Help solving linear 2nd order ode

Feb 2010
45
0
Hi, I had this question on a test and I'm pretty sure I got it wrong and I just wanted to see if someone could help me through it.

Find the general solution of the equation:\(\displaystyle $x^2y'' - 9xy' + 25y = 0, y_1=x^5$\) is a solution.

So I found the second solution with Abel's formula: \(\displaystyle y_2 = y_1 * \int \frac{W}{y_1^2}dx, W = C_0*\exp^\psi, \psi = -\int p(x)\)

Skipping a few steps, I got \(\displaystyle \psi=9*\int \frac{1}{x}dx = \ln(x^9) + 9*C_1\) and \(\displaystyle W = C_0*x^9*\exp^{9*C_1} = C*x^9, C=C_0*\exp^{9*C_1}\)

I eventually found \(\displaystyle y_2\) as \(\displaystyle y_2 = C*x^5*\ln(x) + C*x^5*C_2\)

At this point, I know I need to apply variation of parameters. But I feel like I'm missing something because applying variation of parameters leads to a pretty complicated result.

Is there an easier way to solve this problem? Is it possible to use the characteristic equation to solve it? Any help is appreciated.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey SterlingM.

A quick google search turned up this:

First and Second Order Differential Equations

Note that there is a section on Euler-Cauchy equations which has the exact same form as your DE. It also looks a lot easier to solve as well.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Rather than memorize Abel's formula, it is better to work this out using the way Abel's formula was developed to begin with.
Knowing that \(\displaystyle y= x^5\) satisfies this equation we look for a solution of the form \(\displaystyle y= x^5u\) where u is some unknown function of u.

Then \(\displaystyle y'= 5x^4u+ x^5u'\) and \(\displaystyle y''= 20x^3u+ 10x^4u'+ x^5u''\).

\(\displaystyle x^2y''= 20x^5u+ 10x^6u'+ x^7u''\)
\(\displaystyle -9xy'= -45x^5u- 9x^6u'\) and
\(\displaystyle 25y= 25x^5u\)
so that \(\displaystyle x^2y''- 9xy'+ 25y= x^6u'+ x^7u''= 0\)

\(\displaystyle u'+ xu''= 0\). Letting v= u', \(\displaystyle xv'= -v\), \(\displaystyle dv/v= -dx/x\), \(\displaystyle ln(v)= -ln(x)+ c\), \(\displaystyle u'= v= C/x\),
\(\displaystyle u(x)= C ln(x)+ D\)

So \(\displaystyle y(x)= x^5u= Cx^5 ln(x)+ Dx^5\)
 
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