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Math Help - Last Diff Equation for today

  1. #1
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    Last Diff Equation for today

    I am asked to find the general solution, using the method of separating the variables of the following DE

    dy/dx = (2xcos x)/y where y > 0

    Using the method of seperating the variables I get:

    int y.dy = int 2xcosx dx

    which basically comes to (after a little working)

    = 2 (cos x + xsinx)

    but where do I go from here to get an integration constant?

    2) If y = 2 when x = 0, find y in terms of x

    Could someone help me on this one

    3) Explain why your answer may not be used for x=pi. Comment in relation to the solution curve through (0,2).
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  2. #2
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    Quote Originally Posted by Natasha1
    I am asked to find the general solution, using the method of separating the variables of the following DE

    dy/dx = (2xcos x)/y where y > 0

    Using the method of seperating the variables I get:

    int y.dy = int 2xcosx dx

    which basically comes to (after a little working)

    = 2 (cos x + xsinx)

    but where do I go from here to get a integration constant?

    2) If y = 2 when x = 0, find y in terms of x

    Could someone help me on this one

    3) Explain why your answer may not be used for x=pi. Comment in relation to the solution curve through (0,2).
    Thus,
    \frac{1}{2}y^2=2(\cos x+x\sin x)+C
    But (x,y)=(0,2)
    Thus,
    2=2(1+0)+C
    Thus,
    C=0
    Which gives,
    \frac{1}{2}y^2=2\cos x+2x\sin x
    Thus,
    y=2\sqrt{ \cos x+x\sin x}

    You cannot use x=\pi because you get a negative in the radical, which brings non-real numbers.
    Last edited by ThePerfectHacker; April 30th 2006 at 11:48 AM.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    \frac{1}{2}y^2=2(\cos x+x\sin x)+C
    But (x,y)=(0,2)
    Thus,
    2=2(1+0)+C
    Thus,
    C=0
    Which gives,
    \frac{1}{2}y^2=2\cos x+2x\sin x
    Thus,
    y=2\sqrt{ \cos x+x\sin x}

    You cannot use x=\pi because you get a negative in the radical, which brings non-real numbers.
    What about the relation to the solution curve through (0,2)?
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  4. #4
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    Quote Originally Posted by Natasha1
    What about the relation to the solution curve through (0,2)?
    PH found the general soltion, then substituted the initial condition
    into the general solution to obtain his final solution.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    PH found the general soltion, then substituted the initial condition
    into the general solution to obtain his final solution.

    RonL

    If the general solution is y = 2 SQU ROOT OF cosx + xsinsx

    and subsitution the initial condition y = 2 and x = 0

    we get

    2 = 2 SQU ROOT OF cos 0 + 0sin0
    2=2

    ?
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  6. #6
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    Quote Originally Posted by Natasha1
    If the general solution is y = 2 SQU ROOT OF cosx + xsinsx

    and subsitution the initial condition y = 2 and x = 0

    we get

    2 = 2 SQU ROOT OF cos 0 + 0sin0
    2=2

    ?
    Without checking in detail, my reading of PH's posting is that the general
    solution is:

    <br />
\frac{1}{2}y^2=2(\cos x+x\sin x)+C<br />
,

    into which he substitutes y=2 when x=0, to get

    <br />
\frac{1}{2}y^2=2\cos x+2x\sin x<br />

    or:

    <br />
y=2\sqrt{ \cos x+x\sin x}<br />

    RonL
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