\[\text{Solve the differential equation}\]\[{(3xy+y^2)+(x^2+xy)y'=0}\]\[\text{using the integration factor}\]\[R(x,y)=\frac{1}{xy(2x+y)}\]

\[\text{is of the form}\quad M(x,y)\text dx+N(x,y)\text dy=0\ \[M(x,y)=(3xy+y^2)\]\[N(x,y)=(x^2+xy)\]\[\frac{\partial M}{\partial y}=3x+2y\]\[\frac{\partial N}{\partial y}=2x+y\] \[\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}\]\[\text{The Equation is not exact}\]\[\text{using the integration factor}\]\[R(x,y)=\frac{1}{xy(2x+y)}\]\[{R(x,y)(3xy+y^2)+R(x,y)(x^2+xy)y'=0}\]\[{\frac{(3xy+y^2)}{xy(2x+y)}+\frac{(x^2+xy)}{xy(2x+y)}y'=0}\]\[{\frac{y(3x+y)}{xy(2x+y)}+\frac{x(x+y)}{xy(2x+y)}y'=0}\]\[{\frac{(3x+y)}{x(2x+y)}+\frac{(x+y)}{y(2x+y)}y'=0}\]\[RM=\frac{(3x+y)}{x(2x+y)}\]\[\frac{\partial (RM)}{\partial y}=-\frac{1}{(2x+y)^2}\]\[RN=\frac{(x+y)}{y(2x+y)}\]\[\frac{\partial (RN)}{\partial x}=-\frac{1}{(2x+y)^2}\]\[\frac{\partial (RM)}{\partial y}=\frac{\partial (RN)}{\partial x}\]\[\frac{\partial \psi(x,y)}{\partial x}=R(x,y)M=-\frac{1}{(2x+y)^2}\]\[\psi(x,y)=\int -\frac{1}{(2x+y)^2} \text dx\]\[=\frac{1}{2(2x+y)}+c\]

** \[\psi(x,y)=\int -\frac{1}{(2x+y)^2} \text dx+g(y)\]\[=\frac{1}{2(2x+y)}+g'(y)\]

am i doing this right ?

Looks right to me.

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