Answer
a) $${y}=c_{1}e^{-(\frac{b-\sqrt{b^2-4ac}}{2a})x}+c_{2}e^{-(\frac{b+\sqrt{b^2-4ac}}{2a})x}$$
b) $${a}Y''+bY'+cY=d$$
Work Step by Step
$$ay''+by'+cy=d$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$a{\lambda}^2+b{\lambda}+c=0$ ($\because$ equilibrium solution causes LHS=0)
$$\lambda_{1,2}=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$$
$$\therefore{y}=c_{1}e^{-(\frac{b-\sqrt{b^2-4ac}}{2a})x}+c_{2}e^{-(\frac{b+\sqrt{b^2-4ac}}{2a})x}$$
Substituting $Y$ into the original equation,
$$a(y-y_{e})''+b(y-y_{e})'+c(y-y_{e})=d$$
$$ay''+by'+cy'-(ay_{e}''+by_{e}'+cy_{e})=d$$
Since the equilibrium solution causes the sum of the a, b and c terms to become zero, as mentioned earlier, we are left with the original equation.
Replacing $y-y_{e}$ with $Y$,
$${a}Y''+bY'+cY=d$$
