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Math Help - 2nd O.D.E.

  1. #1
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    2nd O.D.E.

    Hey all

    How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is \frac{d^2y}{dx^2} + 2x(\frac{dy}{dx})^2 = 0

    *_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*


    Thanks in advance
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  2. #2
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    Quote Originally Posted by tsal15 View Post
    Hey all

    How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is \frac{d^2y}{dx^2} + 2x(\frac{dy}{dx})^2 = 0

    *_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*


    Thanks in advance
    Let  u = \frac{dy}{dx}

    And hence  \frac{du}{dx} = \frac{d^2y}{dx^2}

    You can then rearrange to get:

     \frac{du}{u^2} = -2xdx

    Which is solveable by some fairly simple integration , don't forget to sub back into y.

    Explicit just means you write it in the form  y(x) = blah blah

    This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.
    Last edited by Mush; December 21st 2008 at 04:43 AM.
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  3. #3
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    Quote Originally Posted by Mush View Post
    Let  u = \frac{dy}{dx}

    And hence  \frac{du}{dx} = \frac{d^2y}{dx^2}

    You can then rearrange to get:

     \frac{du}{u^2} = -2xdx

    Which is solveable by some fairly simple integration , don't forget to sub back into y.

    Explicit just means you write it in the form  y(x) = blah blah

    This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.
    Great thanks, Mush.

    I'll have a go at it again, and I'll get back to you

    Thanks again
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  4. #4
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    Quote Originally Posted by Mush View Post
    Let  u = \frac{dy}{dx}

    And hence  \frac{du}{dx} = \frac{d^2y}{dx^2}

    You can then rearrange to get:

     \frac{du}{u^2} = -2xdx

    Which is solveable by some fairly simple integration , don't forget to sub back into y.

    Explicit just means you write it in the form  y(x) = blah blah

    This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.

    Hey, Mush.

    Apparently I'm supposed to find the general equation first. How would I go about doing this?
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  5. #5
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    Quote Originally Posted by tsal15 View Post
    Hey all

    How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is \frac{d^2y}{dx^2} + 2x(\frac{dy}{dx})^2 = 0

    *_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*


    Thanks in advance
    If you let  y' = u , the ODE separates giving

     \frac{-u'}{u^2} = x .

    Integrating, applying the IC and integrating again with the second IC gives the solution

     y = tan^{-1} x .
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  6. #6
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    Quote Originally Posted by danny arrigo View Post
    If you let  y' = u , the ODE separates giving

     \frac{-u'}{u^2} = x .

    Integrating, applying the IC and integrating again with the second IC gives the solution

     y = tan^{-1} x .
    I presume that IC = Initial Condition ?

    Also whats the general solution? - is it \frac{u'}{u^2} = x , if not, how do find the general solution ?

    Thanks Danny,

    tsal15
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  7. #7
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    Quote Originally Posted by tsal15 View Post
    I presume that IC = Initial Condition ?

    Also whats the general solution? - is it \frac{u'}{u^2} = x , if not, how do find the general solution ?

    Thanks Danny,

    tsal15
    Yes, IC is the initial condition. Let me provide some more details

     -\frac{u'}{u^2} = 2x \;\;\; \Rightarrow\;\;\;-\int \frac{u'}{u^2} = \int 2x +c_1

    so

     \frac{1}{u} = x^2+c_1 \;\;\; \Rightarrow\;\;(with \;y'(0)\;=\;u(0)=1)\;\;\frac{1}{u} = x^2+1

    so that

     u = y' = \frac{1}{x^2+1}

    Integrating gives

     y = tan^{-1} x + c_2

    and with the IC  y(0) = 0 gives our result.
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  8. #8
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    Quote Originally Posted by tsal15 View Post
    Hey, Mush.

    Apparently I'm supposed to find the general equation first. How would I go about doing this?
    You would surely use the given boundary conditions as you go along.

    It's ridiculous to carry an arbitrary constant into the second integration when it's possible to get its value. Not the least reason being that the technique required for doing the second integration will depend on what the value of that first arbitrary constant is ....
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  9. #9
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    Finding the general solution of the 2ODE, was the process we have been discussing so far?
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