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Solving an equation

  1. May 29, 2005 #1
    [tex]y' + 2y = 6x + 5[/tex]

    Can I solve it by...

    [tex]\frac{dy}{dx} + 2y = 6x + 5[/tex]

    [tex]dy = (6x + 5 - 2y)dx[/tex]

    [tex]y = 3x^2 + 5x - 2yx[/tex]

    [tex]y + 2yx = 3x^2 + 5x[/tex]

    [tex]y = \frac{3x^2 + 5x}{1 + 2x}[/tex]
     
  2. jcsd
  3. May 29, 2005 #2

    arildno

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    No you can't.
    1) Why do you think this works?
    2) Have you checked if it works?
     
  4. May 29, 2005 #3

    dextercioby

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    Nope.It's a nonhomogenous I-st order linear equation with constant coefficients.Find an integrating factor.

    Daniel.
     
  5. May 29, 2005 #4
    Damn......
     
  6. May 29, 2005 #5
    I think it would be [itex] e^{2t} [/tex], right?
     
  7. May 29, 2005 #6

    arildno

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    That would be the integrating factor, yes.
     
  8. May 29, 2005 #7

    HallsofIvy

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    Well, e2x, actually. [tex]\frac{d(e^{2x}y)}{dx}= 2e^{2x}y+ e^{2x}y'[/tex] so multiplying the entire equation by e2x gives e2xy'+ 2e2xy= (e2xy)'= (6x+5)e2x. Integrating the left side gives e2xy and the right side can be integrated by parts.

    By the way, the reason your first method fails is that the integral of -2ydx is not -2xy because y is itself a function of x.
     
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