Question

Solve the differential equation
\[(3xy+y^2)+(x^2+xy)y'=0\]
*using the integrating factor
\[R(x,y)=\frac1{xy(2x+y)}\]
please check my solution:

Answer 1

\[\newcommand
\p \newcommand
\p \dd [1] { \,\mathrm d#1 }
\p \de [2] { \frac{ \mathrm d #1}{\mathrm d#2} }
\p \pa [2] { \frac{\partial #1}{\partial #2} }
\tiny\begin{align*} &(3xy+y^2)+(x^2+xy)y'=0 \\
%(3xy+y^2)\text d x+(x^2+xy)\text d y=0
\\
M&=3xy+y^2 & N&=x^2+xy \\
M_y&=3x+2y & N_x&=2x+y \\
\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad R=\frac1{xy(2x+y)} \\
\\
\overline M &=\frac{3x+y}{x(2x+y)} &\overline N &=\frac{x+y}{y(2x+y)} \\
\overline M_y &=\frac{x(2x+y)\times1-x\times(3x+y)}{x^2(2x+y)^2} &\overline N_x &=\frac{y(2x+y)\times1-(x+y)\times2y}{y^2(2x+y)^2} \\
&=\frac{(2x+y)-(3x+y)}{x(2x+y)^2} & &=\frac{2x+y-2(x+y)}{y(2x+y)^2} \\
&=\frac{-1}{(2x+y)^2} & &=\frac{-1}{(2x+y)^2} \\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\overline M_y=\overline N_x\text{ exact} \\
f &=\int\frac{3x+y}{x(2x+y)}\dd x \\
&=\int\frac{x+(2x+y)}{x(2x+y)}\dd x \\
&=\frac12\int\frac2{2x+y}\dd x+\int\frac1x\dd x \\
&=\tfrac12\ln|2x+y|+\ln |x|+g(y) \\
\\
\pa fy &=\frac1{2(2x+y)}+g'(y) =\frac{x+y}{y(2x+y)}
\\%=\frac{x}{y(2x+y)}+\frac1{2x+y} \\
&\qquad\qquad\qquad\quad g'(y)=\frac{x}{y(2x+y)}+\frac1{2(2x+y)} \\
&&\frac{x}{y(2x+y)}&=\frac A{y}-\frac B{2x+y} \\
&&x&=A(2x+y)+By \\
&&x&=2Ax+(A+B)y \\
&&A&=1/2\qquad A+B=0 \\
&&&\qquad\qquad\qquad B=-1/2 \\
&\qquad\qquad\qquad\quad g'(y)=\frac 1{2y}-\frac1{2(2x+y)}+\frac1{2(2x+y)} \\
&\qquad\qquad\qquad\quad g'(y)=\frac 1{2y} \\
\\
&\qquad\qquad\qquad\quad g(y)=\tfrac12\ln|y|+c \\
\\
f(x,y)&=\tfrac12\ln|2x+y|+\ln |x|+\tfrac12\ln|y|+c \\
(2x+y)x^2y &=k
\end{align*}
\]

Answer 2

can you read it?

Answer 3

I don't quite see how you made it exact. I see where you're at, but your equation, \(R\)

Answer 4

oh yeah sorry , the integrating factor R was given in the question

Answer 5

When you're solving for \(g'(y)\) shouldn't your terms be only in \(y\)? \(x\)'s should cancel out.

Answer 6

g'(y)=1/(2y)

Answer 7

This is all too much work for me to follow right now. OMG.

Answer 8

:P

Answer 9

But it's very neatly done! Give me a good minute to fully absorb this o.o
its like 1 am

Answer 10

Isn't your solution square root?

Answer 11

ah yeah, i changed \(e^{-2c} =k\)

Answer 12

LOL your solution is bothering me. Just the form, but it looks right.

Answer 13

\[2x^3y + x^2y^2=k\]

Answer 14

yeah i guess thats looks a little better

Answer 15

i can just squeeze that onto the page,

Answer 16

omg, stahp! Lol this problem is getting to crazy now.

Answer 17

dont worry , there will be more like this