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

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Homework Help Overview

The discussion revolves around a differential equation of the form x y' + x^2 y'' = k^2 y, where y is a function of x and k is a constant. The original poster seeks to establish whether the functions of the form x^r, where r is a real number, form a basis for the solution space of this equation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the linear independence of the proposed solutions and question how to demonstrate that they span the solution space. There are attempts to manipulate the equation through substitutions and transformations to explore the implications of these functions.

Discussion Status

The conversation includes various attempts to clarify the original poster's intent and to explore the implications of their proposed solutions. Some participants suggest specific substitutions and transformations, while others express confusion about the goal of the proof. There is an ongoing exploration of the relationships between the functions and the differential equation.

Contextual Notes

Participants are navigating the requirements of proving the spanning of the solution space without providing a complete solution. There is an emphasis on understanding the nature of the solutions and their implications within the context of the differential equation.

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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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