Frobenius solution to a diff-eq

[IniquiTrance]

Homework Statement

Find 2 Frobenius series solutions to the following differential equation:

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

Homework Equations

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!

 

[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.

 

[IniquiTrance]

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.

<br /> <br /> It is a power series, I pulled out a radical so as to make it fit the form of sinh&#039;s power series.<br /> <br /> Do you disagree with my recurrence relation?<br /> <br /> Thanks for the response.

 

[gabbagabbahey]

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&#039;(x)=\frac{-a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{a_0}{2x}\cosh(\sqrt{2x})

\implies y&#039;&#039;(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&#039;&#039;(x)+3y_1&#039;(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)=0And 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 \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]

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[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})

 

[HallsofIvy]

Oh, dear, oh, dear!

 

[IniquiTrance]

Thanks!

http://img507.imageshack.us/img507/7167/71268782.jpg

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.

 

[gabbagabbahey]

Thanks!

http://img507.imageshack.us/img507/7167/71268782.jpg

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&#039;&#039;[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]

 

[IniquiTrance]

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

 

[gabbagabbahey]

You're Welcome!