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Math Help - differentiation

  1. #1
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    differentiation

    1, find s in terms of t if ds/dt=3t-8/tsquared and s=1and1/2 when t= 1

    1, square root of (2-3x)dx

    2, given that dy/dx=3/x-2 and y =10 when x=0, find an expression for y in terms of x

    1, find( 2e^3x-e^-x) dx
    Thanks!
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    Quote Originally Posted by afan17 View Post
    1, find s in terms of t if ds/dt=3t-8/tsquared and s=1and1/2 when t= 1

    1, square root of (2-3x)dx

    2, given that dy/dx=3/x-2 and y =10 when x=0, find an expression for y in terms of x

    1, find( 2e^3x-e^-x) dx
    Thanks!
    These questions are written badly.

    The first one is, I presume, to solve
    \frac{ds}{dt} = 3t - \frac{8}{t^2}, s(1) = \frac{3}{2}

    This equation is separable:
    ds = \left ( 3t - \frac{8}{t^2} \right )~dt

    Upon integrating
    s = \frac{3}{2}t^2 + \frac{8}{t} + C

    We know that s(1) = 3/2, so
    \frac{3}{2} = \frac{3}{2} \cdot 1^2 + \frac{8}{1} + C

    Solve for C.
    ============================================

    For the second one I presume you are trying to integrate?
    \int \sqrt{2 - 3x}~dx

    Let y = 2 - 3x \implies dy = -3~dx

    Then your integral becomes
    \int \sqrt{y}~\frac{dy}{-3}
    =============================================

    For the third I presume this is
    \frac{dy}{dx} = \frac{3}{x - 2}, y(0) = 10
    (Please use parenthesis!)

    This is, again, a separable equation:
    dy = \frac{3}{x - 2}~dx

    Integrate both sides by using the substitution u = x - 2.
    =============================================

    I presume the last one is an integration as well and is
    \int (2e^{3x}-e^{-x})~ dx
    (Again, please use parenthesis!)

    Split this into two parts:
    \int 2e^{3x}~dx - \int e^{-x}~dx

    For the first integral let y = 3x and the second let u = -x.

    -Dan
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