# Frobenius solution to a diff-eq

1. May 17, 2009

### IniquiTrance

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

Find 2 Frobenius series solutions to the following differential equation:

2xy'' + 3y' - y = 0

2. Relevant equations

3. The attempt at a solution

I got r = -1/2 and 0 as roots.

Recurrence relation:

I got the following solutions:

According to mathematica, these aren't the right solutions.

What am I doing wrong?

Thanks!

Last edited by a moderator: Apr 24, 2017
2. May 18, 2009

### HallsofIvy

It's easy to see by plugging into the equation that those are NOT solutions to equation. Further since one of the solutions for r is 0, I don't see how you can get $\sqrt{2x}$ in both solutions- one should be a standard power series.

Last edited by a moderator: May 18, 2009
3. May 18, 2009

### IniquiTrance

It is a power series, I pulled out a radical so as to make it fit the form of sinh's power series.

Do you disagree with my recurrence relation?

Thanks for the response.

4. May 18, 2009

### gabbagabbahey

Really?

$$y_1(x)=\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})\implies y_1'(x)=\frac{-a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{a_0}{2x}\cosh(\sqrt{2x})$$

$$\implies y''(x)=\frac{-3a_0}{4x^2}\cosh(\sqrt{2x})+\frac{3a_0}{4\sqrt{2}x^{5/2}}\sinh(\sqrt{2x})+\frac{a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})$$

$$\implies 2xy_1''(x)+3y_1'(x)-y_1(x)=\left(\frac{-3a_0}{2x}\cosh(\sqrt{2x})+\frac{3a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})\right)$$

$$+\left(\frac{-3a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{3a_0}{2x}\cosh(\sqrt{2x})\right)-\left(\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})\right)=0$$

And y_2(x) also satisfies the DE

$$y_1(x)=\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})=a_0\left(1+\frac{x}{3}+\frac{x^2}{630}+\ldots\right)$$

Is a regular power series (just multiply the series for $sinh(\sqrt{2x})$ by $\frac{a_0}{\sqrt{2x}}$ )

5. May 18, 2009

### gabbagabbahey

Mathematica probably gave the total general solution as $$y(x)=\frac{e^{\sqrt{2} \sqrt{x}} C[1]}{\sqrt{x}}-\frac{e^{-\sqrt{2} \sqrt{x}} C[2]}{\sqrt{2} \sqrt{x}}$$....correct?

Which is actually equivalent to the linear combination of your two solutions with $C[1]=\frac{a_0}{2\sqrt{2}}+\frac{b_0}{2}$ and $C[2]=\frac{b_0}{\sqrt{2}}-\frac{a_0}{2}$, since, by definition, $\sinh(u)=\frac{1}{2}(e^u-e^{-u})$ and $\cosh(u)=\frac{1}{2}(e^u+e^{-u})$

Last edited by a moderator: Apr 24, 2017
6. May 18, 2009

### HallsofIvy

Oh, dear, oh, dear!

7. May 18, 2009

### IniquiTrance

Thanks!

http://img507.imageshack.us/img507/7167/71268782.jpg [Broken]

I couldn't make heads or tails of that, and couldn't get the confirmation I needed.

I then tried plugging the result straight into the diff-eq, and mathematica likewise wasn't returning 0 as the answer, which spooked me.

Last edited by a moderator: May 4, 2017
8. May 18, 2009

### gabbagabbahey

In the above DSolve command, you need to use a "*" between x and y''[x] (i.e. $2x*y''[x]$), Mathematica interprets xy''[x] as the second derivative of a function called "xy"

As for checking that your solution satisfies the DE, try using FullSimplify[2x*y''[x]+3y'[x]-y[x]] after defining your y[x]

Last edited by a moderator: May 4, 2017
9. May 18, 2009

### IniquiTrance

Ah, now that makes perfect sense! Thanks for your help.

10. May 18, 2009

### gabbagabbahey

You're Welcome!