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RolyPoly
Group Title
\[ (2y^2  6xy)dx + (3xy4x^2)dy =0\]
Find an integrating factor of the form \(x^ny^m\) and solve the equation...
 one year ago
 one year ago
RolyPoly Group Title
\[ (2y^2  6xy)dx + (3xy4x^2)dy =0\] Find an integrating factor of the form \(x^ny^m\) and solve the equation...
 one year ago
 one year ago

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abb0t Group TitleBest ResponseYou've already chosen the best response.0
1st check to make sure that they are exact.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\(M = 2y^2  6xy \) \(N= 3xy4x^2\) \(\frac{\partial M}{\partial y}=4y  6x\) \(\frac{\partial N}{\partial x}=3y8x\) \[\frac{\frac{\partial N}{\partial x}\frac{\partial M}{\partial y}}{M}=\frac{3y8x(4y  6x)}{2y(y3x)}\]Doesn't go right..
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ ∂M }{ ∂x }=\frac{ ∂N }{ ∂y }\]
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
check your partial derivatives. they should match up
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
either that or your teacher is horrible and wants you to solve this another way ll
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
integrating factor!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
But I couldn't find a suitable integrating factor... :'(
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
ah yes this is not an exact solution one
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
UnkleR is right. Find the integrating factor to make it exact.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
That's why I tried that partial..  partial.. / ... to find one..
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
is that a hint... try getting one in such that it looks like above
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
You were on the right path.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ M_yN_x }{ M }\]
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
That's your integrating factor.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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..
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\] \[R=R(x,y)=x^ny^m\] \[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\] \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
How... does... that... work...?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
which bit
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\]And \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
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 )
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
so you know \[R,M,N\]substitute these in
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
take the partial derivatives
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Is R the integrating factor?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
So, I should multiply the equation by R, right?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\[R=x^ny^m\]\[M=2y^26xy\]\[N=3xy4x^2\] \[\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R\] \[LS\]\[=mx^ny^{m1}(2y^26xy) + (4y6x)x^ny^m\]\[=2mx^ny^{m+1}6mx^{n+1}y^m + 4y^{m+1}x^n – 6x^{n+1}y^m\]\[(2m+4)x^ny^{m+1}(6m+1)x^{n+1}y^m\] \[RS\]\[=nx^{n1}y^m(3xy4x^2) + (3y8x)x^ny^m\]\[=3nx^ny^{m+1}4nx^{n+1}y^m+3x^ny^{m+1} – 8x^{n+1}y^m\]\[(3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m\] 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 :\
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
hmm, i cannot see any error in your working , how do you know its not right?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Because the book whispered me the answer :(
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[\tiny \text{can you whisper it to me too please , }\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\[\tiny \color{white}{\mu =xy} \]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
i see
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\huge \color{red}{\mu =xy}\) now, i see also.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
well we were close \[x^2y^{5/2}(2y^2  6xy)\text dx + x^2y^{5/2}(3xy4x^2)\text dy =0\] \[(2x^2y^{9/2}  6x^3y^{7/2})\text dx +(3x^3y^{7/2}4x^4y^{5/2})\text dy =0\] \[\frac{\partial MR}{\partial y}=9x^2y^{7/2}21x^3y^{5/2}\] \[\frac{\partial NR}{\partial x}=9x^2y^{7/2}16x^3y^{5/2}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[21\sim16\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I.... don't .... understand.... :(
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[R=\mu\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Hmm... I should do it all over again?!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
xy vs x^2 y^(5/2) They look too different to me!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
life would be so much easier if we could use substitution...and not IF.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Then, this, again, proves that life is not easy :(
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
well i used 6 bits of paper , and got the same answer we had before, bother
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
6... bits?! of paper?! :(
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(LS=(2m+4)x^ny^{m+1}(6m+6)x^{n+1}y^m\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
checking RS
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
2m+4 = 3n+3 6m+6 = 4n+8 m=1 n=1
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
just one silly mistale!
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
i see it now , such a tiny tiny mistale
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
thank you so much @UnkleRhaukus , i learned a new method today...R=x^m y^n
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
OK, so now we have the integrating factor \[R(x,y)=\mu(x,y)=xy\], lets integrate!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
actually , i wanted to see the look at @RolyPoly face, when he/she finds out how \(\tiny tiny\) the error was :P
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[\int \mu N\text dx\]\[=\int (2xy^3−6x^2y^2)\text dx\]\[=x^2y^32x^3y^2+g(y)\] \[\int\mu M\text dy\]\[=\int(3x^2y^2−4x^3y)\text dy\]\[=x^2y^32x^3y^2+h(x)\] \[\implies g(y)=h(x)=0\] \[f(x,y)=x^2y^32x^3y^2=c\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
or the other way \[f(x,y)=\int \mu N\text dx=x^2y^32x^3y^2+g(y)\] \[\frac{f(x,y)}{\text dy}=3x^2y^24x^3y+g'(y)=\mu M=3x^2y^2−4x^3y\] \[\implies g'(y)=0\]\[g(y)=c_1\] \[f(x,y)=x^2y^32x^3y^2+c_1=0\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[c_1=c\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
wasn't that fun?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\[ \begin{array}l\color{red}{\text{y}}\color{orange}{\text{e}}\color{#e6e600}{\text{s}}\color{green}{\text{,}}\color{blue}{\text{ }}\color{purple}{\text{o}}\color{purple}{\text{f}}\color{red}{\text{c}}\color{orange}{\text{o}}\color{#e6e600}{\text{u}}\color{green}{\text{r}}\color{blue}{\text{s}}\color{purple}{\text{e}}\color{purple}{\text{ }}\color{red}{\text{:}}\color{orange}{\text{)}}\color{#e6e600}{\text{}}\end{array} \]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Mummy!!!! I want to die :'(
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
plz don't die.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
She! It's a she!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Thanks for rescue!!!! I'm sorry for my silly mistake!!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#e6e600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
*mistale
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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 :(
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes, it is pain, but there is different kind of pleasure when we arrive at correct answer after all that!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I must take this pain then.. Thanks again for all of your help!! Much appreciated!! (Btw, who wants the medal?)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
if you were wondering what the solution 'looks' like ,...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
solutions*
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Ugly :( How do you get the plot?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
i used a graphing program ,
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
That is..?
 one year ago
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