Find a solution to the ODE?

Question

$dy/dx = x\cos2x/3y^2$ So far I've rearranged.

$dy3y^2 = x\cos2x dx$

Then do I just solve for $y$? If so how do I do that? I'm just a little confused on the next steps.

Thanks for any help.

I gave you a hint (answer) down below. Make sure you vote up and approve it if it answered your question, so that the thread goes down as answered. — Rebellos, May 07 '16 at 01:57

score 0 · Accepted Answer · answered May 07 '16 at 01:56

0

Suppose that we write it as : $y'(x) = \frac{xcos(2x)}{y^(x)} $

Hint : First multiply both sides with : $3y^2(x)$ and then integrate both sides with respect to x. You will then find $y(x)$.

answered May 07 '16 at 01:56

Rebellos

After doing that i got $y^3 = 1/3 x^3cos(2) +C$ i feel that is wrong – someguy May 07 '16 at 01:59
You integrated wrong. The answer is : $y^3(x) = \frac{1}{4}\cos(2x) + \frac{1}{2}x\sin(2x) + c$ – Rebellos May 07 '16 at 02:00
Please, be careful and integrate correctly, that's an easy integral to evaluate. – Rebellos May 07 '16 at 02:00
Oh i see what i did! then all i do is take cube root of other side? – someguy May 07 '16 at 02:01
Yes, but you need to detect 3 different cases, depending on the value of the "thing" under the cubic root. – Rebellos May 07 '16 at 02:03
Fillippatos so say if i had an initial condtion of $y(0)=0$ and i had to find $y(\pi)$ how would i do that, i tried and got 3cube root $\pi$ – someguy May 07 '16 at 02:08
You just plug $ x=0$ and find $c$ and then plug $x=π$ and find the value. I don't know, I won't do arithmetic calculations now, you can do that with a calculator in 1 second. – Rebellos May 07 '16 at 02:11

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