Is c1 + c2 x^.5 a Solution of y y'' + (y')^.5 = 0?

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    Fundamental Ode
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Homework Help Overview

The discussion revolves around verifying solutions to the differential equation y y'' + (y')^.5 = 0, specifically examining whether the function c1 + c2 x^.5 can be considered a solution based on the properties of the equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to verify that y1(x) = 1 and y2(x) = x^.5 are solutions, and questions why c1 + c2 x^.5 is not generally a solution. Participants discuss the implications of linearity in relation to the fundamental set of solutions.

Discussion Status

Participants are exploring the implications of the non-linearity of the differential equation and its effect on the validity of combining solutions. There is an ongoing examination of theorems related to linearity and fundamental solutions.

Contextual Notes

There is a mention of the Wronskian being non-zero, which typically indicates a fundamental set of solutions, but the discussion raises questions about the necessity of linearity for this to hold true.

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Homework Statement



Verify that y1(x) = 1 and y2(x) = x^.5 are solutions of the following y y'' + (y')^.5 = 0. Then show that c1 + c2 x^.5 is not in general a solution of this equation.

Homework Equations





The Attempt at a Solution



I was able to show that both y1 and y2 are solutions to the DE. I found the Wronskian to be 1/(2 sqrt(x)) which is not equal to zero, so I was under the impression that this would mean that the two solutions would form a fundamental set of solution. Does anyone see why c1 + c2 x^.5 isn't a solution?
 
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Your ODE isn't linear.
 
Does it matter if it's not linear in general? Boyce/Diprima's theorems don't seem to make note of whether or not the ODE must be linear for a set of fundamental solutions to be valid.
 
I'd have to see the theorem, but linearity is the property that tells you if y1 and y2 are solutions to an ODE then so is c1*y1+c2*y2. I think you need it.
 

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