ODE Euler-Cauchy

  • Thread starter etotheix
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  • #1
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Homework Statement



Find a general solution of the differential equation
xy'' − (1 + x2 )y' = 0.

Homework Equations



Euler-Cauchy general form : xnyn+xn-1yn-1 ... +y=g(x)

The Attempt at a Solution



At first I tried using Euler-Cauchy but by multiplying by x (to get the x2 in front of y''), the term in front of y' becomes (x+x3) and I don't know how to deal with that. I looked in my book and could not find any similar example.

I tried with power series but no luck, and since they are not in the midterm it means we have to use some other method to solve this problem.

According to wolframalpha the answer is supposed to be y(x) = c_1 e^(x^2/2)+c_2
http://www.wolframalpha.com/input/?i=x*y''-(1%2Bx^2)y'%3D0

Any ideas on how to solve this?
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi etotheix! welcome to pf! :smile:

hint: separation of variables :wink:
 
  • #3
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Got it, thanks a lot tiny-tim!

For those that might be interested, you have to make the substitution u=y' and then use separation
of variables.

Another question if you don't mind, would it be possible to solve this problem using power series?
I have to admit that I don't really understand what conditions have to be met in order to use
power series.

This is what I did :

x∑n(n-1)anxn-2-(1+x2)∑nanxn-1=0

Then I arrive at :

-a1 = 0 for n=0,1

an+1=(an-1(n-1))/(n2-1) = an-1/(n+1) for n=2,3,...

y(x) = a0 + a2(x2+x4/4+x6/24+x8/192+...)

And I know that the series expansion of ex2/2 =/= x2+x4/4+x6/24+x8/192+...
it is actually 1+ (1/2) * (x2+x4/4+x6/24+x8/192+...)
 
  • #4
tiny-tim
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an+1=(an-1(n-1))/(n2-1) = an-1/(n+1) for n=2,3,...

that's ok … eg a8 = a6/8 = a48.6 = a28.6.4

( 2.4.6…2n = 2n1.2.3…n = 2nn!, ∑x2n/2nn! = ∑(x2/2)n/n! = ex2/2 ) :smile:
 
  • #5
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I don't understand why you can do that.

I though that since a8=a2/192, then we will have (a2/192)*x^8 in the answer, but in your case we would have (a2/384)*x^8. Why is there a multiplication of the denominator by 2, and where does it come from?

Thanks again for taking the time to answer my questions.
 
  • #6
tiny-tim
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hi etotheix! :smile:

(just got up :zzz: …)

because, for example, 2.4.6.8.10.12 = 1.2.2.2.3.2.4.2.5.2.6.2 = 1.2.3.4.5.6.26 = 6!26 :wink:
 
  • #7
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Hi!
My first post,

I have a similar and the textbook and my notes provide no help.
(1+x^2)y''-2xy'+2y=0
I think that I'm supposed to the euler-cauchy to solve this. I don't get how to deal with the constant multiplier of y''
 
  • #8
tiny-tim
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welcome to pf!

hi ville! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)

use the same method as etotheix did in post #3 …

show us how far you get :smile:
 
  • #9
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Thanks!

If you mean a substitution u=y' there is a term y so this doesn't help?

If you mean the series expansion method I have a big problem since the second part of my real analysis course where we go through the power series is starting next week :)

So if this is the case I'll have to skip this
 
  • #10
tiny-tim
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yes, i meant the series expansion method

maybe next week! :wink:
 

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