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Math Help - Separable Differential Equations-Please help!

  1. #1
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    Separable Differential Equations-Please help!

    I've got a few of these separable differential questions to ask. The thing I'm most confused about is what to do with the constant when you are given what y and x equal.

    Solve each of the initial value problems-

    1) e^x.y.(dy/dx) = 1 y = 0 when x = 0

    2) xy.(dy/dx) = 1 + y^2 y = 0 when x = 2

    Find the particular solution-

    3) dy/dx = (1/3) + y^2 y = 1/2 when x = pi/4

    4) dh/dt = 2h(3-h) where h = 1 when t = 0

    Also these ones.

    Solve:

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

    Sorry about all the questions, but im really confused with these
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  2. #2
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    Quote Originally Posted by hats_06 View Post
    I've got a few of these separable differential questions to ask. The thing I'm most confused about is what to do with the constant when you are given what y and x equal.

    Solve each of the initial value problems-

    1) e^x.y.(dy/dx) = 1 y = 0 when x = 0

    2) xy.(dy/dx) = 1 + y^2 y = 0 when x = 2

    Find the particular solution-

    3) dy/dx = (1/3) + y^2 y = 1/2 when x = pi/4

    4) dh/dt = 2h(3-h) where h = 1 when t = 0

    Also these ones.

    Solve:

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

    Sorry about all the questions, but im really confused with these
    How much experience do you actually have in solving differential equations? I'll do one.

    2) xy.(dy/dx) = 1 + y^2 y = 0 when x = 2.

    \frac{dy}{dx} = \frac{1 + y^2}{xy} \Rightarrow \frac{y}{1 + y^2} \, dy = \frac{1}{x} \, dx

    \Rightarrow \int \frac{y}{1 + y^2} \, dy = \int \frac{1}{x} \, dx

    \Rightarrow \frac{1}{2} \ln | 1 + y^2 | = \ln |x | + C.

    Substitute y = 0 when x = 2: \frac{1}{2} \ln | 1 | = \ln | 2 | + C \Rightarrow C = - \ln 2.

    Therefore:

    \frac{1}{2} \ln | 1 + y^2 | = \ln |x | - \ln 2

    \Rightarrow \ln \sqrt{1 + y^2} = \ln \left| \frac{x}{2} \right|

    \Rightarrow \sqrt{1 + y^2} = \frac{x}{2} .

    Therefore y = .....
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    Thanks for that!
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