InterviewSolution
Saved Bookmarks
| 1. |
Solve `2(y+3)-x y(dy)/(dx)=0`, given that `y(1)=-2` |
|
Answer» Given that, `" "2(y-3)-xy(dy)/(dx)=0` `rArr" "2(y+3)=xy(dy)/(dx)` `rArr" "2(dx)/(x)=((y)/(y+3))dy` `rArr" "2*(dx)/(x)=((y+3-3)/(y+3))dy` `rArr" "2*(dx)/(x)=(1-(3)/(y+3))dy` On integrating both sides, we get `" "2logx=y-3log(y+3)+C" "...(i)` When x= and y=-2, then `" "2log1=-2-3log (-2+3)+C` `rArr" "2*0=-2-3*0+C` `rArr" "C=2` On subsituting the value of C in Eq. (i), we get `" "2logx=y-3log(y-3)+2` `rArr" "2logx+3log(y+3)=y+2` `rArr" "logx^(2)+log(y+3)^(3)=(y+2)` `rArr" "logx^(2)(y+3)^(3)=(y+2)` `rArr" "x^(2)(y+3)^(3)=e^(y+2)` |
|