Determine which of the following equations is exact. If it is exact, solve it. \[\frac{x\text dx}{(x^2+y^2)^{3/2}}+\frac{y\text dy}{(x^2+y^2)^{3/2}}=0\]

\[\text{let }M=\frac{x}{(x^2+y^2)^{3/2}}\]\[\qquad N=\frac{y}{(x^2+y^2)^{3/2}}\]

\[\text{the equation is exact iff}\]\[ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]

\[\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left(\frac{x}{(x^2+y^2)^{3/2}}\right)\]\[\qquad=\frac{-3xy}{(x^2+y^2)^{5/2}}\] \[\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left(\frac{y}{(x^2+y^2)^{3/2}}\right)\]\[\qquad=\frac{-3xy}{(x^2+y^2)^{5/2}}\] \[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\qquad\text{the equation is exact}\]

\[\psi (x,y)=\int M\text dx+g(y)\]\[=\int \frac{x}{(x^2+y^2)^{3/2}}dx+g(y)\]

\[=-\frac{1}{\sqrt{x^2+y^2}}+g(y)\]

\[\frac{\partial \psi (x,y)}{\partial y}=\frac{y}{(x^2+y^2)^{3/2}}+g'(y)=N(x,y)\]\[g'(y)=0\]

\[g(y)=c\]

\[\psi (x,y)=-\frac{1}{\sqrt{x^2+y^2}}+c\]

can someone look over this please /

\[ \frac{-1}{(x^2+y^2)^{3/2} } + f(y) = 0\] \[ f'(y) = \frac{ydy}{(x^2+y^2)^{3/2}}\] seems you are right!!

oh cool

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