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Math Help - Seemingly easy differential equation, but I got confused

  1. #1
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    Seemingly easy differential equation, but I got confused

    Solve:

    2 (dy/dt)=2y+3t

    So I divide through by 2:

    (dy/dt)=y+(3/2)t

    I take the homogenous yh=e^t

    Then I make it

    (dy/dt)-y=(3/2)t

    So I guess a*t

    Plug that in to get:

    a-at=(3/2)t
    So that

    a(1-t)=(3/2)t

    and then

    a=(3/2)t/(1-t)

    And multiplying that my t and combing with the homogeous I get:

    y(t)=e^t+(3/2)t^2/(1-t)

    but I've never seen an answer like this since usually it is a more complicated function with es and cosines....

    Is this correct or where did I make a mistake?

    Thanks.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    Solve:

    2 (dy/dt)=2y+3t

    So I divide through by 2:

    (dy/dt)=y+(3/2)t

    I take the homogenous yh=e^t

    Then I make it

    (dy/dt)-y=(3/2)t

    So I guess a*t

    Plug that in to get:

    a-at=(3/2)t
    So that

    a(1-t)=(3/2)t

    and then

    a=(3/2)t/(1-t)

    And multiplying that my t and combing with the homogeous I get:

    y(t)=e^t+(3/2)t^2/(1-t)

    but I've never seen an answer like this since usually it is a more complicated function with es and cosines....

    Is this correct or where did I make a mistake?

    Thanks.
    First off, a is a constant so you can't solve for a as a function of t.

    Let y_p = at + b so

     <br />
a -at -b = \frac{1}{2}t<br />
    so a-b = 0, -a = \frac{1}{2} so a = b = - \frac{1}{2} and y_p = -\frac{1}{2} t - \frac{1}{2}
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