Question

Find the integrating factor to make the following equation exact , and then solve;

\[\left(3x+\frac6y\right)\mathrm dx+\left(\frac{x^2}y+\frac{3y}x\right)\mathrm dy\]

Answer 1

\[\newcommand\p \newcommand
\p \pat [2] { \tfrac{\partial #1}{\partial #2} }

M =x+6y^{-1} \qquad N=x^2y^{-1}+3x^{-1}y \\
M_y =-6y^{-2} \qquad N_x=2xy^{-1}-3x^{-2}y\\
\pat My\neq\pat Nx\implies\text{ not exact}\quad\]

Answer 2

\[\newcommand\p \newcommand
\p \pat [2] { \tfrac{\partial #1}{\partial #2} }
\text{let}\quad R =R(x,y)=x^my^n \\
\qquad\qquad R_x =mx^{m-1}y^n \\
\qquad\qquad R_y =nx^my^{n-1} \\
\\
\overline M= R(x,y)M\qquad\qquad\qquad\overline N= R(x,y)N \\
\\
\,\\
\pat {\overline M}y=\pat {\overline N}x\quad\text{ for exactness}\]

Answer 3

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

Answer 4

equating coefficients i get

n-m=2
n=1
m=1

but i think ive made a mistake because n-m=2 should be n+m=2

the correct integrating factor should be R(x,y)=xy

Answer 5

oh, right i forget the 3 for some reason ....

Answer 6

\[M=\color{red}3x+6y^{−1}\]

Answer 7

...
\[(3n-m-2)xy^{-1}+6(n-1)y^{-2} =0\\
\,\\
(3n-m-2)=0\qquad n-1=0 \\
\qquad\qquad\qquad\qquad\qquad n=1 \\
1-m=0 \\
m=1\]

Answer 8

thats better

Answer 9

well spotted @UnkleRhaukus, thanks again, i hope you can finish from here..

Answer 10

\[\begin{align*} \overline M&=xy(3x+6y^{-1})&\overline N &= xy(x^2y^{-1}+3x^{-1}y) \\
&=3x^2y+6x & &=x^3+3y^2 \\
\\
\overline M_y &=3x^2 &\overline N_x &=3x^2
\end{align*}\]

exactness

Answer 11

i always did say there was nothing wrong with talking to your self,
ikr.

Answer 12

Now integrate M-bar dx

Answer 13

\[\newcommand\p \newcommand
\p \dd [1] { \,\mathrm d#1 }
\p \pa [2] { \frac{\partial #1}{\partial #2} }
\\
\begin{align*}f &=\int(3x^2y+6x)\dd x \\
&=x^3y+2x^2+g(y) \\
\\
\pa fy&=x^3+g'(y) =x^3+3y^2 \\
&\qquad g'(y) =3y^2 \\
&\qquad g(y) =y^3+c \\
\\
f(x,y) &=x^3y+2x^2+y^3+c
\end{align*}\]

Answer 14

there is another mistakes somewhere

Answer 15

\[\large \newcommand\p \newcommand \p \dd [1] { \,\mathrm d#1 } \p \pa [2] { \frac{\partial #1}{\partial #2} } \\ \begin{align*}f &=\int\limits(3x^2y+6x)\dd x \\ &=x^3y+\color{red}{2}x^2+g(y)\end{align*}\]
Was this the mistake maybe? :o
Looks like that 2 should be a 3.

Answer 16

ah yes. thankyou @zepdrix

Answer 17

Hmm this looks like a fun problem. Grrr I gotta learn my maths :c

Answer 18

You can has a medal also :3 You got through 98% of the problem without any trouble lol

Answer 19

Differential Equations are good fun.

Answer 20

now i am getting
\[f(x,y)= x^3y+3x^2+y^3+c\\\,\\
x^3y+3x^2+y^3=k\]__________________
the solution in the back of the book say
\[x^3y+3x^2y^3=c\]
i guess they made a typo

Answer 21

if that's not fun i don't know what is