I was trying to find the equation for a dampened oscillator using this equation

$$ F = -kx - bv $$

Which becomes the differential equation

$$ m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + kx = 0$$

I know this can be more easily solved by finding the roots of the characteristic equation.

But just to see what would happen I introduced the momentum. $p = m \frac{dx}{dt}$ which, using the chain rule and rearranging, becomes $\frac{dp}{dt} = \frac p m \frac{dp}{dx} $ and after substituting this in the original DE (the Force is the time derivative of momentum and $v=p/m$) we get the DE

$$ p \frac{dp}{dx} + bp = -mkx $$

How is this solved? Is it a non-linear ODE?

I have learnt to solve first order ODEs by separating variables, homogenous ODEs, ODEs of the form $y'=\frac{\text{linear}}{\text{linear}}$, and first order linear ODEs and Bernoulli DE. Does this come under any of the forms I know?

  • $\begingroup$ You seems a very normal person like me with this avatar xDDD $\endgroup$
    – Masacroso
    Jul 17, 2015 at 10:33
  • $\begingroup$ Divide lhs and rhs by $p$ and I think that you'll easily recognize of which form is this equation. $\endgroup$
    – Evgeny
    Jul 17, 2015 at 10:53

1 Answer 1


It's my favourite kind of questions (according to stats of MSE).

Ok, let's start. Divide your equation by $p$: $$ \frac{dp}{dx} + b + mk\frac{x}{p} = 0 $$ ($p\equiv 0$ is not a solution). Let $p=xq$: $$ q + x\frac{dq}{dx} + b + \frac{mk}{q} = 0. $$ Multiply by $q$: $$ q^2 + xq\frac{dq}{dx} + bq + mk = 0, $$ and variables separates: $$ \frac{q\,dq}{q^2+bq+mk} + \frac{dx}{x} = 0. $$ However, we can find $q$ only implicitly from this (in general case).

  • $\begingroup$ I get it now. It's a homogenous ODE. Thanks. :) $\endgroup$ Jul 17, 2015 at 15:41

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