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Math Help - dy/dx=x^2+y^(1/2), y(0)=1

  1. #1
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    dy/dx=x^2+y^(1/2), y(0)=1

    http://www.wolframalpha.com/input/?i=dy%2Fdx%3Dx^2*%28y^%281%2F2%29%29%2Cy%280%29%3D 1&asynchronous=false&equal=Submit

    Alright, so I have wolfram alpha out and am trying to figure this problem out. It's one I missed on my review.

    The problem: dy/dx=x^2*y^(1/2)

    My answer y=x^6/36+c^2, C=1

    My review sheet's answer: 2*y*(1/2) = 1/3 * x^3 + 2.

    I'm in the center of it quite confused and would like to see the problem worked step by step to the review sheet answer, and beyond if any of you are feeling really nice. Thanks!
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  2. #2
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    pickslides's Avatar
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    The two answers look very similar. Are they the same thing?

    This equation looks like it separates, show us your working.
    Last edited by pickslides; March 30th 2011 at 01:42 PM. Reason: Bad Grammar.
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Wolvenmoon View Post
    http://www.wolframalpha.com/input/?i=dy%2Fdx%3Dx^2*%28y^%281%2F2%29%29%2Cy%280%29%3D 1&asynchronous=false&equal=Submit

    Alright, so I have wolfram alpha out and am trying to figure this problem out. It's one I missed on my review.

    The problem: dy/dx=x^2*y^(1/2)

    My answer y=x^6/36+c^2, C=1

    My review sheet's answer: 2*y*(1/2) = 1/3 * x^3 + 2.

    I'm in the center of it quite confused and would like to see the problem worked step by step to the review sheet answer, and beyond if any of you are feeling really nice. Thanks!
    First note that the problem in the title is different than the one in your post.

    The one in the post is

    \displaystyle \frac{dy}{dx}=x^2\sqrt{y}

    This equation is separable

    \displaystyle \frac{dy}{\sqrt{y}}=x^2dx \iff 2\sqrt{y}=\frac{1}{3}x^3+c

    Now using the initial condition

    \displaystyle 2\sqrt{1}=\frac{1}{3}(0)^3+c \implies c=2

    So the solution is

    \displaystyle 2\sqrt{y}=\frac{1}{3}x^3+2
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