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

This Question is Closed

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.01st check to make sure that they are exact.

anonymous
 3 years ago
Best 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..

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ ∂M }{ ∂x }=\frac{ ∂N }{ ∂y }\]

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0check your partial derivatives. they should match up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0either that or your teacher is horrible and wants you to solve this another way ll

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2integrating factor!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But I couldn't find a suitable integrating factor... :'(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah yes this is not an exact solution one

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0UnkleR is right. Find the integrating factor to make it exact.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's why I tried that partial..  partial.. / ... to find one..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that a hint... try getting one in such that it looks like above

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0You were on the right path.

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ M_yN_x }{ M }\]

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0That's your integrating factor.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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..

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How... does... that... work...?

anonymous
 3 years ago
Best 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 \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2well 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 )

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2so you know \[R,M,N\]substitute these in

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2take the partial derivatives

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is R the integrating factor?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, I should multiply the equation by R, right?

anonymous
 3 years ago
Best 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 :\

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because the book whispered me the answer :(

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\tiny \color{white}{\mu =xy} \]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I.... don't .... understand.... :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... I should do it all over again?!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0xy vs x^2 y^(5/2) They look too different to me!

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1life would be so much easier if we could use substitution...and not IF.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then, this, again, proves that life is not easy :(

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.06... bits?! of paper?! :(

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.12m+4 = 3n+3 6m+6 = 4n+8 m=1 n=1

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2i see it now , such a tiny tiny mistale

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

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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mummy!!!! I want to die :'(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for rescue!!!! I'm sorry for my silly mistake!!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm.. 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 :(

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I must take this pain then.. Thanks again for all of your help!! Much appreciated!! (Btw, who wants the medal?)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2if you were wondering what the solution 'looks' like ,...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ugly :( How do you get the plot?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2i used a graphing program ,
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