How to Approach Solving a Nonlinear Second Order ODE with a Quadratic Term?

[Safinaz]

Homework Statement

How to solvebthis second-order ODE:

Relevant Equations

$\frac{\partial^2 x}{ \partial t^2} + b \frac{\partial x}{ \partial t} + C x - D x^2 =0$

Or:

$\ddot{x} + b \dot{x} + C x - D x^2 =0$
Where

$b, C, D$ are constants.

I know how to solve similar ODEs like

$\frac{\partial^2 x}{ \partial t^2} + b \frac{\partial x}{ \partial t} + C x =0$

Where one can let $x = e^{rt}$, and the equation becomes
$r^2 + b r + C =0$

Which can be solved as a quadratic equation.

But now the problem is that there is $x^2$ term, so if one used that substitution, we left by:
$r^2 + b r + C + D e^{rt} =0$

So any help to find the solution of the ODE

 

[anuttarasammyak]

Where one can let x=ert, and the equation becomes
r2+br+C=0

Which can be solved as a quadratic equation.

You have got general solution of homogeneous differential equation. Then you have to find a particular solution to add that for inhomogeneous differential equation with x^2 term. Have you investigated x=constant ?

 

[erobz]

You have got homogeneous general solution. Then you have to find a special solution to add that. Have you investigated x=constant ?

The OP's equation is a non linear ODE?

 

[anuttarasammyak]

The OP's equation is a non linear ODE?

My bad. Thanks. By choosing sign of constants, the equation is interpreted as oscillation of a body in a viscous medium with harmonic if D=0 and inharmonic with D potential. x = 0 is stable, x= C/D is unstable point for small oscillation around.

 

[Mark44]

$\frac{\partial^2 x}{ \partial t^2} + b \frac{\partial x}{ \partial t} + C x - D x^2 =0$
Or:
$\ddot{x} + b \dot{x} + C x - D x^2 =0$

The second version of your DE, using the notation with dots, suggests that x is a function of t alone. In that case the first version of the DE should be written without partials.
Like so:
$\frac{d^2 x}{dt^2} + B\frac{dx}{dt} + Cx - Dx^2 = 0$
Also, since C and D are uppercase, B should probably be uppercase as well.

The OP's equation is a non linear ODE?

Yes. I'm sure your question was rhetorical.