Answer
${y}=3e^{\frac{x}{2}}-e^{x},\quad{y}\rightarrow0\quad{as}\quad{x}\rightarrow-\infty$
Work Step by Step
$$2y''-3y'+y=0,\quad{y}(0)=2,\quad{y'}(0)=\frac{1}{2}$$
Let $y=e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$$2{\lambda}^2-3{\lambda}+1=0$$
$$(2\lambda-1)(\lambda-1)=0$$
$$\lambda_{1,2}=\frac{1}{2},1$$
The general solution is ${y}=c_{1}e^{\frac{x}{2}}+c_{2}e^{x}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=2$ and $\frac{c_{1}}{2}+c_{2}=\frac{1}{2}\Rightarrow\frac{c_{1}}{2}=\frac{3}{2}\Rightarrow{c}_{1}=3\Rightarrow{c}_{2}=-1$
$$\therefore{y}=3e^{\frac{x}{2}}-e^{x}$$
$\lim_{x\to\infty}y=+\infty$. The solution tends to zero when $x\rightarrow-\infty$.
