The equation of one of the curves whose slope of tangent at any point is equal to `y+2x` is

Options

A. `y = 2(e^(x)+x-1)`
B. `y=2(e^(x)-x-1)`
C. `y = 2(e^(x)-x+1)`
D. ` y = 2 (e^(x)+x+1)`

Answer

b

Explanation

Given , `(dy)/(dx) = y + 2x`
` rArr (dy)/(dx) -y = 2x`
This is linear differential equation
`:. IF = e^((int)-1//dx)= e^(-x)`
` :. ` Solution of the differential equation is
` y * e^(-x) int xe^(-x) dx=2 (-xe^(-x)-e^(-x))+C`
[ integration by parts ]
` rArr y = 2e^(x) (-xe^(-x) - e^(-x)) +Ce^(x)`
` rArr y = -2 x - 2 + Ce^(x)`
For C =2 we get
` y = 2 (e^(x) -x-1)`
` y = 2(e^(x) - x - 1)`