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Math Help - Differential Equations questions

  1. #1
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    Differential Equations questions

    (a) Verify that all members of the family y = (5)^{1/2} (c - x^2)^{-1/2} are solutions of the differential equation.

    (b) Find a solution of the initial-value problem.

     <br />
\frac {dy}{dx} = \frac {xy^3}{5} , y(0) = 4<br />

    I do not know how to solve a. However, This is how I solved b. Please give me a hint to solve a and check my answer for b.

    (b)

    \frac {dy}{dx} = \frac {xy^3}{5}

    Move the ys to one side:

    \frac {dy}{y^3} = \frac{xdx}{5}

    Integrate:

    \int \frac {dy}{y^3} = \int \frac{xdx}{5}

    \frac {-2}{y^2} = \frac {x^2}{10} + C

    y = \sqrt{\frac{-5}{x^2}} + C

    I dont think my answer is correct.

    Edit: Added the C.
    Last edited by mr fantastic; June 26th 2009 at 05:07 AM. Reason: Fixed powers in latex. Use ^{}, NOT ^(). And replaced x2 with x^2.
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  2. #2
    MHF Contributor chisigma's Avatar
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    It's all right till to...

    \int \frac{dy}{y^{3}} = \int \frac{x}{5} \cdot dx

    Now you have to remember the integration rule of...

     \int y^{n}\cdot dy

    ... and the fact that an indefinite integral contains an arbitrary constant usually indicated with c...

    Kind regards

    \chi \sigma
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  3. #3
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    Yes, I forgot the C. However, to integrate: y^n isn't it a simple:

     <br />
\frac{y^{n+1}}{n+1}<br />


    PS. How do you raise y to a n+1 with latex?
    Last edited by calc101; June 25th 2009 at 06:48 AM.
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  4. #4
    Newbie vinay's Avatar
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    final answer

    so finding c using y = 4 at x = 0
    finally we get
    y(x) = sqrt ( 80 / (5-4x^2) )
    Last edited by vinay; June 25th 2009 at 11:01 PM. Reason: correcting the answer
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  5. #5
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    Quote Originally Posted by calc101 View Post
    Yes, I forgot the C. However, to integrate: y^n isn't it a simple:

     <br />
\frac{y^ny^1}{n+1}<br />


    PS. How do you raise y to a n+1 with latex?
    Use { } around the n+1:
    \frac{y^{n+1}}{n+1}
    Click on that to see the code.
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  6. #6
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    Quote Originally Posted by calc101 View Post
    (a) Verify that all members of the family y = (5)^(1/2) (c - x2)^(-1/2) are solutions of the differential equation.
    A solution to what differential equation? To the equation in (b)?
    If y= \sqrt{5}(c- x^2)^{-\frac{1}{2}} then \frac{dy}{dx}= \sqrt{5}(-\frac{1}{2})(c- x^2)^{-3/2}(-2x) while y^3= 5\sqrt{5}(c- x^2)^{-\frac{3}{2}} so \frac{xy^3}{5}= \sqrt{5}x(c-x^2)^{-\frac{3}{2}}.

    (b) Find a solution of the initial-value problem.

     <br />
\frac {dy}{dx} = \frac {xy^3}{5} , y(0) = 4<br />

    I do not know how to solve a. However, This is how I solved b. Please give me a hint to solve a and check my answer for b.

    (b)

    \frac {dy}{dx} = \frac {xy^3}{5}

    Move the ys to one side:

    \frac {dy}{y^3} = \frac{xdx}{5}

    Integrate:

    \int \frac {dy}{y^3} = \int \frac{xdx}{5}

    \frac {-2}{y^2} = \frac {x^2}{10} + C

    y = \sqrt{\frac{-5}{x^2}}

    I dont think my answer is correct.

    Edit: Added the C.
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  7. #7
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    Quote Originally Posted by vinay View Post
    so finding c using y = 4 at x = 0
    finally we get
    y(x) = 20 / (5-x^2)

    Can you please show your work?

    Thanks,
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  8. #8
    Newbie vinay's Avatar
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    Move the ys to one side:



    Integrate:



    --------------------- eq 1

    so put x = 0, and y(0) = 4

    which gives you value of c

    c = -2/4^2 = -1/8

    now again subsitute c in main eq 1

    -2/y^2 = x^2/10 -1/8

    which will give the final answer as,

    y(x) = sqrt ( 80 / (5-4x^2) )


    -
    Last edited by vinay; June 25th 2009 at 11:02 PM. Reason: correcting the albebra
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  9. #9
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    Thanks Vinay

    y(x) = \sqrt { \frac {80} {5-4x^2} }
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