InterviewSolution
Saved Bookmarks
| 1. |
Solve: `(xcosy-ysiny)dy+(x siny+ycosy)dx=0` |
|
Answer» We have `(xcosy-ysiny)(dy)/(dx)+(xsiny+ycosy)=0`.............(1) Let `xsiny+ycosy=t` Differentiating w.r.t. `x`, we get `xcosy(dy)/(dx)+siny+cosy(dy)/(dx)-ysiny(dy)/(dx)=(dt)/(dx)` `therefore (xcosy-ysiny)(dy)/(dx)+siny+cosy(dy)/(dx)=(dt)/(dx)` So, from (1), we have `(dt)/(dx)-(siny+(d(siny))/(dx))+t=0` `rArr (d(t-siny))/(dx) = -(tsiny)` `rArr (d(t-siny))/(t-siny)=-dx` `rArr int(d(t-siny))/(t-siny) =-intdx` `rArr log_(e)(t-siny)=-x+log_(e)C` `rArr t-siny=Ce^(-x)` `rArr xsiny+ycosy-siny=Ce^(-x)` |
|