2nd Order ODE

• Jul 16th 2012, 10:53 AM
theod9
2nd Order ODE
Any ideas how can I solve this equation f''=a*f^0.4?
a=real number

Thank you
• Jul 16th 2012, 03:28 PM
Ivanator27
Re: 2nd Order ODE
I'm not certain on this but I did the following:

http://latex.codecogs.com/gif.latex?f''(x)=af(x)^{0.4}
http://latex.codecogs.com/gif.latex?...x)f(x)^{0.4}dx

Dealing with each side of the equation individually:

[IMG]http://latex.codecogs.com/gif.latex?\displaystyle \begin{align*} \int f'(x)f''(x)dx \\ &= \int f''(x)dx \times f'(x)-\iint f''(x)dx \times Dx(f'(x)) \\ &= f'(x)^2 - \int f'(x)f''(x)dx \\ \\ \Rightarrow \int f'(x)f''(x)dx=\frac{1}{2}f'(x)^2 \end{align*}[/IMG]

And

[IMG]http://latex.codecogs.com/gif.latex?\begin{align*} a \int f'(x)f(x)^{0.4}dx \\ &=a\left ( \int f'(x)dx\times f(x)^{0.4}-\iint f'(x)dx \times Dx(f(x)^{0.4})dx \right ) \\ &=a\left ( f(x)^{1.4}-\int f(x) \times 0.4f'(x)f(x)^{-0.6}dx \right ) \\ &=a\left ( f(x)^{1.4}-0.4\int f'(x)f(x)^{0.4}dx \right ) \\ \Rightarrow a \int f'(x)f(x)^{0.4}dx = \frac{5a}{7}f(x)^{1.4} \end{align*}[/IMG]

Giving:

[IMG]http://latex.codecogs.com/gif.latex?\displaystyle \begin{align*} \frac{1}{2}f'(x)^2 = \frac{5a}{7}f(x)^{1.4} \end{align*}[/IMG]

Integrating one more time should be relatively straightforward now. But please don't work straight off this answer - Do everything above and make sure I haven't made a mistake - it's 2am right now. =D

EDIT: Sorry, I can't get the latex to display properly - just copy/paste everything between the image tags into your browser and you'll see the equations. If anyone can tell me why the equations aren't displaying I'd be grateful.
• Jul 16th 2012, 10:02 PM
Prove It
Re: 2nd Order ODE
Quote:

Originally Posted by theod9
Any ideas how can I solve this equation f''=a*f^0.4?
a=real number

Thank you

\displaystyle \displaystyle \begin{align*} \frac{d^2f}{dx^2} &= a\,f^{0.4} \\ \frac{df}{dx}\,\frac{d^2f}{dx^2} &= a\,f^{0.4}\,\frac{df}{dx} \end{align*}

Now if we let \displaystyle \displaystyle \begin{align*} u = \frac{df}{dx} \end{align*}, then we can write

\displaystyle \displaystyle \begin{align*} u\,\frac{du}{dx} &= a\,f^{0.4}\,\frac{df}{dx} \\ \int{u\,\frac{du}{dx}\,dx} &= \int{a\,f^{0.4}\,\frac{df}{dx}\,dx} \\ \int{u\,du} &= \int{a\,f^{0.4}\,df} \\ \frac{u^2}{2} + C_1 &= \frac{a\,f^{1.4}}{1.4} + C_2 \\ \frac{u^2}{2} &= \frac{5a\,f^{1.4}}{7} + C_2 - C_1 \\ u^2 &= \frac{10a\,f^{1.4} + C}{7} \textrm{ where } C = 14\left(C_2 - C_1\right) \\ u &= \pm \sqrt{\frac{10a\,f^{1.4} + C}{7}} \\ \frac{df}{dx} &= \pm \frac{\sqrt{10a\,f^{1.4} + C}}{\sqrt{7}} \\ \pm \frac{\sqrt{7}}{\sqrt{10a\,f^{1.4} + C}}\,\frac{df}{dx} &= 1 \\ \int{\pm \frac{\sqrt{7}}{\sqrt{10a\,f^{1.4} + C}} \, \frac{df}{dx}\,dx} &= \int{1\,dx} \\ \int{\pm \frac{\sqrt{7}}{\sqrt{10a\,f^{1.4} + C}}\,df} &= x + C_3 \end{align*}

To solve this integral you need to use the Hypergeometric Function as shown here.
• Jul 16th 2012, 10:06 PM
Prove It
Re: 2nd Order ODE
Quote:

Originally Posted by Ivanator27
\displaystyle \displaystyle \begin{align*} \frac{1}{2}f'(x)^2 = \frac{5a}{7}f(x)^{1.4} \end{align*}

Integrating one more time should be relatively straightforward now. But please don't work straight off this answer - Do everything above and make sure I haven't made a mistake - it's 2am right now. =D

Sorry but this equation here is missing an integration constant, which makes the integral NOT straightforward.
• Jul 17th 2012, 07:52 AM
Ivanator27
Re: 2nd Order ODE
Ah - stupid of me to forget that. My apologies.

On another note - any idea why my latex didn't display properly?