Mathematics
OpenStudy (unklerhaukus):

$\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)}$

OpenStudy (unklerhaukus):

$\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$

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

am i doing this right ?

OpenStudy (blockcolder):

Looks right to me.