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
