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Math Help - Dif EQ Prob

  1. #1
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    Dif EQ Prob

    Determine whether the points (0,1) and (2,1) are on the graph of the solution of:

    (t - 1) dy + 2y dt = 0, y(0) = 1.

    MY WORK:

    (t - 1) dy = -2y dt
    -1/(2y) dy = 1/(t - 1) dt

    Take integral of both sides:

    -ln(|y|)/2 = ln(|t - 1|) + C

    Plug in 0 for t and plug in 1 for y:

    -ln(|1|)/2 = ln(|0 - 1|) + C

    => 0 = 0 + C

    => C = 0

    Now I'm not sure where to go.

    Thanks for the help.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    Quote Originally Posted by Ideasman View Post
    Determine whether the points (0,1) and (2,1) are on the graph of the solution of:

    (t - 1) dy + 2y dt = 0, y(0) = 1.

    MY WORK:

    (t - 1) dy = -2y dt
    -1/(2y) dy = 1/(t - 1) dt

    Take integral of both sides:

    -ln(|y|)/2 = ln(|t - 1|) + C

    Plug in 0 for t and plug in 1 for y:

    -ln(|1|)/2 = ln(|0 - 1|) + C

    => 0 = 0 + C

    => C = 0

    Now I'm not sure where to go.

    Thanks for the help.
    C=0, so,
    -(1/2)ln|y| = ln|t-1|
    ln|y^(-1/2)| = ln|t-1|
    So,
    |y^(-1/2)| = |t -1|
    |1/(y^(1/2))| = |t-1|
    Square both sides,
    1/y = (t-1)^2 ----------(i)

    Test (0,1) on (i),
    1/1 =? (0-1)^2
    1 =? 1
    Yes, so, (0,1) is on the graph of (i).

    Test (2,1) on (i),
    1/1 =? (2-1)^2
    1 =? 1
    Yes, so, (2,1) is also on the graph of (i).
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