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

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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

 
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By solving it! :smile:
Try the substitution y(x)=u(x)\,x^k
 
i didn't get what exactly you want to prove! do you want to solve it for x?
 
When I do that I get (2k+1)u'(x)+x u''(x)=0. Is that a contradiction?
 
No it isn't! Now let u'(x)=a(x),\,u''(x)=a'(x) which yields a separable 1st order ODE.
 
astrosona said:
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}, r \in \mathbb{R} spans the entire solution space of that equation.
 
Then I get a(x)=\frac{C}{x^{2k+1}}, where C is a constant. Very nice Rainbow Child!
 
oh... i see, thank you
 
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