If`(dy)/(dx) = 1 +x +y +xy` and `y(-1)= 0` , then function `y` is :A. `e^(((1+x)^(2))/(2))-1`B. `e^(((1-x)^(2))/(2))`C. `log(1+x)-1`D. `log(1-x)`

If`(dy)/(dx) = 1 +x +y +xy` and `y(-1)= 0` , then function `y` is :A.

1.	If`(dy)/(dx) = 1 +x +y +xy` and `y(-1)= 0` , then function `y` is :A. `e^(((1+x)^(2))/(2))-1`B. `e^(((1-x)^(2))/(2))`C. `log(1+x)-1`D. `log(1-x)`
Answer» Correct Answer - A Let `(dy)/(dx)=1+x+y+xy` `rArr (dy)/(dx)=(1+x)(1+y)` `rArr (dy)/(1+y)=dx(1+x)` `rArrint(1)/(1+y)dy=int(1+x)dx` `rArr log(1_y)=x+(x^(2))/(2)+c` Given At `x=-1, y=0` `rArr log 1=-1+(1)/(2)+c` `rArr c=(1)/(2)` `therefore" "log(1+y)=x+(x^(2))/(2)+(1)/(2)=((1+x)^(2))/(2)` `rArr" "1+y=e^(((1+x)^(2))/(2))` `rArr" "y=e^(((1+x)^(2))/(2))-1`