## chikaa 2 years ago how to find the general solution: dP/dt=kP cos(rt-v)^2 k, r and v are positive constants

1. myininaya

Separation of variables is needed.

2. myininaya

Put the problem in this form and the integrate both sides. $g(p) dp=f(t) dt$

3. myininaya

You can picture those k, r , and v's being like 1,2, and 3's. They are just constants. You need your P's together and your t's together. I don't care and you shouldn't care where this puts the constants.

4. myininaya

You can actually leave k on the side it is on or divide both sides by k. It doesn't matter.

5. chikaa

this is what I got: $P=\exp(\frac{ k }{ 2r }\sin(2rt-2v)+\frac{ kt }{ 2 })$ However, this is what I should be getting according to the answer n my tutorial: $P=\exp((\frac{ k }{ 2r }\sin(2rt-2v)+\frac{ kt }{ 2 }$$+\frac{ k }{ 2r }\sin(2v))$

6. chikaa

I don't get where the last bit came from

7. myininaya

So did you set if up like I told you to?

8. myininaya

And then integrate?

9. myininaya

$\frac{1}{p} dp=k \cos^2(rt-v) dt$

10. myininaya

$\ln|p|=k \int\limits_{}^{} \frac{1}{2}(\cos(2[rt-v]+1) dt$

11. chikaa

yeah that's what I did

12. chikaa

i'm just not getting the last bit k/2r sin(2v)

13. chikaa

help needed urgently!!!

14. KenLJW

dP/dt=kP cos(rt-v)^2 dP/P=kcos(rt-v)^2dt ln(P)=(k/3r)cos(rt-v)^3 check with taking the Derivative with chain rule P=C1EXP[(k/3r)cos(rt-v)^3