1.

If `ye^(y) dx = (y^(3) + 2xe^(y))dy, y(0) = 1`, then the value of x when y = 0 isA. `-1`B. 0C. 1D. 2

Answer» Correct Answer - B
`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


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