# Homework Help: Nonlinear second order differential equation

1. Nov 11, 2012

### mumaga

I am having a problem finding the solution for this eq:

y''(x)+(2/x)y'(x)+(w^2)y(x)=0

I couldnt find examples in the text book that goes on a similar line, and have been searching the internet as well, but no use. I am thinking of using substitution v=y' but not sure how to do that in the presence of y??

any help would be much appreciated!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 11, 2012

### Dustinsfl

Is w a constant? You could always multiple by x and you would have a Cauchy-Euler type.

Last edited: Nov 11, 2012
3. Nov 11, 2012

### mumaga

Yes w is a constant! thank you VERY much Dustinsfl! I am now reading about the Cauchy-euler equations and how to solve them!

very appreciated!!! :)

4. Nov 11, 2012

### mumaga

ahh, just checked the cauchy-euler equation, and from what i understood it does not apply in this case! as the x is in the denominator and only in the second term!

5. Nov 11, 2012

### Dustinsfl

Multiple by x.

6. Nov 11, 2012

### mumaga

I did, but substituting for trial solutions does not deal to cancelling x, for example y=x^m, you will have x left,,

7. Nov 11, 2012

### Dustinsfl

May be you can try reduction of order.

8. Nov 11, 2012

### mumaga

tried that, no implicit solution,,,

9. Nov 12, 2012

### HallsofIvy

I should point out that this is NOT a "non-linear" equation. It is a linear equation with variable coefficients. The standard method for such equations is to use a "power series"

I would start by multiplying both sides by x, as Dustinsfl suggested, but that does NOT give a "Cauchy Euler equation", we get, rather, $xy''+ 2y'+ \omega^2 xy= 0$. That is not a "Cauchy-Euler equation" because the coefficient of y'' is x, not x2. We could get that by multipling both sides by x2 but then the coefficient of y is $x^2$ also.

Because that "1/x" is undefined at 0, and, after multipying by x the coefficient of y'' is 0 at x= 0, x= 0 is a "regular singular point" and should use "Frobenius' method" rather than a regular power seties. That is, we look for a solution of the form $y= \sum_{n= 0}^\infty a_nx^{n+ c}$ where "c" is not necessarily a positive integer (not necessariy an integer).

Differentiating term by term, $y'= \sum_{n=0}^\infty(c+n)a_nx^{n+c- 1}$ and $y''= \sum_{n=0}^\infty(n+c)(n+ c- 1)a_nx^{n+ c- 2}$
Putting those into the equation, we have
$\sum_{n=0}^\infty (n+c)(n+c-1)a_nx^{n+ c- 1}+ \sum_{n=0}^\infty 2(n+ c)x^{n+ c- 1}+ \sum_{n=0}^\infty \omega^2 a_n x^{n+c+ 1}$.

The lowest power will occur when n= 0 which would give power $x^{c- 1}$ in the first and second terms, $x^{c+1}$ for the last term. That is, the lowest power wil be $x^{c- 1}$ and its coefficient will be $c(c-1)a_0+ 2ca_0= (c(c- 1)+ 2c)a_0= (c^2+ c)a_0= 0$. . We could choose values of c so that the first, "i" terms were 0 but, in order to be specific we we choose so that the very first term, $a_0$ is NOT 0. In order for that to be true we must have $c^2+ c= 0$. This is called the "indicial equation". Obviously c= 0 and c= -1 satisfy this.

Putting c= 0 and then c= -1 into the equations for the coefficients leads to recursive equations for $a_n$. Does any of that sound familiar to you?