# Math Help - ODE solved using Euler's equation

1. ## 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.

2. 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

3. Excellent, I'll try that out and see how it works. Thanks

edit: solved it

4. I can tell you that it works pretty well !

5. Originally Posted by stevedave
edit: solved it
Could you post your result ?

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

$y=-x^{-3}cos(2ln(-x))$

7. Originally Posted by stevedave
$y=-x^{-3}cos(2ln(-x))$
I find the same as you