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

[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?

## The Attempt at a Solution

By solving it!
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.

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