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How to integrate this function?

  1. Jan 10, 2014 #1
    1. The problem statement, all variables and given/known data
    $$xy'' = y' + (y')^{3}$$


    2. Relevant equations
    $$ y'' = p' = \frac{dp}{dx}$$
    $$ y' = p$$



    3. The attempt at a solution
    $$xy'' = y' + (y')^{3}$$
    $$ x*\frac{dp}{dx} = p + p^{3} $$
    $$ \frac{dx}{x} = \frac{dp}{p+p^3} $$
    $$ ln x + C = ∫\frac{dp}{p(1+p^2)} $$
    $$ u = p^{2} $$ $$du = 2p dp$$
    $$ ln x + C = ∫\frac{1}{p(1+p^2)}*\frac{du}{2p} $$
    $$ ln x + C = ∫\frac{du}{2p^2(1+p^2)} $$
    $$ ln x + C = ∫\frac{du}{2u(1+u)} $$
    $$ln x + C = \frac{1}{2}∫\frac{du}{u} - \frac{1}{2}∫\frac{du}{1+u} $$
    $$ln x + C = \frac{1}{2}ln (\frac{p^2}{p^2+1}) $$
    $$ Dx^2 = \frac{p^2}{p^2+1} $$
    $$ p^2 = \frac{Dx^2}{1-Dx^2} $$
    $$ y' = \frac{√(Dx^2)}{√(1-Dx^2)} $$

    How do I integrate $$ y' = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$
     
  2. jcsd
  3. Jan 10, 2014 #2

    LCKurtz

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    You don't want a u substitution here. Try partial fractions$$
    \frac 1 {p(1+p^2)}=\frac A p + \frac {Bp+C}{1+p^2}$$
     
  4. Jan 10, 2014 #3
    I used partial fractions after u-substitution.
    $$ ∫\frac{dx}{x} = ∫\frac{du}{2u(1+u)} $$
    $$ \frac{A}{2u} + \frac{B}{1+u} = \frac{1}{2u(1+u)} $$
    $$ A(1+u) + B(2u) = 1 $$
    $$ A = 1 $$ and $$ B = -\frac{1}{2} $$
    $$ ∫\frac{dx}{x} = ∫\frac{du}{2u} - \frac{1}{2}∫\frac{du}{1+u} $$
    $$ ∫\frac{dx}{x} = \frac{1}{2} ∫ \frac{du}{u} - \frac{1}{2} ∫ \frac{du}{1+u} $$
    $$ ln x + C = \frac{1}{2} ln(p^2) -\frac{1}{2}ln(p^2+1) $$
    Therefore, $$ Dx^2 = \frac{p^2}{p^2+1} $$
    Since $$ ∫\frac{du}{2u(1+u)} \equiv∫\frac{dp}{p(1+p^2)}$$
    $$ du = 2p dp → u = p^2 $$
    $$\frac{du}{2u(1+u)} \equiv \frac{2pdp}{2p^2(1+p^2)} \equiv \frac{dp}{p(1+p^2)} $$
     
    Last edited: Jan 10, 2014
  5. Jan 10, 2014 #4
    Can someone show me the steps to integrating $$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$
    Attempt:
    $$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} $$
    $$ dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
    $$ y = ∫\frac{√(D)x}{√(1-Dx^2)} dx $$
    Should I integrate by parts?
     
    Last edited: Jan 10, 2014
  6. Jan 10, 2014 #5
    Is this correct?
    $$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} $$
    $$ dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
    $$ y = ∫\frac{√(D)x}{√(1-Dx^2)} dx $$
    $$ y = √(D)∫\frac{x}{√(1-Dx^2)} dx $$
    $$ u = 1-Dx^2 $$
    $$ du = -2Dx $$
    $$ y = -\frac{1}{2√(D)}∫\frac{du}{√(u)} $$
    $$ y = -\frac{1}{2√(D)}* 2√(u) + E $$
    $$ y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E $$
    $$ y = E -\frac{2√(1-Dx^2)}{2√(D)} $$
    $$ y = E -\frac{√(1-Dx^2)}{√(D)} $$
     
    Last edited: Jan 10, 2014
  7. Jan 10, 2014 #6

    haruspex

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    How about x = Bsin(θ) for some suitable B?
     
  8. Jan 10, 2014 #7

    Mark44

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    The above should be du = -2Dx dx.
    Because of the factor of x in the numerator, an ordinary substitution works here. Other than the minor error I pointed out, I don't see anything else wrong, but I didn't check that closely. At this point, you can verify that your work is correct by differentiating to verify that your solution satisifies the differential equation. You don't need us to do that for you.
     
  9. Jan 10, 2014 #8

    Mark44

    Staff: Mentor

    The above should be du = -2Dx dx.
    Because of the factor of x in the numerator, an ordinary substitution works here. Other than the minor error I pointed out, I don't see anything else wrong, but I didn't check that closely. At this point, you can verify that your work is correct by differentiating to verify that your solution satisifies the differential equation. You don't need us to do that for you.
     
  10. Jan 10, 2014 #9
    Thank you for pointing this out for me Mark44 :-)
     
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