InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Solve: `dy/dx = sin(x+y) + cos(x+y)` |
|
Answer» Given, `" "(dy)/(dx)=cos(x+y)+sin(x+y)" "`...(i) Put `" "x+y=z` `rArr" "1+(dy)/(dx)=(dz)/(dx)` On substituting these values in Eq. (i), we get `" "((dz)/(dx)-1)=cosz+sinz` `rArr" "(dz)/(dx)=(cosz+sinz+1)` `rArr" "(dz)/(cosz+sinz+1)=dx` On integrating both sides, we get `" "int(dz)/(cosz+sinz+1)=int1dx` `rArr" "int(dz)/((1-tan^(2)z//2)/(1+tan^(2)z//2)+(2tanz//2)/(1+tan^(2)z//2)+1)=intdx` `rArr" "int(dz)/((1-tan^(2)z//2+2tanz//2+1+tan^(2)z//2)/((1+tan^(2)z//2)))=intdx` `rArr" "int((1+tan^(2)z//2)dz)/(2+2tan^(2)z//2)=intdx` `rArr" "int(sec^(2)z//2dz)/(2(1+tanz//2))=intdx` Put `1+tanz//2 =t` `rArr" "((1)/(2)sec^(2)z//2)dz=dt` `rArr" "int(dt)/(t)=intdx` `rArr" "lot|t|=x+C` `rArr" "log|1+tanz//2|=x+C` `rArr" "log|1+tan""((x+y))/(2)|=x+C` |
|