mathstresser
Junior Member
- Joined
- Jan 28, 2006
- Messages
- 134
Solve the exact equation
(1-ysinx)dx + cosx dy = 0
We know that it's an exact equation because Pdy = (-sin x) and Qdx= -sin x.
But, I don't know what to do with an exact equation.
dy/dx= (ysinx-1)/cosx
dy/dx= ysinx/cosx - 1/cosx
dy/dx= (1/cosx)(yxsinx-1)
I don't know what to do. I can't separate it. I know it's linear, but I don't know what to do.
(1-ysinx)dx + cosx dy = 0
We know that it's an exact equation because Pdy = (-sin x) and Qdx= -sin x.
But, I don't know what to do with an exact equation.
dy/dx= (ysinx-1)/cosx
dy/dx= ysinx/cosx - 1/cosx
dy/dx= (1/cosx)(yxsinx-1)
I don't know what to do. I can't separate it. I know it's linear, but I don't know what to do.