Frobenius solution to a diff-eq

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In summary: 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.
  • #1
IniquiTrance
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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!
 
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  • #2
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 [itex]\sqrt{2x}[/itex] in both solutions- one should be a standard power series.
 
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  • #3
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 [itex]\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.
 
  • #4
HallsofIvy said:
It's easy to see by plugging into the equation that those are NOT solutions to equation.

Really?

[tex]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}) [/tex]

[tex] \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})[/tex]

[tex]\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)[/tex]

[tex]+\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[/tex]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}[/itex] in both solutions- one should be a standard power series.

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

Is a regular power series:wink: (just multiply the series for [itex]sinh(\sqrt{2x})[/itex] by [itex]\frac{a_0}{\sqrt{2x}}[/itex] )
 
  • #5
IniquiTrance said:
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 [tex]y(x)=\frac{e^{\sqrt{2} \sqrt{x}} C[1]}{\sqrt{x}}-\frac{e^{-\sqrt{2} \sqrt{x}} C[2]}{\sqrt{2} \sqrt{x}}[/tex]...correct?

Which is actually equivalent to the linear combination of your two solutions with [itex]C[1]=\frac{a_0}{2\sqrt{2}}+\frac{b_0}{2}[/itex] and [itex]C[2]=\frac{b_0}{\sqrt{2}}-\frac{a_0}{2}[/itex], since, by definition, [itex]\sinh(u)=\frac{1}{2}(e^u-e^{-u})[/itex] and [itex]\cosh(u)=\frac{1}{2}(e^u+e^{-u})[/itex]
 
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  • #6
Oh, dear, oh, dear!:blushing:
 
  • #7
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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  • #8
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. [itex]2x*y''[x][/itex]), 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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  • #9
Ah, now that makes perfect sense! Thanks for your help.
 
  • #10
You're Welcome!:smile:
 

What is the Frobenius solution to a differential equation?

The Frobenius solution is a method used to find a power series solution to a differential equation. It is particularly useful for finding solutions to equations with singular points.

When is the Frobenius method commonly used?

The Frobenius method is commonly used when attempting to solve a differential equation with a singular point at one or both of its boundaries. It is also useful for finding solutions to equations with variable coefficients.

How does the Frobenius method work?

The Frobenius method involves assuming a solution in the form of a power series, substituting it into the differential equation, and solving for the coefficients of the series. This process is repeated until the desired accuracy is achieved.

What are the limitations of the Frobenius method?

The Frobenius method may not work for all types of differential equations, particularly those with non-constant coefficients. It also requires the existence of a singular point at one or both boundaries, which may not always be the case.

Are there any alternative methods to the Frobenius method for solving differential equations?

Yes, there are other techniques such as the power series method, the Laplace transform method, and the method of undetermined coefficients that can also be used to solve differential equations. The choice of method depends on the specific type of equation and its characteristics.

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