How to find the solution of a differential equation

Question

m.(d²x/dt²) + R(dx/dt) -kx=0

 

when m= x² , R= x-x² , k=1

 

I have no idea of what method to use to get the solution.

Answer 1

Using x=e^(rt) as a trail solution is only valid when dealing with constant coefficients. Remember, x(t) is the function you are solving for. If x appears as a coefficient in the characteristic equation, that means you'll end up with answers that look like X(t)=Ae^{({function of X})}, which doesn't make sense.

 

The coefficients involve x(t) itself; the ODE is nonlinear. This would be classified as a homogeneous second-order nonlinear autonomous ODE. Using power series methods is as close to impossible as you can get. This needs to be solved numerically (computationally) by recasting it as a system of differential equations.

 

(x²)(x'') + (x-x²)(x') +(-k)(x)=0

 

Introduce these variables:

• F(t)=x(t)
• G(t)=x'(t)=F'
• H(t)=x''(t)=G'=F''

 

The original ODE becomes F²H+(F-F²)G-kF=0, which we can solve for H(F,G). At this point, because H=G', we no longer require H explicitly. We can express F' and G' as functions of F and G. Hence we have a system of two ODEs, which can be written as

1. G'=[kF-(F-F²)G]/[F²]
   
2. F'=G

 

I'm not sure what type of numerical solvers you're accustomed to using, but this is a fairly straightforward system for Matlab or something similar. As the original ODE was written, it is very similar to what's called the Van der Pol equation. The VdP equation is used pretty extensively as an introductory example to nonlinear ODEs. Researching that may give you a few tips on proceeding.

 

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As was mentioned before, the ODE system in this case is nonlinear. A linear system could have been written as something like

1. F'=aF+bG
2. G'=cF+dG

 

where a,b,c,d are constants. Problems like this are amenable to linear algebra, and we could have found solutions as linear combinations of eigenvectors. Not at all relevant to the problem at hand, but just an FYI if this pops up in the future!

 

Hope that helped!

Answer 2

Let's rewrite the differential equation using easier notation.  This looks like a second-order homogenous differential. Therefore, we will have a second derivative term.

 

mx'' + Rx' - kx = 0

 

 

Plug in all the known values into this equation.

 

x²x'' + (x - x²)x' - x = 0

 

 

When solving these types of differential equations, we look for the characteristic equation.

 

Let  x = e^rt

 

      x' = re^rt

 

      x'' = r²e^rt

 

 

Plug in these values into the equation.

 

 

x²(r²e^rt) + (x - x²)(re^rt) - e^rt = 0

 

 

e^rt(x²r² + r(x - x²) - 1) = 0

 

e^rt(x²r² + rx - rx² - 1) = 0

 

 

Set the term in parenthesis equal to zero.  Then solve for r.

 

x²r² + rx - rx² - 1 = 0

 

 

Solve for r. The general form of the solution to this differential equation is

 

x = C₁e^r1t + C₂e^r2t

 

 

where C₁ and C₂ are constants.  r1 and r2 is what you want to solve for.