Question

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

Answer 1

\[{(3xy+y^2)+(x^2+xy)y'=0}\]\[{(3xy+y^2)\text d x+(x^2+xy)\text d y=0}\]\[M=(3xy+y^2)\qquad\qquad N=(x^2+xy)\]\[\frac{\partial M}{\partial y}=3x+2y\qquad\qquad\frac{\partial N}{\partial x}=2x+y\] \[\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}\]

Answer 2

\[R=R(x,y)=\frac{1}{xy(2x+y)}\]\[\frac{\partial R(x,y)M}{\partial y}=\frac{\partial R(x,y)N}{\partial x}\]\[{R(x,y)(3xy+y^2)+R(x,y)(x^2+xy)y'=0}\]\[{\frac{(3x+y)}{x(2x+y)}+\frac{(x+y)}{y(2x+y)}y'=0}\]

Answer 3

\[\frac{\partial R(x,y)M}{\partial y}=-\frac{1}{(2x+y)^2}\]\[\frac{\partial R(x,y)N}{\partial x}=-\frac{1}{(2x+y)^2}\]\[\frac{\partial R(x,y)M}{\partial y}=\frac{\partial R(x,y)N}{\partial x}\]

Answer 4

\[\frac{\partial \psi(x,y)}{\partial x}=R(x,y)M\]\[\psi(x,y)=\int R(x,y)M\text dx+g(y)\]\[\psi(x,y)=\int {\frac{(3x+y)}{x(2x+y)}\text dx+g(y)}\]\[\psi(x,y)=\int {\frac{1}{(2x+y)}+\frac{1}{x}\text dx+g(y)}\] \[\psi(x,y)=\frac 12 \log(2x+y)+\log(x)+g(y)\]
\[\frac{\partial \psi(x,y)}{\partial y}=\frac 1{4x+2y}+g'(y)\]
\[R(x,y)N=\frac{(x+y)}{y(2x+y)}\]

Answer 5

that's all i got so far ,

Answer 6

\[\psi(x,y)=\frac 12\int\frac{1}{2x+y}+\frac 1y\text dy +h(x)\]\[\qquad\qquad=\frac 12\left(\log(2x+y)+\log(y)\right)+h(x)\]\[\frac{\partial\psi(x,y)}{\partial x}=\frac{2}{2x+y}+h'(x)\]\[R(x,y)M=\frac{(3x+y)}{x(2x+y)}\]

Answer 7

how did you find IF??

Answer 8

the integrating factor R(x,y) or IF was given in the question

Answer 9

Oh ... thank goodness ... your last problem bugged me for a day!!

Answer 10

\[ {\frac{(x+y)}{y(2x+y)}= } \frac{\partial \psi(x,y)}{\partial y}=\frac 1{4x+2y}+g'(y) \]
Find g(y)

Answer 11

\[ g(y) = \int \frac{2(x+y) - y}{2y(2x+y)} dy\]

Answer 12

\[R(x,y)N=\frac{x}{y(2x+y)}+\frac{1}{(2x+y)}\]\[\frac 1{2(2x+y)}+g'(y)=\frac{x}{y(2x+y)}+\frac{1}{(2x+y)}\]\[g'(y)=\frac{x}{y(2x+y)}+\frac{1}{(2x+y)}-\frac 1{2(2x+y)}\]\[g'(y)=\frac{x}{y(2x+y)}+\frac{1}{2(2x+y)}\]

Answer 13

?>

Answer 14

\[ \psi(x,y)=\frac 12 \log(2x+y)+\log(x)+g(y)\]
\[ \frac{\partial \psi(x,y)}{\partial y}=\frac 1{4x+2y}+g'(y) \]
\[ R(x,y)N = \frac 1{4x+2y}+g'(y)\]
Since you now R(x,y) and N, put those values , and find g(y) integrating with dy

Answer 15

\[\frac 1{2(2x+y)}+g'(y)=\frac{x}{y(2x+y)}+\frac{1}{(2x+y)}\]

Answer 16

find g(y) ... g'(y) = d(g(y))/dy

Answer 17

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

\[g'(y)=\frac{x}{y(2x+y)}+\frac{1}{2(2x+y)}\]

Answer 18

\[ g(y)= \int \left ( \frac{x}{y(2x+y)}+\frac{1}{2(2x+y)} \right ) dy \]

Answer 19

\[g(y)= \int \left ( \frac{x}{y(2x+y)}+\frac{1}{2(2x+y)} \right ) dy\]\[g(y)=\frac 12\left(\log(y)-\log(2x+y)+\log(x+y)\right)+c_1\]\[g(x)=\frac{\log(y)}{2}\]

Answer 20

*\(g(y)\)

\[\psi(x,y)=\frac 12 \log(2x+y)+\log(x)+g(y)\]
\[\psi(x,y)=\frac 12 \log(2x+y)+\log(x) +\frac{\log(y)}{2} \]

Answer 21

seems ok

Answer 22

to me ..

Answer 23

\[\psi(x,y)=\frac 12 \log(2x+y)+\log(x) +\frac{\log(y)}{2}=c\]