Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is `y = 2(e^(x) - x-1)`.

Explanation

We have , `(dy)/(dx) = y+2x`
`rArr (dy)/(dx) - y = 2x`
`IF = e^(-int dx) = e^(-x)`
Multiplying the equation by IF and integrating
`y xx e^(-x) = int 2xe^(-x) dx + k`, k being the constant of integration.
`= 2[x int e^(x) dx - int 1.(-e^(x))dx] + k`, using integration by parts
`= 2-xe^(-x) - 2e^(x) + k` ...(i)
As curve (i) passes through (0, 0)
0 = 0 - 2 + k
`:. k = 2`
Thus the curve is
`ye^(-x) = -2xe^(-x) - 2e^(-x) + 2`
`:. y = -2x - 2 + 2e^(x)`
`:. y = 2(e^(x) - x-1)`