Find the integral.Solve the initial value problem.

$\frac{d^{2}y}{d{x}^2}=4-10x,y^{\text{'}}(0)=8,y(0)=2$

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

B$.y=2$

C$.y=4x^2+10x^3+8x+2$

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

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(2\sqrt[3]{x^4}-x^{-3})dx$

A$.\frac{6}{7}{x}^{\frac{7}{3}}-2x^{-2}+c$

B$.\frac{4}{3}{x}^{\frac{7}{3}}-3x^{-2}+c$

c$.\frac{6}{7}{x}^{\frac{6}{7}}+\frac{1}{2x^2}+c$

D$.\frac{6}{7}{x}^{\frac{7}{3}}+\frac{1}{2x^2}+C$