# ODE solved using Euler's equation

• January 20th 2009, 09:37 AM
stevedave
ODE solved using Euler's equation
This problem is getting messy fast and I'm not sure if I am doing it correctly. Any advice will be very appreciated.

y is a function of x

$x^2y'' + 7xy'+13y=0$; $y(-1) = 1$; $y'(-1)=3$

using $x = e^t$, $t = ln(x)$ and $y(x) = Y(t)$ this becomes

$Y'' +6Y'+13Y = 0$

The characteristic equation has roots $-3\pm{i2}$

so $Y = c_1e^{-3t}e^{i2t} + c_2e^{-3t}e^{-i2t}$

back substitution yields:

$y = (c_1+c_2)x^{-3}cos(2ln(x)) +j(c_1-c_2)x^{-3}sin(2ln(x))$

This is about where I ran out of steam. I had started to try and choose my constants to give me two separate equations which should be linearly independent which gives me the following:

$y = k_{1}x^{-3}cos(2ln(x))+k_{2}x^{-3}sin(2ln(x))$

but plugging in my initial conditions gets so messy taking the log of negative numbers giving me imaginary arguments to sinusoids which gives hyperbolic functions.

None of my other problems have been this tough and I'm guessing I'm doing something wrong. Thanks for any help.
• January 20th 2009, 09:57 AM
running-gag
Hi

This is normal : when using $x = e^t$, you assume that $x > 0$ which is not the case since obviously $y(-1)$ and $y'(-1)$ are defined

Use $x = -e^t$ can help you
• January 20th 2009, 10:04 AM
stevedave
Excellent, I'll try that out and see how it works. Thanks

edit: solved it :)
• January 20th 2009, 10:09 AM
running-gag
I can tell you that it works pretty well ! (Wink)
• January 20th 2009, 10:57 AM
running-gag
Quote:

Originally Posted by stevedave
edit: solved it :)

Could you post your result ?
• January 20th 2009, 11:20 AM
stevedave
Quote:

Originally Posted by running-gag
Could you post your result ?

$y=-x^{-3}cos(2ln(-x))$
• January 20th 2009, 11:35 AM
running-gag
Quote:

Originally Posted by stevedave
$y=-x^{-3}cos(2ln(-x))$

I find the same as you (Clapping)