Author Topic: TUT0801 Quiz5  (Read 760 times)

XiaolongZhao

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TUT0801 Quiz5
« on: November 01, 2019, 02:12:20 PM »
Q:Find the general solution of y^''+4y^'+4y=t^(-2) e^(-2t)
A:

Firstly, find its homo solution

   r^2 + 4r + 4 = 0

   r = -2 (repeated solution)

   y_c (t) = c_1 e^(-2t) + c_2 te^(-2t)

   y_1 (t) = e^(-2t), y_2 (t) = te^(-2t)

Secondly, find the solution Y(t)

   y_1' (t) = -2e^(-2t),  y_2' (t)=e^(-2t) - 2te^(-2t)

   W(t) = e^(-4t)

   W_1 (t) = -te^(-2t)

   W_2 (t) = e^(-2t)

   g(t) = t^(-2) e^(-2t)

   Y(t) = e^(-2t) ∫(-te^(-2t) t^(-2) e^(-2t))/e^(-4t)  dt + te^(-2t) ∫(e^(-2t) t^(-2) e^(-2t))/e^(-4t)  dt

         = -e^(-2t) ln(t) - e^(-2t)

Thus, the general solution is:

   y(t) = y_c (t) + Y(t) = c_1 e^(-2t) + c_2 te^(-2t) -e^(-2t) ln(t) - e^(-2t)

The clearer answer is shown in the picture below