# Differential Equation from a Physics Problem

• May 27th 2012, 04:19 AM
Ivanator27
Differential Equation from a Physics Problem
In order to solve a problem I need to solve the following differential equation:

http://latex.codecogs.com/gif.latex?...2}}+v^2x^{{x}}

I'll figure out how to do any integration or differentiation that's necessary, but I don't know how to get it into a form that I can work with. If anyone could please show me how to get to a point from which I can integrate or differentiate?
• May 27th 2012, 04:58 AM
Electric
Re: Differential Equation from a Physics Problem
Hi mate,

I think that this one needs integrating factors.
What I would do is divide everything by mv, and then take the (vx^x)/m term to the other side. This will give you the standard form of a non-separable first order ode (google this if you are unsure of the method). Your integrating factor is then e^(-int (x^x)/m) - but this is very hard to integrate. You could then rearrange and solve for v(x).
• May 27th 2012, 05:01 AM
Prove It
Re: Differential Equation from a Physics Problem
This is not going to be pretty...

\displaystyle \displaystyle \begin{align*} m\,v\,\frac{dv}{dx} &= \frac{1}{x^2} + v^2x^x \\ m\,v\,\frac{dv}{dx} - v^2x^x &= \frac{1}{x^2} \\ \frac{dv}{dx} - v\,\frac{x^x}{m} &= v^{-1}\,\frac{1}{m\,x^2} \end{align*}

This is a Bernoulli DE, so let

\displaystyle \displaystyle \begin{align*} y &= v^2 \\ v &= y^{\frac{1}{2}} \\ \frac{dv}{dx} &= \frac{1}{2}y^{-\frac{1}{2}}\,\frac{dy}{dx} \end{align*}

and substituting into the DE gives

\displaystyle \displaystyle \begin{align*} \frac{1}{2}y^{-\frac{1}{2}}\,\frac{dy}{dx} - y^{\frac{1}{2}}\,\frac{x^x}{m} &= y^{-\frac{1}{2}}\,\frac{1}{m\,x^2} \\ \frac{dy}{dx} - y\,\frac{2x^x}{m} &= \frac{2}{m\,x^2} \end{align*}

which is first order linear, so multiplying by the Integrating Factor \displaystyle \displaystyle \begin{align*} e^{-\int{\frac{2x^x}{m}\,dx}} \end{align*} gives

\displaystyle \displaystyle \begin{align*} e^{-\int{\frac{2x^x}{m}\,dx}} \, \frac{dy}{dx} - y \, \frac{2x^x}{m} \, e^{-\int{\frac{2x^x}{m}\,dx}} &= \frac{2}{m \, x^2} \, e^{-\int{\frac{2x^x}{m}\,dx}} \\ \frac{d}{dx} \left( y \, e^{-\int{\frac{2x^x}{m}\,dx}} \right) &= \frac{2}{m \, x^2} \, e^{-\int{\frac{2x^x}{m}\,dx}} \\ y \, e^{-\int{\frac{2x^x}{m} \, dx}} &= \int{\frac{2}{m \, x^2} \, e^{-\int{\frac{2x^x}{m} \, dx}} \, dx} \\ y &= e^{\int{\frac{2x^x}{m} \, dx}}\int{\frac{2}{m \, x^2} \, e^{-\int{\frac{2x^x}{m} \, dx}} \, dx} \\ v^2 &= e^{\int{\frac{2x^x}{m} \, dx}}\int{\frac{2}{m \, x^2} \, e^{-\int{\frac{2x^x}{m} \, dx}} \, dx} \\ v &= \pm \sqrt{e^{\int{\frac{2x^x}{m} \, dx}}\int{\frac{2}{m \, x^2} \, e^{-\int{\frac{2x^x}{m} \, dx}} \, dx}} \end{align*}

Unfortunately, these integrals do not have closed-form solutions, so I don't know how much use they will be to you...
• May 27th 2012, 05:16 AM
Ivanator27
Re: Differential Equation from a Physics Problem
I don't need to integrate indefinitely, so I can use a Riemann sum or something. I don't know anything about integrating factors and different type of differential equations so I'm not surprised I couldn't do it. Thanks so much! You really are amazing 'Prove It'. =)
• May 27th 2012, 06:05 AM
Ivanator27
Re: Differential Equation from a Physics Problem
Nevermind, I got it. :)
• May 27th 2012, 06:27 AM
Prove It
Re: Differential Equation from a Physics Problem
It's recognising that you have a product rule expansion for a derivative in reverse - I also realise I've made a mistake in that step, as the integrating factor should actually have a negative in it. Doesn't change the result much though...