If `ye^(y) dx = (y^(3) + 2xe^(y))dy, y(0) = 1`, then the value of x when y = 0 is

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

B

Explanation

`ye^(y) dx = (y^(3)+2xe^(y))dy`
`rArr" "(dx)/(dy)=(y^(2))/(e^(y))+(2x)/(y)`
`rArr" "(dx)/(dy)-(2)/(y)x=(y^(2))/(e^(y))`
I.F. `= e^(-2 Iny) = (1)/(y^(2))`
General solution `x (1)/(y^(2))= int (y^(2))/(e^(y))-(1)/(y^(2)) dy + c`
`(x)/(y^(2))=e^(-y) + c = -e^(-y) + c`
`x = 0, y = 1 rArr c = e^(-1)`
`rArr" "x = -y^(2)(e^(-y) - e^(-1))`
If y = 0, then x = 0