The equation of one of the curves whose slope of tangent at any point is equal to `y+2x` is `y=2(e^x+x-1)` `y=2(e^x-x-1)` `y=2(e^x-x+1)` `y=2(e^x+x+1)` (5) `y=e^x-x-1`

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

`(dy)/(dx)=y+2ximplies(dy)/(dx)-y=2x`
IF `=e^(int-1dx)=e^(-x)`
`:.` Solution of differential equation is
`"y e"^(-x)=2intxe^(-x)dx`
`impliesye^(-x)=2(-xe^(-x)-e^(-x))+c`
`impliesy=2e^(x)(-xe^(-x)-e^(-x))+ce^(x)`
`impliesy=-2x-2+ce^(x)`
For c = 2
we get `y=2(e^(x)-x-1)`