Question

Determine if exact and if it is, find the general equation.
2x(1+(x^2-y)^(1/2))dx - (x^2-y)^(1/2)dy=0

Answer 1

\[2x\left(1+(x^2-y)^{1/2}\right)\text dx - \left(x^2-y\right)^{1/2}\text dy=0\]

Answer 2

The equation is of the from\[M\text dx+N\text dy=0\]
The equation is exact if
\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]

Answer 3

\[\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left(2x\left(1+(x^2-y)^{1/2}\right)\right)\]\[=2x\frac{\partial }{\partial y}\left(1+(x^2-y)^{1/2}\right)\]\[=2x\left(-\frac {(x^2-y)^{-1/2}}2\right)\]\[=\frac{-x}{\sqrt{x^2-y}}\]

\[\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left(- \left(x^2-y\right)^{1/2}\right)\]\[=-2x\frac{\left(x^2-y\right)^{-1/2}}{2}\]\[=\frac{-x}{\sqrt{x^2-y}}\]

Answer 4

\[\psi(x,y)=\int M\text dx+g(y)=\int N\text dy+h(x)\]