Question

how to find the general solution:
dP/dt=kP cos(rt-v)^2

k, r and v are positive constants

Answer 1

Separation of variables is needed.

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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:
\[P=\exp((\frac{ k }{ 2r }\sin(2rt-2v)+\frac{ kt }{ 2 }\]\[+\frac{ k }{ 2r }\sin(2v))\]

Answer 6

I don't get where the last bit came from

Answer 7

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

Answer 8

And then integrate?

Answer 9

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

Answer 10

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

Answer 11

yeah that's what I did

Answer 12

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

Answer 13

help needed urgently!!!

Answer 14

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