# Frobenius solution to a diff-eq

## Homework Statement

Find 2 Frobenius series solutions to the following differential equation:

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

## The Attempt at a Solution

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

Recurrence relation:

http://image.cramster.com/answer-board/image/cramster-equation-200951755626337813656283812509812.gif

I got the following solutions:

http://image.cramster.com/answer-board/image/cramster-equation-2009517550266337813622621312505598.gif
http://image.cramster.com/answer-board/image/cramster-equation-2009517551276337813628747875004691.gif

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

What am I doing wrong?

Thanks!

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## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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.

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

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

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

gabbagabbahey
Homework Helper
Gold Member
I got the following solutions:

http://image.cramster.com/answer-board/image/cramster-equation-2009517550266337813622621312505598.gif
http://image.cramster.com/answer-board/image/cramster-equation-2009517551276337813628747875004691.gif

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

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

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

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HallsofIvy
Science Advisor
Homework Helper
Oh, dear, oh, dear! 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.

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gabbagabbahey
Homework Helper
Gold Member
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.

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]

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

gabbagabbahey
Homework Helper
Gold Member
You're Welcome! 