What is the general solution to the 2nd order linear ODE xy''+2y'+4xy=0?

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Homework Help Overview

The problem involves finding the general solution to a second-order linear ordinary differential equation (ODE) of the form xy'' + 2y' + 4xy = 0. The subject area pertains to differential equations, specifically methods for solving linear ODEs.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the Frobenius method and mentions roots of the characteristic equation. They express uncertainty about the next steps after finding the roots and question the validity of their derived solution involving trigonometric functions. Other participants question the correctness of the proposed solution and suggest that exponentials may be more appropriate.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made by the original poster. Some guidance has been offered regarding the need to verify the proposed solution against the original equation.

Contextual Notes

The original poster indicates a preference for avoiding reduction of order due to its complexity. There is also mention of a discrepancy between their solution and that provided by Wolfram Alpha, highlighting potential confusion regarding the method of solution.

sydneyfranke
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Homework Statement


Find general solution to:

xy''+2y'+4xy=0

Homework Equations



Frobenius Method or Bessel's Equation

The Attempt at a Solution



I know how to get the roots for this problem (which are r1 = 0 & r2 = -1). But not I don't know what to do with these roots. I know that this is considered "Case 3", but I seriously don't know what to do from here.

Furthermore, I have attempted to solve for y1(x) and came up with the answer:

y1(x) = a0(x^-1)cos(2x) + ((a1)/2)(x^-1)sin(2x)

If this is right, then I'm not sure what the next step is. Do I do reduction of order? I hope not because that would take FOREVER.

Wolfram Alpha has the general solution as

y= (c1(exp(-2ix)))/x - (ic2(exp(2ix)))/4x

I'm totally lost. Any help would be appreciated.
 
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sydneyfranke said:
y1(x) = a0(x^-1)cos(2x) + ((a1)/2)(x^-1)sin(2x)
Did you check that satisfies the original equation? (It doesn't look right to me. I get exponentials, not trig.)
 
No I didn't, but exponentials match the wolfram alpha solution. Can you maybe explain how you got the exponentials?
 
The way it works on these fora is that you post your working and others try to spot where you went wrong.
 

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