MHB Differential equation - distance needed to achieve target speed

[Jonter]

 

[MarkFL]

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

$$\d{u}{x}-\frac{B}{m}u=\frac{A}{m}u^{-1}$$

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

$$u\d{u}{x}-\frac{B}{m}u^2=\frac{A}{m}$$

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

$$\d{v}{x}-\frac{B}{2m}v=\frac{A}{2m}$$

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