InterviewSolution
Saved Bookmarks
| 1. |
The solution of equation `(2y-1)dx-(2x+3)dy=0` isA. `(2x-1)/(2y+3)=k`B. `(2y+1)/(2x-3)=k`C. `(2x+3)/(2y-1)=k`D. `(2x-1)/(2y-1)=k` |
|
Answer» Given that, `" "(2y-1)dx-(2x+3)dy=0` `rArr" "(2y-1)dx=(2x+3)dy` `rArr" "(dx)/(2x+3)=(dy)/(2y-1)` On integrating both sides, we get `" "(1)/(2) log(2x+3)=(1)/(2)log(2y-1)+logC` `rArr" "(1)/(2)[log*(2x+3)-log(2y-1)]=logC` `rArr" "(1)/(2)log((2x+3)/(2y-1))=logC` `" "((2x+3)/(2y-1))^(1//2)=C` `rArr" "(2x+3)/(2y-1)=C^(2)` `rArr" "(2x+3)/(2y-1)=k`, where `K=C^(2)` |
|