Question

Solve:
$\frac{dy}{dx}=x^2(x-2)$, given $y=2$ where $x=0$.

Answer 1

We have,

$\frac{dy}{dx}=x^2(x-2)$

$dy=x^2(x-2)dx$ ……. (1)

On integrating both sides, we get

$\int dy=\int x^2(x-2)dx$

$\int dy=\int (x^3-2x^2)dx$

$y=\frac{x^4}{4}-\frac{2x^3}{3}+C$

Since, $y=2,x=0$

Therefore,

$2=0-0+C$

$C=2$

Thus,

$y=\frac{x^4}{4}-\frac{2x^3}{3}+2$

Hence, this is the answer.