Question

Differential equation problem

How did they get v(x)?

Answer 1

You can find it by integrating twice:

$\displaystyle\int v''(x)\,\mathrm dx=\int(-6x+2)\,\mathrm dx\iff v'(x)=-3x^2+2x+C_1$
$\displaystyle\int v'(x)\,\mathrm dx=\int(-3x^2+2x+C_1)\,\mathrm dx\iff v(x)=-x^3+x^2+C_1x+C_2$

Given that $v(0)=0$, you have

$0=-0^3+0^2+0C_1+C_2\implies C_2=0$

and given $v(1)=-1$,

$-1=-1^3+1^2+C_1\implies C_1=-1$

$\implies v(x)=-x^3+x^2-x$