ODE with integrating factor

  • Thread starter iRaid
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  • #1
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Problem:
[tex]xy'+2y=3x[/tex]
Attempt:
Divide by x...
[tex]y'+\frac{2y}{x}=3[/tex]
I think I find the integrating factor by doing:
[tex]e^{\int \frac{2}{x}dx}[/tex]

Not sure if that's right but if it is then the solution to the integral is just 2x.

Any help is appreciated
 

Answers and Replies

  • #2
rock.freak667
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You will need to add in the constant of integration to the right side before you do any simplification.
 
  • #3
HallsofIvy
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Problem:
[tex]xy'+2y=3x[/tex]
Attempt:
Divide by x...
[tex]y'+\frac{2y}{x}=3[/tex]
I think I find the integrating factor by doing:
[tex]e^{\int \frac{2}{x}dx}[/tex]

Not sure if that's right but if it is then the solution to the integral is just 2x.

Any help is appreciated
Yes, the integrating factor is [itex]e^{\int (2/x)dx}[/itex] but then the integral is 2 ln(x) so the integrating factor is [itex]e^{2ln(x)}= e^{ln(x^2)}= x^2[/itex], not 2x.

Multiplying the equation by that gives [itex]x^2y'+ 2xy= (x^2y)'= 3x^2[/itex]
 
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  • #4
559
8
Yeah I figured it out ends up:
[tex]x^{2}y'+2xy=3x^{2}[/tex]
Take integral/derivative of integrating factor:
[tex]\int \frac{d}{dx} x^{2}y=\int 3x^{2}dx[/tex]
Simplifies to:
[tex]x^{2}y=x^{3}+C \implies y=\frac{x^{3}+C}{x^{2}}[/tex]
(can be simplified a little more too)

Thanks for the help.
 

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