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Differential Equations: Separable Equations

  1. Jan 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the equation

    dy/dx = x/(y^2√(1+x))


    2. Relevant equations



    3. The attempt at a solution
    I separated them:
    y^2 dy = dx/√(1+x)

    I then integrated the dy side, I got (1/3)y^3 + C. I am stuck at integrating the dx side. Thanks in advance!
     
  2. jcsd
  3. Jan 23, 2014 #2

    Curious3141

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    Shouldn't that be ##\int y^2dy = \int \frac{x}{\sqrt{1+x}}dx##?

    For the RHS, try ##u^2 = 1 + x##.
     
  4. Jan 23, 2014 #3

    SammyS

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    You dropped an x from the numerator on the right side.
     
  5. Jan 23, 2014 #4
    Yes that should be, thank you. How does the u^2 work?
     
  6. Jan 23, 2014 #5

    Curious3141

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    Did you try it? Post what you have. Latex formatted please.
     
  7. Jan 24, 2014 #6
    Well, I set [itex]\sqrt{1+x}[/itex] = u. Then I set u2 = 1 + x. But I don't understand how to get du from that or how that helps with getting rid of the x on top.
     
  8. Jan 24, 2014 #7

    pasmith

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    [tex]
    \frac{x}{\sqrt{1 +x}} = \frac{u^2 - 1}{u} = u - \frac1u \\
    [/tex]
     
  9. Jan 24, 2014 #8

    HallsofIvy

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    From [itex]u^2= 1+ x[/itex], [itex]x= u^2- 1[/itex] and [itex]2u du= dx[/itex] so
    [tex]\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du[/tex]
     
  10. Jan 24, 2014 #9
    Thank you so much!
     
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