Solution of the differential equation cosxdy=y(sinx-y)dx ,0<x<π/2

Dear Student,
Please find below the solution to the asked query:

We have, cosx dy=ysinx-ydxdydx=y tanx-y2secx1y2dydx-1ytanx=-secx        .....1Let 1y=t-1y2dydx=dtdx, so equation 1 becomes,-dtdx-t tanx=-secxdtdx+t tanx=secxNow,I.F.=etanx dx=elogsecx=secxSo,t×I.F.=secx×I.F. dxt secx=sec2xdxt secx=tanx +C1y secx=tanx +Csecx=ytanx +C

Hope this information will clear your doubts about this topic.
If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards
 
 

  • 6
What are you looking for?