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Math Help - Differential Equation Problem

  1. #1
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    Differential Equation Problem

    Find the solution of the differential equation that satisfies the given initial condition. xy' + y = y^2 , y(1) = -2

    I've been attempting to solve it by separable equations, by making the equation xy'=y(y-1), but my answer has not been working correctly.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MelanieC View Post
    Find the solution of the differential equation that satisfies the given initial condition. xy' + y = y^2 , y(1) = -2

    I've been attempting to solve it by separable equations, by making the equation xy'=y(y-1), but my answer has not been working correctly.
    ...you're not done. you should get it to \frac 1{y(y - 1)}~dy = \frac 1x~dx

    now integrate both sides and continue
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  3. #3
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    I've done thiat already, and by using partial fractions, I come up with:
    ln(y-1/y) =lnx +c
    y-1/y= x + C2
    1 - 1/y = x + C2
    -1/y = x + C3
    y = 1/(-x + C3)

    Using my initial condition, I get
    -2 = 1/ (-1 + C3)
    Therefore, C3 = 1/2

    So my solution should be

    y= 1/(-x + 1/2)

    However, when I tried to check my answer by substituting it back into the original given differential equation, my answer did not work out. Can anyone help me?
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  4. #4
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    Quote Originally Posted by MelanieC View Post
    I've done thiat already, and by using partial fractions, I come up with:
    ln(y-1/y) =lnx +c
    y-1/y= x + C2
    No. Taking the exponential of both sides gives (y-1)/y= C2 x where C2= e^c

    1 - 1/y = x + C2
    -1/y = x + C3
    y = 1/(-x + C3)

    Using my initial condition, I get
    -2 = 1/ (-1 + C3)
    Therefore, C3 = 1/2

    So my solution should be

    y= 1/(-x + 1/2)

    However, when I tried to check my answer by substituting it back into the original given differential equation, my answer did not work out. Can anyone help me?
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  5. #5
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    I know that C2 = e^c, but since e^c is a constant, i just renamed it to be C2, to make it easier to work with
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  6. #6
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    Quote Originally Posted by MelanieC View Post
    I've done thiat already, and by using partial fractions, I come up with:
    ln(y-1/y) =lnx +c
    y-1/y= x + C2 ... this step is incorrect.
    \frac{y-1}{y} = e^{\ln{x}+C} = C_2 \cdot x , not x + C_2
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