Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Really need help with a simple ODE

  1. Jan 25, 2012 #1
    The equation I'm trying to solve is

    [itex]\frac{dy}{dx} = \frac{y^2 - 1}{x^2-1}[/itex], given y(2) = 2

    The methods I'm somewhat familiar with are separation of variables, integrating factor, and exact. I tried this:

    [itex]\frac{dy}{dx} = \frac{y^2 - 1}{x^2-1}[/itex]

    [itex](x^2 - 1)dy = (y^2-1)dx[/itex]

    [itex](x^2 - 1)dy - (y^2-1)dx= 0[/itex]

    So, now it's an exact equation, right?

    I tried integrating each part:

    [itex]\int (x^2 - 1)dy = (x^2-1)y+c1(x)[/itex]

    [itex]\int (y^2 - 1)dx = (y^2-1)x+c1(y)[/itex]

    But now I'm confused what I'm supposed to do! If I just let the constants of integration be zero, then I have:

    [itex](x^2-1)y[/itex]

    [itex](y^2-1)x[/itex]

    But what do I do with those?

    I'm really confused :(
     
  2. jcsd
  3. Jan 25, 2012 #2
    Hello,
    Try this:
    [tex]dy/(y^2-1)=dx/(x^2-1)[/tex]
    You can apply partial fractions
     
  4. Jan 25, 2012 #3
    Hi, I don't see how I can apply partial fractions to [itex]\frac {dy}{y^2-1}[/itex] ?
     
  5. Jan 25, 2012 #4
    [tex]1/(y^2-1)=1/[(y+1)(y-1)]=A/(y+1)+B/(y-1)[/tex]
    You would then be required to solve for A and B.
     
  6. Jan 26, 2012 #5
    Thanks for your help, I really appreciate it.

    Here's what I tried:

    [itex]\frac{1}{y^2-1} = \frac{1}{(y+1)(y-1)} = \frac {A}{(y+1)} + \frac{B}{(y-1)}[/itex]

    [itex]1 = A(y-1) + B(y+1)[/itex]

    [itex]y = 1: 1 = 0 + B(2) \iff B = \frac{1}{2}[/itex]

    [itex]y = -1: 1 = A(-2) + 0 \iff A = -\frac{1}{2}[/itex]

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

    [itex]-\frac{1}{2}ln|y+1| + \frac{1}{2}ln|y-1| = -\frac{1}{2}ln|x+1| + \frac{1}{2}ln|x-1| + C_1[/itex] , x > 1, y > 1

    [itex]ln(y+1)^{-\frac{1}{2}} + ln(y-1)^{\frac{1}{2}} = ln(x+1)^{-\frac{1}{2}} + ln(x-1)^{\frac{1}{2}} + C_1[/itex], raise by e

    [itex](y+1)^{-\frac{1}{2}} + (y-1)^{\frac{1}{2}} = (x+1)^{-\frac{1}{2}} + (x-1)^{\frac{1}{2}}+ C_2, C_2 = e^{C_1}[/itex]

    But I don't see how I can solve for y. Any hints?
     
  7. Jan 26, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you required to solve for y?
     
  8. Jan 26, 2012 #7
    Most of the time, it is enough to give the final equation in implicit form. To find the constant, you plug in the initial conditions: ie, y(2) = 2 means that x=2 and y=2. If the initial conditions were y(5) = 3, then it would be y=3 and x=5.

    A minor nitpick but there's an algebra mistake in the last part:

    Hint: [itex]ln(A) + ln(B) = ln(AB)[/itex]
     
  9. Jan 26, 2012 #8
    I'm supposed to "solve" the differential equation. Does that not require solving for y?
     
  10. Jan 26, 2012 #9
    doesn't y = x + c satisfy this ODE?

    y(2) = 2

    2 = 2 + c

    c = 0

    so checking y = x via differentiation:

    dy/dx = 1

    then substitution:

    dy/dx = (y^2 -1) / (x^2 - 1) = (x^2 - 1) / (x^2 - 1) = 1

    sorry if I'm a little off base or this doesn't help much, I'm just used to solving PDEs and when you get them to ODEs just using some ansatz math.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Really need help with a simple ODE
  1. ODE and BVP need help (Replies: 2)

  2. Some ODE help needed (Replies: 1)

Loading...