Answer
$${y}=12\exp^{\frac{1}{3}x}-8\exp^{\frac{1}{2}x}$$
Work Step by Step
$$6y''-5y'+y=0,\quad{y}(0)=4,\quad{y'}(0)=0$$Let $y=e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$$6{\lambda}^2-5{\lambda}+1=0$$
$$(3\lambda-1)(2\lambda-1)=0$$
$$\lambda_{1,2}=\frac{1}{3},\frac{1}{2}$$
The general solution is ${y}=c_{1}\exp^{\frac{1}{3}x}+c_{2}\exp^{\frac{1}{2}x}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=4$ and $\frac{c_{1}}{3}+\frac{c_{2}}{2}=0\Rightarrow{c}_{1}=-\frac{3c_{2}}{2}$
$$\therefore-\frac{3c_{2}}{2}+c_{2}=-\frac{c_{2}}{2}=4\Rightarrow{c_{2}}=-8$$
$$\therefore{c_{1}}=4-(-8)=12$$
$$\therefore{y}=12\exp^{\frac{1}{3}x}-8\exp^{\frac{1}{2}x}$$
