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Math Help - This Bernouilli ODE is killing me

  1. #1
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    This Bernouilli ODE is killing me

    I've been working on this one differential equation for an hour and a half now and I can't for the life of me figure out where I'm going wrong: xy' -15 y = (9 x^2 -5 x + 2) y^{\frac{4}{5}},\qquad y(1)=0

    First I divide by x. Then by  y^{\frac{4}{5}}.

    This leaves me with:
    f(x) = (9x^ - 5x + 2)

    y^(-4/5)y' - (15/x)y^(1/5) = f(x)/x

    I substitute:
    v = y^(1/5)
    v' = (y^(-4/5)y')/5

    ==> 5v' - (15/x)*v = f(x)/x
    ==> v' - (3/x)*v = f(x)/(5x) [Linear Eqn]

    Find integrating factor:
    mu = e^(I(-3/x)) = x^-3

    So I have:
    (v*x^-3)' = 9/(5x^2) - (x^-3) + 2/(5x^4)

    Integrate:
    v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c
    v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)
    y^(1/5) = g(x)
    y = g(x) ^ 5

    Now to solve the IVP I simply y = 0 and x = 1[

    Leaves me with:
    0 = -9/5 + 1/2 -2/15 + C
    C = -43/30

    y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

    Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by BigC View Post
    I've been working on this one differential equation for an hour and a half now and I can't for the life of me figure out where I'm going wrong: xy' -15 y = (9 x^2 -5 x + 2) y^{\frac{4}{5}},\qquad y(1)=0

    First I divide by x. Then by  y^{\frac{4}{5}}.

    This leaves me with:
    f(x) = (9x^ - 5x + 2)

    y^(-4/5)y' - (15/x)y^(1/5) = f(x)/x

    I substitute:
    v = y^(1/5)
    v' = (y^(-4/5)y')/5

    ==> 5v' - (15/x)*v = f(x)/x
    ==> v' - (3/x)*v = f(x)/(5x) [Linear Eqn]

    Find integrating factor:
    mu = e^(I(-3/x)) = x^-3

    So I have:
    (v*x^-3)' = 9/(5x^2) - (x^-3) + 2/(5x^4)

    Integrate:
    v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c
    v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)
    y^(1/5) = g(x)
    y = g(x) ^ 5

    Now to solve the IVP I simply y = 0 and x = 1[

    Leaves me with:
    0 = -9/5 + 1/2 -2/15 + C
    C = -43/30

    y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

    Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.
    Divide through by x to get \frac{\,dy}{\,dx}-\frac{15}{x}y=\left(9x-5+\frac{2}{x}\right)y^{\frac{4}{5}}

    Now apply the substitution z=y^{1-\frac{4}{5}}=y^{\frac{1}{5}}\implies z^{5}=y

    Thus, \frac{\,dy}{\,dx}=\frac{\,dy}{\,dz}\cdot\frac{\,dz  }{\,dx}=5z^{4}\frac{\,dz}{\,dx}

    Thus, the DE becomes:

    5z^{4}\frac{\,dz}{\,dx}-\frac{15}{x}z^{5}=\left(9x-5+\frac{2}{x}\right)z^{4}\implies\frac{\,dz}{\,dx}-\frac{3}{x}z=\left(\tfrac{9}{5}x-1+\tfrac{2}{5}x^{-1}\right)

    Now we see its linear.

    The integrating factor is \varrho\!\left(x\right)=x^{-3}

    Thus the DE becomes \frac{\,d}{\,dx}\left[x^{-3}z\right]=\left(\tfrac{9}{5}x^{-2}-x^{-3}+\tfrac{2}{5}x^{-4}\right)

    Thus, we see that x^{-3}z=-\tfrac{9}{5}x^{-1}+\tfrac{1}{2}x^{-2}-\tfrac{2}{15}x^{-3}+C\implies z=-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+C

    Now apply the initial condition to find that 0=-\tfrac{9}{5}+\tfrac{1}{2}-\tfrac{2}{15}+C\implies C=\tfrac{43}{30}

    Thus, z=-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+\tfrac{43}{30}

    Now, since z=y^{\frac{1}{5}} we see that the Solution is y^{\frac{1}{5}}=-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+\tfrac{43}{30} \implies \color{red}\boxed{y=\left(-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+\tfrac{43}{30}\right)^5}

    Does this make sense?
    Last edited by Chris L T521; January 30th 2009 at 08:31 PM.
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  3. #3
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    Your solution is virtually identical to mine.

    I don't understand how the final solution is ^3 and not to ^5
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by BigC View Post
    Your solution is virtually identical to mine.

    I don't understand how the final solution is ^3 and not to ^5
    Thanks for catching that. I did mean ^5, not ^3

    Note that there is a difference in our solutions!!! I don't have \frac{43}{30} multiplied to x^{3}!!
    Last edited by Chris L T521; January 30th 2009 at 08:33 PM.
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  5. #5
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    Quote Originally Posted by BigC View Post
    Integrate:
    v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c
    v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)
    y^(1/5) = g(x)
    y = g(x) ^ 5

    Now to solve the IVP I simply y = 0 and x = 1[

    Leaves me with:
    0 = -9/5 + 1/2 -2/15 + C
    C = -43/30
    I think your problem is here. c = +43/30

    Quote Originally Posted by BigC View Post
    y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

    Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.
    Try

    y = \left(\frac{43}{30}x^3 - \frac{9}{5}x^2 + \frac{1}{2}x - \frac{2}{15}\right)^5
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