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Math Help - Solve non-linear differential analytically

  1. #1
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    Solve non-linear differential analytically

    Hi,

    I am trying to solve the following equation analytically. I think the solution shouldnt be that hard but I'm really rusty on these kind of things.

    V'' - k*J*V^-1/2 = 0

    I have then said u = V' and therefore u' = u(du/dV)

    u(du/dV) - k*J*V^-1/2 = 0

    now you can use separation of variables but assuming I did it correctly you get a u in the answer and since u = V' you have to start all over.

    The goal of this is to prove J proportional to V^3/2

    Any help on how to do this is appreciated, got to turn this in tomorrow!
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  2. #2
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    Re: Solve non-linear differential analytically

    Quote Originally Posted by pdizzle0 View Post
    Hi,

    I am trying to solve the following equation analytically. I think the solution shouldnt be that hard but I'm really rusty on these kind of things.

    V'' - k*J*V^-1/2 = 0

    I have then said u = V' and therefore u' = u(du/dV)

    u(du/dV) - k*J*V^-1/2 = 0

    now you can use separation of variables but assuming I did it correctly you get a u in the answer and since u = V' you have to start all over.

    The goal of this is to prove J proportional to V^3/2

    Any help on how to do this is appreciated, got to turn this in tomorrow!
    Are you using \displaystyle V'' to represent \displaystyle \frac{d^2V}{dJ^2}? In other words, is J your independent variable?
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  3. #3
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    Re: Solve non-linear differential analytically

    No, sorry for not making that clear. J is constant. k*J essentially could be considered 1 constant in this context just for the final answer J proportional to V^3/2

    The independent variable in this case is x -> differentiating with respect to space
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Solve non-linear differential analytically

    Quote Originally Posted by pdizzle0 View Post
    Hi,

    I am trying to solve the following equation analytically. I think the solution shouldnt be that hard but I'm really rusty on these kind of things.

    V'' - k*J*V^-1/2 = 0

    I have then said u = V' and therefore u' = u(du/dV)

    u(du/dV) - k*J*V^-1/2 = 0

    now you can use separation of variables but assuming I did it correctly you get a u in the answer and since u = V' you have to start all over.

    The goal of this is to prove J proportional to V^3/2

    Any help on how to do this is appreciated, got to turn this in tomorrow!
    For semplicity we write the DE as...

    y^{''}= \frac{a}{\sqrt{y}} (1)

    ... where a is a constant. Setting y^{'}=u we have...

    y^{''}= \frac{d u}{d x}= \frac{d u}{d y}\ \frac{dy}{dx}= u\ \frac{d u}{dy} (2)

    ... so that the (1) becomes...

    u\ \frac{du}{dy}= \frac{a}{\sqrt{y}} (3)

    In (3) the variables are separable so that the solution is easily found...

    u^{2}= 4\ a\ \sqrt{y} + c_{1} \implies y^{'}= \pm \sqrt{4\ a\ \sqrt{y} + c_{1}} (4)

    The (4) is a firdt order DE the solution of which leads us to y... are You able to proceed?...

    Kind regards

    \chi \sigma
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