Solve the initial value problem.

$\frac{d^{2}y}{d{x}^2}=1-5x,y^{\text{'}}(0)=8,y(0)=2$

A$.y=2$

B$.y=\frac{1}{2}{x}^2-\frac{5}{6}{x}^3+8x+2$

C$.y=x^2+5x^3+8x+2$

D$.y=\frac{1}{2}{x}^2+\frac{5}{6}{x}^3-8x-2$