Question

Using this to explain a DiffEQ problem to someone

Answer 1

You have:
\[\frac{dv}{dt}=150-5v-v^2 \implies \frac{dv}{dt}\frac{1}{150-5v-v^2}=1\]

Answer 2

v is a function of t, right?

Answer 3

\[\frac{dv}{v^2+5v-150}=-dt\]

Answer 4

you want to integrate ?

Answer 5

Yes, I know how to work it I'm just waiting for someone to come see this so I'm typing it now

Answer 6

Factoring the denominator you have:
\[\frac{dv}{(v+15)(v-10)}=-dt\]
\[\int\limits \frac{dv}{(v-10)(v+15)}=-\int\limits dt\]
Doing partial fractions you get the left integral is equivalent to:
\[\frac{-1}{25}\int\limits \frac{dv}{v+15}+\frac{1}{25}\int\limits \frac{dv}{v-10}=-\int\limits dt\]
Integrating you have:
\[-\frac{1}{25}\ln|v+15|+\frac{1}{25}\ln|v-10|=-t+C\]
Now use properites of logs to solve for v then solve for C. Done.

Answer 7

You don't need to solve for v to solve C. So using v(0)=0 we see.
\[\frac{-1}{25}\ln(15)+\frac{1}{25}\ln(10)=C \implies C=\frac{1}{25}\ln(\frac{10}{15})=\frac{1}{25}\ln(\frac{2}{3})\]