No double-dipping
(a):
We can solve homo firstly:
r^2-3r+2=0\\ (r-2)(r-1)=0\\ r_1=2,r_2=1
Therefore:
y=c_1e^{2t}+c_2e^t
So we can get:
W=\begin{vmatrix} e^{2t} & e^t \\ 2e^{2t} & e^t \end{vmatrix}=-e^{3t}\\ W_1=\begin{vmatrix} 0 & e^{t} \\ 1 & e^{t} \end{vmatrix}=-e^{t}\\ W_2=\begin{vmatrix} e^{2t} & 0 \\ 2e^{2t} & 1 \end{vmatrix}=e^{2t}
So we can get:
Y(t)=e^{2t}\int{\frac{-e^{s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds + e^{t}\int{\frac{e^{2s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds\\ Y(t)=e^{2t}\int{\frac{e^{s}}{e^{2s}+1}}ds - e^{t}\int{\frac{e^{2s}}{e^{2s}+1}}ds\\ Y(t)=e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
Finally:
y(t)=c_1e^{2t}+c_2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
(b):
So we can get y'(t):
y'=2c_1e^{2t}+c_2e^t+2e^{2t}arctan(e^t)+e^{2t}*\frac{e^t}{e^{2t}+1}-0.5e^t*ln(e^{2t}+1)-0.5e^t*\frac{e^{2t}}{e^{2t}+1}
We take y(0)=y'(0)=0,so we can get:
2c_1+2c_2+0.5\pi-ln2=0\\ 2c_1+c_2+0.5\pi-0.5ln2=0
So
c_1=-0.25\pi,c_2=0.5ln2
Therefore:
y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)