Part (a)$\\$
The equation for the Wronskian would be:
$$W = c\times exp[-\int p_{1}(t)dt] $$
Since there is no $y''$ term, $p_{1}(t)$ would be $0$:
$$W = c\times exp[-\int 0 dt]=c\times exp(0)=c$$
Therefore, the Wronskian would be a constant.
Part (b)$\\$
Consider the homogeneous equation:
$$y''' - 3y' + 2y = 0$$
The characteristic equation would be:
$$r^3-3r+2=0$$
Solving this gives:
$$(r-1)^2(r+2)\rightarrow r_{1}=r_{2}=1, r_{3}=-2$$
Therefore, the homogeneous solution would be:
$$y_{c}(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{-2t}\qquad(2)$$
Computing the Wronskian:
$$W=\begin{array}{|c c c|}e^t &te^t &e^{-2t}\\e^t&(t+1)e^t&-2e^{-2t}\\e^t&(t+2)e^t&4e^{-2t}\end{array}=4(t+1)+2(t+2)-t(4+2)+(t+2)-(t+1)=9$$
Therefore, the Wronskian is a constant just as expected based on part (a).
Part (c)$\\$
The particular solution should be of the form:
$$Y(t)=Ae^{-2t}$$
Since $e^{-2t}$ is part of the homogeneous solution, we look for solutions of the form:
$$Y(t)=Ate^{-2t}\qquad(3)$$
Differentiating this:
$$Y'(t)=Ae^{-2t}-2Ate^{-2t} \qquad(4)$$
Differentiating again:
$$Y''(t)=-4Ae^{-2t}+4Ate^{-2t}$$
Differentiating once more:
$$Y'''(t)=12Ae^{-2t}-8Ate^{-2t} \qquad(5)$$
Plugging (3), (4) and (5) into (1):
$$12Ae^{-2t}-8Ate^{-2t}-3Ae^{-2t}+6Ate^{-2t}+2Ate^{-2t}=18e^{-2t}$$
Simplifying gives:
$$9Ae^{-2t}=18e^{-2t}$$
Therefore, $A=2$. Subbing this value of A into (3) and combining it with (2) gives the general solution:
$$y(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{2t}+2te^{-2t}$$
