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Math Help - please help with this differential eq. problem

  1. #1
    Junior Member
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    please help with this differential eq. problem

    hello,

    I am trying to solve

    K = y''/{[1+(y')^2]}^(3/2), K=const.

    i see that there is no y term so i make the substitution y'=p => y''=pp'

    K = pp'/[1+(p^2)]^(3/2)

    then i'm thinking maybe... 1+(p^2)=u => 2pp'=u'

    K/2 = u'/ [u^(3/2)] => (Kx/2)+C = -2/[u^(1/2)] =>

    u = 4/[(Kx/2)+C]^2 = 1+(p^2) => p = {{4/[(Kx/2)+C]^2}-1}^(1/2) = y' =>

    y = integr of {{4/[(Kx/2)+C]^2}-1}^(1/2)

    which is honestly quite unwieldy
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  2. #2
    Super Member
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    Write is as: \frac{dy'}{\left(1+(y')^2\right)^{3/2}}=k. Now, integrate that with respect to y' on the left. I get:

    \frac{y'}{\sqrt{1+(y')^2}}=kx+c_1

    Square it then isolate (y')^2 with long division and solve for (y')^2:

    (y')^2=\frac{1}{1+(kx+c_1)^2}-1

    You can finish that right? Just take square roots, get two equations, then solve those.
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  3. #3
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    Quote Originally Posted by shawsend View Post
    Write is as: \frac{dy'}{\left(1+(y')^2\right)^{3/2}}=k. Now, integrate that with respect to y' on the left. I get:

    \frac{y'}{\sqrt{1+(y')^2}}=kx+c_1

    Square it then isolate (y')^2 with long division and solve for (y')^2:

    (y')^2=\frac{1}{1+(kx+c_1)^2}-1

    You can finish that right? Just take square roots, get two equations, then solve those.
    thanks for the help
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