Answer
$y=x^3-4x^2+5$
Work Step by Step
The anti-derivative for $\dfrac{d^3y}{dx^3}=6$ are:
$\dfrac{d^2y}{dx^2}=6x+c$; $\dfrac{dy}{dx}=3x^2+cx+c'$, $y=x^3+\dfrac{cx^2}{2}+c'x+c''$
Now, apply Initial conditions $y''(0)=-8 ; y'(0)=0,y(0)=5$, we have
This implies, $c=-8$ and $c'=2$ and $c''=5$
Hence, $y=x^3-4x^2+5$
