# Differential equation - distance needed to achieve target speed

#### MarkFL

Staff member
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

##### Well-known member
MHB Math Helper
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 $$\displaystyle C''= e^{C'}$$).