Question

First order exact differential equation with integrating factor R(x,y)
\[\left(3x+\frac 6y\right)\text dx+\left(\frac {x^2}{y}+\frac{3y}{x}\right)\text dy =0\]

Answer 1

\[\left(3x+\frac 6y\right)\text dx+\left(\frac {x^2}{y}+\frac{3y}{x}\right)\text dy =0\]\[M=3x+\frac 6y\qquad\qquad N=\frac {x^2}{y}+\frac{3y}{x}\]\[\frac{\partial M}{\partial y}=\frac {-6}{y^{2}}\qquad\qquad \frac{\partial N}{\partial x}=\frac{2x}{y}-\frac{3y}{x^2}\]\[\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\]

Answer 2

\[R=R(x,y)=x^my^n\]\[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\]\[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]\[R_y(x,y)\left(3x+\frac 6y\right)+R(x,y)\left(\frac {-6}{y^{2}}\right)=R_x(x,y)\left(\frac {x^2}{y}+\frac{3y}{x}\right)+R(x,y)\left(\frac{2x}{y}-\frac{3y}{x^2}\right)\]\[nx^my^{n-1}\left(3x+\frac 6y\right)+x^my^n\left(\frac {-6}{y^{2}}-\frac{2x}{y}+\frac{3y}{x^2}\right)-mx^{m-1}y^n\left(\frac {x^2}{y}+\frac{3y}{x}\right)=0\]\[nx^{m+2}y^{n+1}\left(3xy+6\right)+x^{m}y^n\left(-6x^2-2x^3y+3y^3\right)-mx^{m}y^n\left({x^3y}+3y^3\right)=0\]

Answer 3

agh how do i find
m, and n

Answer 4

hey is that exact??

Answer 5

it should be exact with the integrating factor \( R(x,y)=xy\)
how do i show \(m,n=1\)

Answer 6

how do you know R(x,y) will be of the form x^ny^m??

Answer 7

because i checked the answer int the back of the book,
\[R=R(x,y)=xy\]

Answer 8

lol ....
if it is of the form of polynomial .... simplify those x's an y's

Answer 9

how exactly \[nx^{m+2}y^{n+1}\left(3xy+6\right)+x^{m}y^n\left(-6x^2-2x^3y+3y^3\right)-mx^{m}y^n\left({x^3y}+3y^3\right)=0\]
if i expand the brackets its gonna get (even more) messy

Answer 10

first take x^ny^m common on both sides and cancel them

Answer 11

\[nx^{m+2}y^{n+1}\left(3xy+6\right)+(1-m)x^{m}y^n\left(-6x^2-2x^3y+3y^3{x^3y}+3y^3\right)=0\]

Answer 12

like that?

Answer 13

x^ny^m( ...some junk terms) = x^ny^m(.. other junk term)

Answer 14

\[nx^my^{n-1}\left(3x+\frac 6y\right)+x^my^n\left(\frac {-6}{y^{2}}-\frac{2x}{y}+\frac{3y}{x^2}\right)-mx^{m-1}y^n\left(\frac {x^2}{y}+\frac{3y}{x}\right)=0\]\[x^my^n(ny)\left(3x+\frac6y\right)+x^my^n\left(\frac {-6}{y^{2}}-\frac{2x}{y}+\frac{3y}{x^2}\right)-x^my^n(xm)\left(\frac {x^2}{y}+\frac{3y}{x}\right)=0\]
\[x^my^n\left((ny)\left(3x+\frac6y\right)+\left(\frac {-6}{y^{2}}-\frac{2x}{y}+\frac{3y}{x^2}\right)-(xm)\left(\frac {x^2}{y}+\frac{3y}{x}\right)\right)=0\]

Answer 15

\[x^my^n\left(n\left(3xy+6\right)+\left(\frac {-6}{y^{2}}-\frac{2x}{y}+\frac{3y}{x^2}\right)-m\left(\frac {x^3}{y}+{3y}\right)\right)=0\]

Answer 16

\[x^my^n\left(n\left(3xy+6\right)-m\left(\frac {x^3}{y}+{3y}\right)\right)=x^my^n\left(\frac {6}{y^{2}}+\frac{2x}{y}-\frac{3y}{x^2}\right)\]

Answer 17

\[n\left(3xy+6\right)-m\left(\frac {x^3}{y}+{3y}\right)=\frac {6}{y^{2}}+\frac{2x}{y}-\frac{3y}{x^2}\]

Answer 18

simply those coefficients ..

Answer 19

which ones?

Answer 20

\[ x^2y^2(n\left(3xy+6\right)-m\left(\frac {x^3}{y}+{3y}\right) )=xy(\frac {6}{y^{2}}+\frac{2x}{y}-\frac{3y}{x^2}) \]

Answer 21

sorry ... x^2y^2 on the right one too ... and try comparing coefficients ... not sure if this works

Answer 22

\[nx^2y^2\left(3xy+6\right)-mx^2y\left( {x^3}+{3y^2}\right)=2x^2+2x^3y-3y^3\]

Answer 23

\[x^2y\left(ny\left(3xy+6\right)-m( {x^3}+{3y^2})\right)=2x^2+2x^3y-3y^3\]

Answer 24

now wht\

Answer 25

not sure ... didn't work :(
I am looking ...

Answer 26

@phi

Answer 27

Here's one way