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Math Help - separation of variables (calculus)

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    separation of variables (calculus)

    ok, so how does one attempt this using separation of variables?:

    y' = the square root of ty

    so far, what i have done is set y' equal to t^(1/2) * y^(1/2) since the square root of two variables is just the square root of each individual variable.

    i'm not really sure what to do after that so i would be grateful for any help!
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    Quote Originally Posted by clockingly View Post
    ok, so how does one attempt this using separation of variables?:

    y' = the square root of ty

    so far, what i have done is set y' equal to t^(1/2) * y^(1/2) since the square root of two variables is just the square root of each individual variable.

    i'm not really sure what to do after that so i would be grateful for any help!
    you were right so far. Now just divide both sides to obtain

    y'/y^(1/2) = t^(1/2) or equivalently y^(/1/2) dy = t^(1/2) dt

    now integrate both sides

    2y^(1/2) = (2/3)t^(3/2) + C
    => y = [{(2/3)t^(3/2) + C}^2]/2
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    Quote Originally Posted by Jhevon View Post
    you were right so far. Now just divide both sides to obtain

    y'/y^(1/2) = t^(1/2) or equivalently y^(/1/2) dy = t^(1/2) dt

    now integrate both sides

    2y^(1/2) = (2/3)t^(3/2) + C
    => y = [{(2/3)t^(3/2) + C}^2]/2
    What about y=0
    (Do not worry many make that mistake when solving differencial equations. I do not even think professors mention this for some reason).
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    What about y=0
    (Do not worry many make that mistake when solving differencial equations. I do not even think professors mention this for some reason).
    Yeah, professors usually don't mention that. Perhaps the possibility of y=0 exists within the constant itself though
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    Quote Originally Posted by Jhevon View Post
    Yeah, professors usually don't mention that. Perhaps the possibility of y=0 exists within the constant itself though
    There is actually a stranger situation, that might occur. I addressed it on this site some time ago, when y=0 at some point on the interval. My belief is that y=0 either everywhere on the interval or y not = 0. Meaning y=0 at some points yes and some points no, cannot happen. I never been able to formally prove it, but my idea is that when a curve y not = 0 for some point, then around that neighborhood of points we have a solution (which is of the form you mentioned it) now if it is zero at some other point and you connect the non-trivial solution (of the form you mentioned) with a curve, we get a "rough" point, you know a sharp turn, and therefore it is non-differenciable at that point. But that cannot happen because by definition a solution to some differencial equation must be differenciable. This tells us a very useful fact (assuming it is true), only consider the case y=0 before you divide.
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