Question

\[ (2y^2 - 6xy)dx + (3xy-4x^2)dy =0\]

Find an integrating factor of the form \(x^ny^m\) and solve the equation...

Answer 1

1st check to make sure that they are exact.

Answer 2

\(M = 2y^2 - 6xy \) \(N= 3xy-4x^2\)
\(\frac{\partial M}{\partial y}=4y - 6x\) \(\frac{\partial N}{\partial x}=3y-8x\)
\[\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=\frac{3y-8x-(4y - 6x)}{2y(y-3x)}\]Doesn't go right..

Answer 3

\[\frac{ ∂M }{ ∂x }=\frac{ ∂N }{ ∂y }\]

Answer 4

check your partial derivatives. they should match up

Answer 5

either that or your teacher is horrible and wants you to solve this another way ll

Answer 6

integrating factor!

Answer 7

But I couldn't find a suitable integrating factor... :'(

Answer 8

ah yes this is not an exact solution one

Answer 9

UnkleR is right. Find the integrating factor to make it exact.

Answer 10

That's why I tried that partial.. - partial.. / ... to find one..

Answer 11

is that a hint... try getting one in such that it looks like above

Answer 12

You were on the right path.

Answer 13

\[\frac{ M_y-N_x }{ -M }\]

Answer 14

That's your integrating factor.

Answer 15

The way I learn to get an integrating factor is by showing \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) or \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) equals to a function of x/y .. But for this (and the next question), I got troubles..

Answer 16

\[\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\]
\[R=R(x,y)=x^ny^m\]
\[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\]
\[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]

Answer 17

How... does... that... work...?

Answer 18

which bit

Answer 19

\[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\]And \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]

Answer 20

well we want a integrating factor that will make the partial derivatives equal (which will makes the equation exact)

applying the product rule for derivatives on both sides
we get that last line
i have used mixed notation for derivatives for some reason \[Z_w\leftrightarrow\frac{\partial Z}{\partial w}\](these mean the same thing, just written differently )

Answer 21

so you know
\[R,M,N\]substitute these in

Answer 22

take the partial derivatives

Answer 23

Is R the integrating factor?

Answer 24

yes

Answer 25

So, I should multiply the equation by R, right?

Answer 26

\[R=x^ny^m\]\[M=2y^2-6xy\]\[N=3xy-4x^2\]
\[\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R\]
\[LS\]\[=mx^ny^{m-1}(2y^2-6xy) + (4y-6x)x^ny^m\]\[=2mx^ny^{m+1}-6mx^{n+1}y^m + 4y^{m+1}x^n – 6x^{n+1}y^m\]\[(2m+4)x^ny^{m+1}-(6m+1)x^{n+1}y^m\]
\[RS\]\[=nx^{n-1}y^m(3xy-4x^2) + (3y-8x)x^ny^m\]\[=3nx^ny^{m+1}-4nx^{n+1}y^m+3x^ny^{m+1} – 8x^{n+1}y^m\]\[(3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m\]
So,
\[(2m+4)x^ny^{m+1}-(6m+1)x^{n+1}y^m = (3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m\]
2m+4 = 3n+3
6m+1 = 4n+8

m=5/2 , n=2

But that is not right :\

Answer 27

hmm, i cannot see any error in your working , how do you know its not right?

Answer 28

Because the book whispered me the answer :(

Answer 29

\[\tiny \text{can you whisper it to me too please , }\]

Answer 30

\[\tiny \color{white}{\mu =xy} \]

Answer 31

i see

Answer 32

\(\huge \color{red}{\mu =xy}\)
now, i see also.

