• Support PF! Buy your school textbooks, materials and every day products Here!

Frobenius solution to a diff-eq

  • #1
190
0

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!
 
Last edited by a moderator:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,794
925
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.
 
Last edited by a moderator:
  • #3
190
0
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
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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]
 
Last edited by a moderator:
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,794
925
Oh, dear, oh, dear!:blushing:
 
  • #7
190
0
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:
  • #8
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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]
 
Last edited by a moderator:
  • #9
190
0
Ah, now that makes perfect sense! Thanks for your help.
 
  • #10
gabbagabbahey
Homework Helper
Gold Member
5,002
6
You're Welcome!:smile:
 

Related Threads for: Frobenius solution to a diff-eq

Replies
1
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
941
Replies
3
Views
1K
Replies
2
Views
1K
Replies
16
Views
7K
Replies
1
Views
1K
Top