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A first order DE

  1. Aug 21, 2010 #1

    I have the first order differential equation [tex]y+y' = x[/tex]
    y(0) = 0
    First i tried to assume a solution of the form Ax+b, that didn't quite work.

    Then i tried to use the Integrating factor technique, work follows

    [tex]e^{\int1} = e^x[/tex]

    [tex]\int e^x(y+y') = \int{xe^x}[/tex]

    [tex] ye^x = \int xe^x[/tex]

    let u = x, du=1
    v = e^x dv=e^x

    So i end up with:
    [tex] ye^x = xe^x -e^x[/tex]

    which is wrong...
    Any ideas?
  2. jcsd
  3. Aug 21, 2010 #2
    You forgot your " + C" at the end of your integration. : )
  4. Aug 21, 2010 #3
    The answer is supposed to be [tex]y = ce^{-x} + x-1[/tex]

    But the answer i get when dividing both sides by e^x is y = x-1
    Last edited: Aug 21, 2010
  5. Aug 21, 2010 #4
    Did you read my post?

    Where is the integration constant? Where is the "+ C"? That is important because when you divide by e^x, you'll end up with the e^(-x) you were looking for.
  6. Aug 21, 2010 #5


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    What didn't work when you did this? Is it the same problem where you're missing the e-x term?
  7. Aug 22, 2010 #6

    Put y = Ax + b
    then y' = A

    So i get: Ax + b + A = x
  8. Aug 22, 2010 #7
    What values of A and b make the left hand side also equal to x?

    if y=f(x) is a single solution of the equation y'+y = x the other solutions
    are y = f(x) + C g(x) where y= g(x) is a solution of y' + y = 0.
  9. Aug 22, 2010 #8
    The values A=1, b=-1

    But does this mean i have to guess G(x) ?
  10. Aug 22, 2010 #9


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    No, you have to solve the corresponding homogeneous equation y' + y = 0 and add that solution to the other one, yp=x-1, to get the general solution.
  11. Aug 22, 2010 #10


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    As l'Hopital said, you forgot the "constant of integration", C, in your original solution:
    [tex]ye^x= xe^x- e^x+ C[/itex]

    Now, dividing through by [itex]e^x[/itex] gives [itex]y= x- 1+ Ce^{-x}[/itex]
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