Differential equation x y' +x^2 y'' = k^2 y

  • Thread starter ehrenfest
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  • #1
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[SOLVED] differential equation

Homework Statement


x y' +x^2 y'' = k^2 y

where y=y(x), k is constant.

How do you prove that x^r, where r is a real number form a basis for that differential equation? They are obviously linearly independent. But how do you prove that they span the solution space?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
By solving it! :smile:
Try the substitution [itex]y(x)=u(x)\,x^k[/itex]
 
  • #3
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i didn't get what exactly you want to prove! do you want to solve it for x?
 
  • #4
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When I do that I get (2k+1)u'(x)+x u''(x)=0. Is that a contradiction?
 
  • #5
No it isn't! Now let [itex]u'(x)=a(x),\,u''(x)=a'(x)[/itex] which yields a separable 1st order ODE.
 
  • #6
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i didn't get what exactly you want to prove! do you want to solve it for x?

I want to find out if the set of solutions {x^r}, [tex]r \in \mathbb{R}[/tex] spans the entire solution space of that equation.
 
  • #7
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Then I get [tex]a(x)=\frac{C}{x^{2k+1}}[/tex], where C is a constant. Very nice Rainbow Child!
 
  • #8
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oh... i see, thank you
 

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