016 at 18:22
  • $\begingroup$ @WakaWaka123: Yes. It is not always that simple, then the best that can happen is that separation of variables works, as in the answer of E.H.E. See also "exact differential equation", where one treats essentially the same class of questions. $\endgroup$
    – Lutz Lehmann
    Commented Nov 21, 2016 at 19:52
  • $\begingroup$ and how can I solve the ODE? $\endgroup$
    – ChikChak
    Commented Nov 22, 2016 at 21:02
  • $\begingroup$ $u(t)=u(t_0)\sqrt[3]{\dfrac{t_0\cos t_0}{t\cos t}}$ is the obvious solution. It is of course only valid between two zeros of $t\cos t$. $\endgroup$
    – Lutz Lehmann
    Commented Nov 22, 2016 at 21:06
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by using the separable method

$$\frac{du}{u}=\frac{t\sin t-\cos t}{3t\cos t}dt$$ $$\frac{du}{u}=(\frac{1}{3}\tan t-\frac{1}{3t})dt$$

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