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First Order Differential Equations

  1. Mar 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve the following differential equation using seperation of variables

    [tex] (1+x)^2 y' = (1-y)^2 , y(1) = 2 [/tex]

    2. Relevant equations



    3. The attempt at a solution
    haven't gotten very far in this at all :/

    i've tried dividing both sides by [tex](1+x)^2[/tex], in order to get y' on it's own..

    [tex]y' = \frac{(1-y)^2}{(1+x)^2} [/tex]


    but i don't know how to integrate this...but had a go anyway

    apparently the rule for integrating an expression in brackets is..

    [tex] \frac{(ax + b)^n}{a(n+1)} [/tex]

    so i tried integrating both halves of the fraction seperatley...giving
    [tex]
    \frac{(-y+1)^3 }{-3y}[/tex]
    [tex]
    \frac{(x+1)^2}{3x}
    [/tex]
    putting these together and dividing gave

    [tex]\frac{3x(-y+1)^2}{-3y(x+1)^3} + C[/tex]


    however i don't think this is accurage, as substituting in 1 and 2 for x and y respectivley gmade C some out as -3/48. Can someone shed some light at where I've went wrong?
     
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2

    Dick

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    You haven't gotten y on it's own. Write y'=dy/dx and put the dy with the y's and the dx with the x's.
     
  4. Mar 13, 2007 #3
    are you saying i need to get the y terms on one side of the equals and the x terms on the other side?

    if so then i'm stumped. I'm not sure how to split up the equation in it's initial form.
     
  5. Mar 13, 2007 #4

    Dick

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    How about dy/(1-y)^2=dx/(1+x)^2? Does it look split now?
     
  6. Mar 13, 2007 #5
    ok, right, rearranged it and heres what i have:

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

    [tex]\frac {1}{(1-y)^3}{3} = {1}{(1+x)^3}{3} + C[/tex]

    which, after substituting y(0) = -1 (wrong values on first post BTW) i get

    [tex]\frac{-7}{24} = C[/tex]

    I'm pretty sure this is wrong...but anyway lets sub that back into the integrated function

    [tex] \frac{1}{3} (1-y)^3 = \frac{1}{3} (1+x)^3 - 7/24 [/tex]
    [tex] \frac 8(1-y)^3 = 8(1+x)^3 - 7[/tex]

    the answer in teh book is given as

    [tex] y = \frac{x-1}{1+3x}[/tex]
     
  7. Mar 13, 2007 #6

    Dick

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    It would help a lot if you did the integrations correctly. The integral of 1/x^2 is not 1/x^3.
     
  8. Mar 13, 2007 #7
    er sorry thats a typo, i get the integral of 1/(1+x)^3 to be....

    1/3(1+x)^3
     
  9. Mar 13, 2007 #8

    Dick

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    You want to integrate 1/(1+x)^2. Try again.
     
  10. Mar 13, 2007 #9

    Dick

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    Hint: integral of 1/u^2 du=-1/u.
     
  11. Mar 13, 2007 #10
    so...the integral of 1/(1+x)^2 is just -1 - (1/x) ?


    even that doesn't seem right...though my formula book doesn't have the formula for such an instance.
     
  12. Mar 13, 2007 #11

    Dick

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    Make a substitution u=x+1. Similarly for the y integral. You have done that, right?
     
  13. Mar 13, 2007 #12
    ok so that gives...

    [tex]- \frac{1}{(1-y)} = - \frac{1}{(1+x)} [/tex]
     
  14. Mar 14, 2007 #13

    HallsofIvy

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    I added the integral sign on the right side.

    I think you meant
    [tex]\frac{1}{3(1-y)^3}= \frac{1}{3(1+x)^3}+ C[/tex]
    but you've missed a sign. But only one sign. As Dick said you need to substitute for both 1+ x and 1- y. If u= 1- y, what is du?
     
  15. Mar 14, 2007 #14

    Dick

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    I think Halls was just cleaning up your notation and doesn't mean to imply that cube is the right power. But you did miss the substitution sign.
     
  16. Mar 14, 2007 #15
    i don't see where the missing sign is. unless the sign attatched to the letter comes out? but thats not what i thought you meant when you said

    1/u^2 = -1/u

    so do you mean that i should have...

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

    EDIT: wait that can't be right...yeah i'm lost.
     
    Last edited: Mar 14, 2007
  17. Mar 14, 2007 #16

    Dick

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    If you differentiate the right side of this you get 1/(1+x)^2. That's what you want. If you differentiate the left side you get -1/(1-y)^2. Don't forget the chain rule. That's NOT what you want.
     
  18. Mar 14, 2007 #17

    HallsofIvy

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    He certainly never said that! he said that the anti-derivative of 1/u2 is -1/u.

    If you let u= 1+ x, then du= dx and dx/(1+x)2= du/u2. What's the anti-derivative of that?

    If you let v= 1-y, then dv= -dy and y/(1-y)2= -dv/v2. What's the anti-derivative of that?
     
  19. Mar 14, 2007 #18
    ok so i tried a totally different method and it seems to work.

    instaed of substitution i tried using the standard integral

    [tex] \int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)} [/tex]

    thanks a lot for your help guys!
     
  20. Mar 15, 2007 #19

    Dick

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    Let u=(ax+b), du=a*dx, du/a=dx.

    [tex] \int (ax+b)^n dx= \frac{1}{a} \int (u)^n du = \frac{1}{a}
    \frac{u^(n+1)}{(n+1)} = \frac{(ax+b)^{n+1}}{a(n+1)}[/tex]

    That is a TOTALLY DIFFERENT METHOD!
     
  21. Mar 15, 2007 #20

    Dick

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    Let u=(ax+b), du=a*dx, du/a=dx.

    [tex] \int (ax+b)^n dx= \frac{1}{a} \int (u)^n du = \frac{1}{a}
    \frac{u^{n+1}}{(n+1)} = \frac{(ax+b)^{n+1}}{a(n+1)}[/tex]

    That is a TOTALLY DIFFERENT METHOD!
     
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