# What's the solution of this non-linear (?) differential equation for a dampened harmonic oscillator?

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?

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

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).