Answer
$y_{p}=2t^{2}e^{\frac{t}{2}}$
Work Step by Step
$$4y''-4y'+y=16e^{\frac{t}{2}}$$
Consider the homogeneous LHS. Let $y=e^{\lambda{t}}$ so that $(lny)'=\lambda$.
$$\therefore(2\lambda-1)^2=0$$
$$\lambda=\frac{1}{2}$$
$\therefore{y}=(c_{1}+c_{2}t)e^{\frac{t}{2}}$ is the general solution. Let $u(t)$ be the function associated with the solution in the inhomogeneous case.
$$\therefore{y}=u(t)e^{\frac{t}{2}},\quad{y'}=u'(t)e^{\frac{t}{2}}+\frac{1}{2}u(t)e^{\frac{t}{2}},\quad{y''}=u''(t)e^{\frac{t}{2}}+u'(t)e^{\frac{t}{2}}+\frac{1}{4}u(t)e^{\frac{t}{2}}$$
$$\therefore{4y''}-4y'+y=4u''e^{\frac{t}{2}}=16e^{\frac{t}{2}}$$
By association, $u''(t)=4$
$$\therefore{u(t)}=\int{\int{4}dt}dt=4(\frac{t^2}{2}+c_{1}t+c_{2})\equiv(c_{1}+c_{2}t)e^{\frac{t}{2}}+2t^{2}e^{\frac{t}{2}}\\
{\Rightarrow}y_{p}=2t^{2}e^{\frac{t}{2}}$$
Substituting particular solution $y_{p}$ into the original equation,
$$LHS:4(4e^{\frac{t}{2}}+4te^{\frac{t}{2}}+\frac{t^{2}e^{\frac{t}{2}}}{2})-4(4te^{\frac{t}{2}}+t^{2}e^{\frac{t}{2}})+2t^{2}e^{\frac{t}{2}}=16e^{\frac{t}{2}}=RHS,\quad{t}\in\mathbb{R}$$
