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Math Help - second-order and no independed variable

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation second-order differential equations without independent variable

    Dear friends,

    I have the following question.

    Question. What is the general solution of the differential equation yy^{\prime\prime}-2(y^{\prime})^{2}=y^{2}\ln(y)?

    I have a solution but it involves an indefinite integral, I wonder if one will be able to give an explicit result.

    Thanks.

    bkarpuz
    Last edited by bkarpuz; January 17th 2010 at 03:44 AM. Reason: Corrected the title
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  2. #2
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    Quote Originally Posted by bkarpuz View Post
    Dear friends,

    I have the following question.

    Question. What is the general solution of the differential equation yy^{\prime\prime}-2(y^{\prime})^{2}=y^{2}\ln(y)?

    I have a solution but it involves an indefinite integral, I wonder if one will be able to give an explicit result.

    Thanks.

    bkarpuz
    I remember in my very first homework assignment of my first differential equations course, getting the solution to an integral and then struggling mightely to integrate. When I finally gave up and looked in the back of the book, I found that the answer was given in terms of that integral!

    Are you sure you are required to give the answer in "closed form" and not in terms of an integral?

    In any case, what is the integral you are having trouble with?
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  3. #3
    Senior Member bkarpuz's Avatar
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    Question My solution.

    Quote Originally Posted by bkarpuz View Post
    Dear friends,

    I have the following question.

    Question. What is the general solution of the differential equation yy^{\prime\prime}-2(y^{\prime})^{2}=y^{2}\ln(y)?

    I have a solution but it involves an indefinite integral, I wonder if one will be able to give an explicit result.

    Thanks.

    bkarpuz
    Here is my solution.

    Solution. Let p:=\frac{\mathrm{d}y}{\mathrm{d}x<br />
}, then by the chain rule we have \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}<br />
}=p\frac{\mathrm{d}p}{\mathrm{d}y}.
    Hence the given differential equation reduces to
    yp\mathrm{d}p-\big(2p^{2}+y^{2}\ln(y)\big)\mathrm{d}y=0.........(1)
    Clearly, this equation is not exact since \frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}\neq0, where U(p,y):=yp and V(p,y):=-\big(2p^{2}+y^{2}\ln(y)\big).
    The integrating factor for this equation is
    \mu(y):=\exp\bigg\{-\int\frac{\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}}{U}\mathrm{d}y\bigg\}=\exp\bigg\{-5\int\frac{1}{y}\mathrm{d}y\bigg\}=\frac{1}{y^{5}}  .
    Grouping the terms after multiplying through (1) by \mu, we get
    \frac{1}{2}\bigg(\frac{2py^{4}\mathrm{d}p-4p^{2}y^{3}\mathrm{d}y}{\big(y^{4}\big)^{2}}\bigg)  +\frac{\ln(y)}{y^{3}}\mathrm{d}y=0
    or equivalently
    \frac{1}{2}\mathrm{d}\bigg(\frac{p^{2}}{y^{4}}\big  g)+\frac{\ln(y)}{y^{3}}\mathrm{d}y=0.
    From this, we are lead to
    p=\pm\sqrt{c_{1}+\frac{y^{2}}{2}\big(1+2\ln(y)\big  )},
    which gives us
    x=\pm\int\frac{1}{\sqrt{c_{1}+\frac{y^{2}}{2}\big(  1+2\ln(y)\big)}}\mathrm{d}y+c_{2}
    by using the relation \mathrm{d}x=p\mathrm{d}y.
    This is the desired solution to (1)..... \rule{0.3cm}{0.3cm}

    It seems not possible to solve the last integral.
    I think there may be other solutions for this equation which may give explicit result.
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Here is my solution.

    Solution. Let p:=\frac{\mathrm{d}y}{\mathrm{d}x<br />
}, then by the chain rule we have \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}<br />
}=p\frac{\mathrm{d}p}{\mathrm{d}y}.
    Hence the given differential equation reduces to
    yp\mathrm{d}p-\big(2p^{2}+y^{2}\ln(y)\big)\mathrm{d}y=0.........(1)
    Clearly, this equation is not exact since \frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}\neq0, where U(p,y):=yp and V(p,y):=-\big(2p^{2}+y^{2}\ln(y)\big).
    The integrating factor for this equation is
    \mu(y):=\exp\bigg\{-\int\frac{\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}}{U}\mathrm{d}y\bigg\}=\exp\bigg\{-5\int\frac{1}{y}\mathrm{d}y\bigg\}=\frac{1}{y^{5}}  .
    Grouping the terms after multiplying through (1) by \mu, we get
    \frac{1}{2}\bigg(\frac{2py^{4}\mathrm{d}p-4p^{2}y^{3}\mathrm{d}y}{\big(y^{4}\big)^{2}}\bigg)  +\frac{\ln(y)}{y^{3}}\mathrm{d}y=0
    or equivalently
    \frac{1}{2}\mathrm{d}\bigg(\frac{p^{2}}{y^{4}}\big  g)+\frac{\ln(y)}{y^{3}}\mathrm{d}y=0.
    From this, we are lead to
    p=\pm\sqrt{c_{1}+\frac{y^{2}}{2}\big(1+2\ln(y)\big  )},
    which gives us
    x=\pm\int\frac{1}{\sqrt{c_{1}+\frac{y^{2}}{2}\big(  1+2\ln(y)\big)}}\mathrm{d}y+c_{2}
    by using the relation \mathrm{d}x=p\mathrm{d}y.
    This is the desired solution to (1)..... \rule{0.3cm}{0.3cm}

    It seems not possible to solve the last integral.
    I think there may be other solutions for this equation which may give explicit result.
    I think you should get this

    x = \pm \int\!\frac{dy}{\sqrt{C_1 y^4 - \dfrac{y^2}{2}(1+2\ln y)}}.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by DeMath View Post
    I think you should get this

    x = \pm \int\!\frac{dy}{\sqrt{C_1 y^4 - \dfrac{y^2}{2}(1+2\ln y)}}.
    Yes, it seems I had mistakes.
    But this integral looks so bad and not possible to integrate up to me...
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