Solving a first order linear differential equation by variation of parameters

  • Thread starter bitrex
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  • #1
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Homework Statement



I have to solve the following differential equation by the "variation of parameters" method.


Homework Equations



[tex]\frac{dy}{dx}x +2y = 3x[/tex]



The Attempt at a Solution



The associated homogeneous equation of the initial equation is:

[tex]\frac{dy}{dx} = -2x^{-1}y[/tex]

So

[tex]\frac{1}{y}dy = -2x^{-1}dx[/tex]

[tex]ln(y) = -2ln(x)[/tex]

[tex]ln(y) = ln(x^{-2})[/tex]

[tex]y = x^{-2}[/tex]

Unfortunately, this doesn't satisfy the homogeneous equation.
 

Answers and Replies

  • #2
djeitnstine
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Where did the '3' go to? When you divide by x it is [tex]\frac{dy}{dx}=3-2\frac{y}{x}[/tex] not [tex]-2\frac{y}{x}[/tex]

Then you do the substitution v=y/x, then it comes easy.
 
  • #3
djeitnstine
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Double post
 
  • #4
rock.freak667
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You can use djeitnstine guide OR divide by x and then use an integrating factor. But you said you had a certain method to use.

Since that is the case then you should know that y1=x-2 is a solution of the homogenous equation. So just find the Wronskian of y1, which should be easy and just put it into the formula.
 
  • #5
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I'm sorry I didn't clarify in my earlier post - I'm supposed to take an equation of the form [tex] y' + p(x)y = q(x)[/tex], set the right hand side of my equation to zero, solve that equation by using separation of variables, and then take the homogeneous equation result as yh*u and substitute that back into the original differential equation, and get another separable equation. So I do the following with the answer to the homogeneous equation above and the original equation:

[tex]\frac{dy}{dx}(x^{-2}u) + 2x^{-1}x^{-2}u = 3[/tex]

=

[tex]-2x^{-1}u + x^{-2}\frac{du}{dx} + 2x^{-3}u = 3[/tex]

The way it's supposed to work is that the first term and the third term cancel, leaving me with a linear differential equation in du/dx to solve. However, in this case it doesn't seem to be working out that way...
 
  • #6
HallsofIvy
Science Advisor
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-2-1= -3, not -1.


The derivative of x-2 is -2x-3, not -2x-1.
 
  • #7
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:facepalm: Yeah, that's the problem. Getting my differentiation and integration confused again.:cry:
 

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