(y+cosx)dx+(x+siny)dy=0M=y+cosx∂M∂y=1N=x+siny∂N∂x=1Therefore, since ∂M∂y=∂N∂x,the differential equation is exact.\displaystyle (y+\cos{x})\mathrm{d}x+(x+\sin{y})\mathrm{d}y=0\\ M = y + \cos{x}\\ \frac{\partial M}{\partial y} = 1\\ N = x + \sin{y}\\ \frac{\partial N}{\partial x} = 1\\ \textsf{Therefore, since}\, \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, \\ \textsf{the differential equation is exact.}(y+cosx)dx+(x+siny)dy=0M=y+cosx∂y∂M=1N=x+siny∂x∂N=1Therefore, since∂y∂M=∂x∂N,the differential equation is exact.
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