The equation of the curve whose slope at any point is equal to y + 2 x and which passes through the origin is

We have, $\frac{dy}{dx}=y+2x$

$\Rightarrow \frac{dy}{dx}-y=2x$

Now, $\mathrm{IF}=e^{-\int 1dx}=e^{-x}$

$y·e^{-x}=\int 2x{e}^{-x}dx+k, k$ be the constant of integration

$=2\left[x\int e^{-x}dx-\int 1·\left(-e^{-x}\right)dx\right]+k$  [using integration by parts]

$\Rightarrow y·e^{-x}=-2x{e}^{-x}-2e^{-x}+k ...\left(\mathrm{i}\right)$

As curve (i) passes through $\left(0,0\right)$

$\therefore 0=0-2+k$

$\Rightarrow k=2$

Thus, the curve is

$y{e}^{-x}=-2x{e}^{-x}-2e^{-x}+2$

$\therefore y=2\left(e^x-x-1\right)$

Hence, option (b) is correct.