# Thread: v' = av - b(v^2)

1. ## v' = av - b(v^2)

x=x(t)

I need to solve x''=ax'-b(x')^2 to find x' in terms of t.

I rewrote it using x' = v:

v' = av - b(v^2)

From here I'm stuck. It's mainly the v^2 term that's putting me off.

Any help or hints would be much appreciated.

2. ## Re: v' = av - b(v^2)

I need to solve x''=ax'-b(x')^2 to find x' in terms of t.
assuming $a$ and $b$ are constants ...

$\displaystyle \frac{dv}{dt} = av - bv^2$

$\displaystyle \frac{dv}{av-bv^2} = dt$

$\displaystyle \frac{dv}{v(a-bv)} = dt$

partial fraction decomposition ...

$\displaystyle \frac{1}{a}\left( \frac{1}{v} + \frac{b}{a-bv}\right) \, dv = dt$

$\displaystyle \left( \frac{1}{v} + \frac{b}{a-bv}\right) \, dv = a \, dt$