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`