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Math Help - [SOLVED] Differentiation

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Differentiation

    Solve the initial value problem y′= \frac{x^2}{2y + 1}, y(0)=−1 by separation of variables.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    Solve the initial value problem y′= \frac{x^2}{2y + 1}, y(0)=−1 by separation of variables.
    separate the variables

    (2y + 1) y' = x^2

    \Rightarrow \int (2y + 1)~dy = \int x^2 ~dx

    now finish up
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    Member ronaldo_07's Avatar
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    2yx+x= \frac{x^3}{3} Then rearange to make y=....... then let y{0)=-1 to find the solution right?
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    Quote Originally Posted by ronaldo_07 View Post
    2yx+x= \frac{x^3}{3} Then rearange to make y=....... then let y{0)=-1 to find the solution right?
    ??

    you do realize that on the left side you are integrating with respect to y, right? and what about your +C? you'll need it
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    Member ronaldo_07's Avatar
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    With respect to y, would that be y^2 + y
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    Quote Originally Posted by ronaldo_07 View Post
    With respect to y, would that be y^2 + y
    yes
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  7. #7
    Member ronaldo_07's Avatar
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    y^2+y=\frac{x^3}{3}+c Im not sure how to get y on its own at this stage
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    Quote Originally Posted by ronaldo_07 View Post
    y^2+y=\frac{x^3}{3}+c Im not sure how to get y on its own at this stage
    you have a quadratic equation in y. it's hard to factor of course, but you can always complete the square, or use the quadratic formula (here, your constant term is -(x^3/3 + C))
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  9. #9
    Member ronaldo_07's Avatar
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    Im still not able to solve for c using quadratic formulae
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    Quote Originally Posted by ronaldo_07 View Post
    Im still not able to solve for c using quadratic formulae
    you are not to use the quadratic formula to solve for C, you use y(0) = -1 to do that. the quadratic formula is to solve for y
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  11. #11
    Member ronaldo_07's Avatar
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    c=1? do i need to solve for y to answer this question?
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    Quote Originally Posted by ronaldo_07 View Post
    c=1? do i need to solve for y to answer this question?
    actually, no, C is not 1

    you do not need to solve for y to find C, however, when a problem says "solve the differential equation", they usually expect you to end up with y = ... or if x is your output variable, x = ...
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    Member ronaldo_07's Avatar
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    can you help im not sure how to finish this
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    Quote Originally Posted by ronaldo_07 View Post
    can you help im not sure how to finish this
    y(0) = -1 means when x = 0, y = -1. with this you can find C

    the quadratic formula says the solution to a polynomial equation of the form ay^2 + by + c = 0 is y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}
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  15. #15
    Member ronaldo_07's Avatar
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    when I got c=1, I got it because I let -1=Left side (y^2+y). So I guess I should have done (-1)^2-(-1)= \frac{(0)}{3}^3+C. But this would give C to be 0
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