# Frobenius solution to a diff-eq

• IniquiTrance
It's easy to see by plugging into the equation that those are NOT solutions to equation.Thanks for the response.In summary, the student is trying to find Frobenius series solutions to a differential equation, but is getting incorrect results. He then attempts to use a recurrence relation to find solutions, but gets stuck. He is then helped by the tutor and is able to find the correct solutions using a power series.

## 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:

I got the following solutions:

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

What am I doing wrong?

Thanks!

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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.

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HallsofIvy said:
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. HallsofIvy said: 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 $\frac{a_0}{\sqrt{2x}}$ )

IniquiTrance said:
I got the following solutions:

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

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

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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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IniquiTrance said:
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.

You're Welcome!