# Help solving linear 2nd order ode

#### SterlingM

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

Last edited:
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