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Math Help - [SOLVED] Stumped on this DE

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Stumped on this DE

    Solve.

    \int 7x cos(x) = \int (2y + e^{3y}) y', y(0) = 0

    I'm pretty sure my work is right, but as usual Webassign does not care for my answer.
    <br />
-7 cos(x) - 7x sin(x) + C = y^2 + \frac{e^{3y}}{3}

    *Now, webassign is having me give the answer based on what and how they provide HALF of the answer. So, here is what WA has on the left:

    e^{3y} + 3y^2 = __ my answer________

    So, before I solved for C, I went ahead and multiplied through by the 3 so that I could just work it through in the form it has to be in anyway:

    3y^2 + e^{3y} = 21 (-xsin(x) - cos(x)) + 3C

    yadda yadda C =   \frac{22}{3}

    So, the right side answer that I am getting is:

    -21x sin(x) - 21 cos(x) + \frac{22}{3}

    Anyone? Anyone? Thanks!
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  2. #2
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    \int 7x\cos{x} \, dx = 7(\cos{x} + x\sin{x}) + C

    \int 2y + e^{3y} \, dy = y^2 + \frac{1}{3}e^{3y} + C

    7(\cos{x} + x\sin{x}) + C = y^2 + \frac{1}{3}e^{3y}

    21(\cos{x} + x\sin{x}) + C = 3y^2 + e^{3y}

    y(0) = 0

    21(1) + C = 1

    C = -20

    21(\cos{x} + x\sin{x}) - 20 = 3y^2 + e^{3y}
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by skeeter View Post
    \int 7x\cos{x} \, dx = 7(\cos{x} + x\sin{x}) + C

    \int 2y + e^{3y} \, dy = y^2 + \frac{1}{3}e^{3y} + C

    7(\cos{x} + x\sin{x}) + C = y^2 + \frac{1}{3}e^{3y}

    21(\cos{x} + x\sin{x}) + C = 3y^2 + e^{3y}

    y(0) = 0

    21(1) + C = 1

    C = -20

    21(\cos{x} + x\sin{x}) - 20 = 3y^2 + e^{3y}

    So you don't make C, 3C when you multiply through with that 3?
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    I'm pretty sure my work is right, but as usual Webassign does not care for my answer.
    <br />
-7 cos(x) - 7x sin(x) + C = y^2 + \frac{e^{3y}}{3}
    Not quite.

    \int7x\cos x\,dx

    =\int7x(\sin x)'\,dx

    =7x\sin x-7\int\sin x\,dx

    =7x\sin x+7\cos x+C_0

    =7(x\sin x+\cos x)+C_0.

    3y^2 + e^{3y} = 21 (-xsin(x) - cos(x)) + 3C
    So we should have

    7(x\sin x+\cos x)+C_0=y^2+\frac{e^{3y}}3

    Multiplying by 3, you should get

    21(x\sin x+\cos x)+C=3y^2+e^{3y} (where C=3C_0)

    which gives C=-20.

    So, the right side answer that I am getting is:

    -21x sin(x) - 21 cos(x) + \frac{22}{3}
    Which, after fixing the mistakes, should be

    21(x\sin x+\cos x)-20.


    Edit:
    So you don't make C, 3C when you multiply through with that 3?
    Technically, yes. Skeeter was being a little informal with his notation, but his method is correct.
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  5. #5
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    Quote Originally Posted by mollymcf2009 View Post
    So you don't make C, 3C when you multiply through with that 3?
    no ... 3 times a Constant is just another Constant.
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  6. #6
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    Quote Originally Posted by mollymcf2009 View Post
    So you don't make C, 3C when you multiply through with that 3?
    C is an unknown constant so 3C is just another unknown constant.
    Skeeter just used the same symbol to represent the two different constants.
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  7. #7
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    Quote Originally Posted by Reckoner View Post
    ... Skeeter was being a little informal with his notation, but his method is correct.
    I'll wear my tux next time ...
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  8. #8
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    Quote Originally Posted by skeeter View Post
    I'll wear my tux next time ...
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  9. #9
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by skeeter View Post
    I'll wear my tux next time ...
    Y'all crack me up!
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