Answer 33

well we were close
\[x^2y^{5/2}(2y^2 - 6xy)\text dx + x^2y^{5/2}(3xy-4x^2)\text dy =0\]
\[(2x^2y^{9/2} - 6x^3y^{7/2})\text dx +(3x^3y^{7/2}-4x^4y^{5/2})\text dy =0\]

\[\frac{\partial MR}{\partial y}=9x^2y^{7/2}-21x^3y^{5/2}\]
\[\frac{\partial NR}{\partial x}=9x^2y^{7/2}-16x^3y^{5/2}\]

Answer 34

\[21\sim16\]

Answer 35

I.... don't .... understand.... :(

Answer 36

\[R=\mu\]

Answer 37

Hmm... I should do it all over again?!

Answer 38

xy vs x^2 y^(5/2)
They look too different to me!

Answer 39

life would be so much easier if we could use substitution...and not IF.

Answer 40

Then, this, again, proves that life is not easy :(

Answer 41

well i used 6 bits of paper , and got the same answer we had before,
bother

Answer 42

6... bits?! of paper?!
:(

Answer 43

\(LS=(2m+4)x^ny^{m+1}-(6m+6)x^{n+1}y^m\)

Answer 44

checking RS

Answer 45

2m+4 = 3n+3
6m+6 = 4n+8
m=1
n=1

Answer 46

just one silly mistale!

Answer 47

*mistake

Answer 48

oh

Answer 49

i see it now , such a tiny tiny mistale

Answer 50

thank you so much @UnkleRhaukus , i learned a new method today...R=x^m y^n

Answer 51

OK, so now we have the integrating factor
\[R(x,y)=\mu(x,y)=xy\],
lets integrate!

Answer 52

actually , i wanted to see the look at @RolyPoly face, when he/she finds out how \(\tiny tiny\) the error was :P

Answer 53

\[\int \mu N\text dx\]\[=\int (2xy^3−6x^2y^2)\text dx\]\[=x^2y^3-2x^3y^2+g(y)\]

\[\int\mu M\text dy\]\[=\int(3x^2y^2−4x^3y)\text dy\]\[=x^2y^3-2x^3y^2+h(x)\]

\[\implies g(y)=h(x)=0\]

\[f(x,y)=x^2y^3-2x^3y^2=c\]

Answer 54

or the other way

\[f(x,y)=\int \mu N\text dx=x^2y^3-2x^3y^2+g(y)\]

\[\frac{f(x,y)}{\text dy}=3x^2y^2-4x^3y+g'(y)=\mu M=3x^2y^2−4x^3y\]
\[\implies g'(y)=0\]\[g(y)=c_1\]
\[f(x,y)=x^2y^3-2x^3y^2+c_1=0\]

Answer 55

\[c_1=-c\]

Answer 56

woo!

Answer 57

wasn't that fun?

Answer 58

\[ \begin{array}l\color{red}{\text{y}}\color{orange}{\text{e}}\color{#e6e600}{\text{s}}\color{green}{\text{,}}\color{blue}{\text{ }}\color{purple}{\text{o}}\color{purple}{\text{f}}\color{red}{\text{c}}\color{orange}{\text{o}}\color{#e6e600}{\text{u}}\color{green}{\text{r}}\color{blue}{\text{s}}\color{purple}{\text{e}}\color{purple}{\text{ }}\color{red}{\text{:}}\color{orange}{\text{)}}\color{#e6e600}{\text{}}\end{array} \]

Answer 59

Mummy!!!! I want to die :'(

Answer 60

lol :P

Answer 61

plz don't die.

Answer 62

She! It's a she!

Answer 63

Thanks for rescue!!!! I'm sorry for my silly mistake!!

Answer 64

\[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#e6e600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]

Answer 65

*mistale

Answer 66

Hmmm.. So must I use
\[\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R\] to find R?! You know it's a pain :(

Answer 67

yes, it is pain, but there is different kind of pleasure when we arrive at correct answer after all that!

Answer 68

I must take this pain then..
Thanks again for all of your help!! Much appreciated!!
(Btw, who wants the medal?)

Answer 69

if you were wondering what the solution 'looks' like ,...

Answer 70

solutions*

Answer 71

Ugly :(
How do you get the plot?

Answer 72

i used a graphing program ,

Answer 73

That is..?