Differential equation - distance needed to achieve target speed

[Jonter]

 

[MarkFL]

I would begin by writing the given ODE in the form:

\(\displaystyle \d{u}{x}-\frac{B}{m}u=\frac{A}{m}u^{-1}\)

We see we have a Bernoulli equation. Multiply by \(u\):

\(\displaystyle u\d{u}{x}-\frac{B}{m}u^2=\frac{A}{m}\)

Let \(v=u^2\) hence \(\displaystyle \d{v}{x}=2u\d{u}{x}\) and so we have:

\(\displaystyle \d{v}{x}-\frac{B}{2m}v=\frac{A}{2m}\)

We now have a linear equation, and so can you proceed?

Note: I have moved this thread to our "Differential Equations" forum.

 

[HallsofIvy]

The equation \(\displaystyle mu\frac{du}{dx}= A+ Bu^2\) can be written as \(\displaystyle \frac{mu du}{A+ Bu^2}= dx\). To integrate the left side, let \(\displaystyle v= A+ Bu^2\) so that \(\displaystyle dv= 2Bu du\) or \(\displaystyle udu= \frac{dv}{2B}\). Then \(\displaystyle \frac{mu du}{A+ Bu^2}= \frac{m dv}{2Bv}= dx\). Integrating, \(\displaystyle \frac{m}{2B} ln(v)= x+ C\) or \(\displaystyle ln(v)= \frac{2Bx}{m}+ C'\) (where C'= 2BC/m) and then \(\displaystyle v= A+ Bu^2= C''e^{2Bx/m}\) (where \(\displaystyle C''= e^{C'}\)).