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

1. myininaya Group Title

Separation of variables is needed.

2. myininaya Group Title

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

3. myininaya Group Title

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 Group Title

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

5. chikaa Group Title

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 Group Title

I don't get where the last bit came from

7. myininaya Group Title

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

8. myininaya Group Title

And then integrate?

9. myininaya Group Title

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

10. myininaya Group Title

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

11. chikaa Group Title

yeah that's what I did

12. chikaa Group Title

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

13. chikaa Group Title

help needed urgently!!!

14. KenLJW Group Title

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