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Non-linear Diff Eq

  1. Oct 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the given differential equation by using the substitution u=y'

    3. The attempt at a solution
    If I got to this last integral correctly, I dont know how to solve it. I'm thinking I didn't get to that step correctly.. But I can't find my mistake..
  2. jcsd
  3. Oct 17, 2010 #2
    Your work is absolutely correct, you just missed a little detail

    [tex]\frac{dy}{dx} = (e^{-2y}-1)^{\frac{1}{2}}C_3 \Rightarrow \frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx[/tex]

    You forgot the square root in the left side of the last equation

    [tex]\frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx[/tex]

    I believe you can take it from here right?

    Hint: Trig substitution.
  4. Oct 17, 2010 #3
    Hey thanks for replying Je m'appelle. I'm a little rusty on trig substitution. I know how to evaluate the normal three with x2 as the function but I dont remember how to handle e-x (or -y) function in that case..

    And thanks for noting I didn't use the sqrt, sometimes I forget and move on with that mistake :/. I need to train new habits lol.
  5. Oct 17, 2010 #4
    Ok.. So I went on to the next problem and I can't complete it either..

    #5) x2y''+(y')2=0
    disregard #7, I'm finishing that one up..
  6. Oct 18, 2010 #5
    Use [tex]v=e^{-y}[/tex] and rewrite

    [tex]\frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx[/tex]


    [tex]-\frac{dv}{v\sqrt{(v^2 - 1)}} = C_3dx[/tex]


    [tex]ln(v) = -y[/tex]

    [tex]\frac{-dv}{v} = dy[/tex]

    And remember that

    [tex]\int \frac{dx}{x\sqrt{x^2 - 1}} = arcsec|x| + C[/tex]

    So back to our equation, and using the above we'll get to

    [tex]arcsec|v| + C = -C_3x[/tex]

    [tex]v = sec(C_3x) [/tex]

    [tex]e^{-y} = sec(C_3x)[/tex]

    [tex]y = -ln(sec(C_3x)) [/tex]

    Now find out [tex]C_3[/tex] by replacing y back into the original differential equation.

    Your work is OK up to this point

    [tex]\frac{-1}{u} = \frac{1}{x} + c_1 [/tex]

    Now replace [tex]u = \frac{dy}{dx}[/tex] and rewrite it as

    [tex]\frac{-dx}{dy} = \frac{1}{x} + c_1 [/tex]

    [tex](\frac{1}{x} + c_1 )dx = -dy [/tex]

    Integrate both sides

    [tex]\int (\frac{1}{x} + c_1)dx = -\int dy \Rightarrow \int \frac{1}{x} \ dx + \int c_1 \ dx = -\int dy[/tex]

    And work it out from here.
    Last edited: Oct 18, 2010
  7. Oct 19, 2010 #6
    So how does this look? These two problems are the same 5 and 7 posted above. a->5, b->7

    These look ok?

    Thank you for that!
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