Second Order Differential Equation

**cheesecake91** · November 2nd 2012, 05:49 PM

Solve [(alpha)^2]u''+a*e^u=0 by multiplying it by du/dx and integrating it in x

I just keep going in circles and ending up with the original problem.

**Prove It** · November 2nd 2012, 08:00 PM

Originally Posted by cheesecake91

Solve [(alpha)^2]u''+a*e^u=0 by multiplying it by du/dx and integrating it in x

I just keep going in circles and ending up with the original problem.

$\displaystyle \begin{align*} \alpha ^2 \, \frac{d^2u}{dx^2} + a\, e^u &= 0 \\ \alpha ^2 \, \frac{d^2u}{dx^2} &= -a\, e^u \\ \alpha ^2 \, \frac{du}{dx} \, \frac{d^2u}{dx^2} &= -a\, e^u \, \frac{du}{dx} \end{align*}$

Now, if we let $\displaystyle \begin{align*} w = \frac{du}{dx} \end{align*}$ , then $\displaystyle \begin{align*} \frac{dw}{dx} = \frac{d^2u}{dx^2} \end{align*}$ and the DE becomes

$\displaystyle \begin{align*} \alpha ^2 \, w \, \frac{dw}{dx} &= -a \, e^u \, \frac{du}{dx} \\ \int{ \alpha^2 \, w \, \frac{dw}{dx} \, dx} &= \int{ -a\, e^u \, \frac{du}{dx} \, dx} \\ \int{\alpha^2 \, w \, dw} &= \int{ -a \, e^u \, du} \end{align*}$

Finish it.

**cheesecake91** · November 3rd 2012, 11:13 AM

Thanks! I was thinking it had something to do with substitution but couldn't figure it out.

Thread: Second Order Differential Equation

LinkBack

Thread Tools

Display

Second Order Differential Equation

Re: Second Order Differential Equation

Re: Second Order Differential Equation

Similar Math Help Forum Discussions

First order differential equation

Simplifying an equation containing both a first order and second order differential

2nd order differential equation

Formulating a second-order ordinary differential equation as a first-order system?

Second Order Differential Equation

Search Tags