Second order non-linear ode - am I on the right path?

Question

I've found this to be difficult to solve:

$$ \frac{d^2 x }{dt^2} + (a x + b) \frac{dx}{dt} = 0 $$

I've done some reading, and I guess I could write this as:

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

If I then treat $v(x) = \frac{dx}{dt}$ as an independent variable, I would get:

$$ \frac{dv}{dt} + bv + axv = 0 $$

This is sort of like a nonhomogenous equation. If I take the homogenous solution, I would get:

$$ v(t) = A e^{-bt}$$

I think.... I'm not sure where to go from here though.

Answer 1

We know the solution you have is wrong by putting it back in $$ v' = -bv $$ so we have in your original equation $$ -bv + bv + axv = axv = 0 $$ this is in general not true.

Your issue was not converting the $x$ in terms of $v$.

To give a hint. $$ x'' = \frac{dx}{dt}\frac{dv}{dx} $$ this leads to $$ \frac{dx}{dt}\frac{dv}{dx} + b\frac{dx}{dt} + ax\frac{dx}{dt} = \left(\frac{dv}{dx} + b + ax\right)\frac{dx}{dt} = 0 $$ In general $x' \neq 0$, so we can try to solve $$ \frac{dv}{dx} + b + ax = 0 $$ This is a first order ode with the correct variables to solve nicely.

Answer 2

The substitution $p=\frac{dx}{dt}$, so $\frac{d^2x}{dt^2}=p \frac{dp}{dx}$ and this gives \begin{eqnarray*} dp = -(ax+b) dx \\ \frac{dx}{dt} = -ax^2/2-bx+c. \end{eqnarray*} To integrate further depends upon the discriminant of the quadratic in $x$.

Answer 3

$$\frac{d^2 x }{dt^2} + (a x + b) \frac{dx}{dt} = 0$$ $$x'' + (a x + b) x' = 0$$ $$x'' + a xx' + bx' = 0$$ $$x'' + \frac a 2 (x^2)' + bx' = 0$$ Integrate $$x'+ \frac a 2 x^2 + bx = K$$ $$(x+\frac ba)'+ \frac a 2 (x^2 + \frac {2b}ax+\frac {b^2}{a^2}) = C$$ Substitute $z=x+\frac ba$ $$z'+ \frac a 2 z^2 = C$$ This last equation is separable