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sirm3d Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{ \partial }{ \partial y }(2xy)=2x,\frac{ \partial }{ \partial x }(y^23x^2)=6x\]\[\frac{ (\frac{ \partial M }{ \partial y }  \frac{ \partial N }{ \partial x })}{ 2xy }=\frac{ 4 }{ y }\]the integrating factor that will make the DE exact is \[\large e^{\int\limits_{}^{}\frac{ 4 }{ y }dy}=\frac{ 1 }{ y^4 }\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
if you have not reached this part of the course, an alternate solution is \[\large x=vy\] and \[\large dx = vdy + ydv\] where the original DE is now separable in v and y.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
have you solved this problem @RolyPoly ?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I'm on the way solving it.. Sorry, please give me some more time...
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I'm not quite sure if I have learnt this way before. I have no memory of seeing this, but it appears in my book..
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
homogeneous function, doesn't it ring a bell?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
my comment on your attached file: \[\huge F=\frac{ x^2 }{ y^3 }+g_1(y)\] NOT \[\huge F_x\] and \[\huge F=\frac{1}{y}+\frac{x^2}{y^3}+g_2(x)\] instead of \[\huge F_y\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
:O That's how my teacher writes... \(F_x\) means integrating w.r.t. x, I think :S
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
about that alternate solution \[\large \deg(2xy)=2;\deg(y^2)=2;\deg(3x^2)=2\] the degrees of the terms are identical. a substitution \[\large y = ux \text{ or }x=vy\] will transform the DE that is solvable by separation
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
taking \[x=vy \text{ and } dx=vdy+ydv\]\[\large 2(vy)y(vdy+ydv)+(y^2v^2y^2)dy=0\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I suppose it is separable, but somehow, I can't separate it. I guess I have to check my calculations again first :/
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
collecting terms of dy and dv, \[\large (2v^2y^2+y^2v^2y^2)dy+2vy^3dv=0\]or\[\large y^2(v^2+1)dy+2vy^3dv=0\]divide by \[\large y^3(v^2+1)\] and the DE becomes \[\large \frac{dy}{y}+\frac{2vdv}{v^2+1}=0\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
How come... you can get it but I can't... :'(
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
expand after replacing x by vy and dx by vdy + ydv collect terms of dy and dv, simplify and factor.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I tried w = y/x but it doesn't seem to be working for me :(
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Sorry, I got to go now.. I'd keep trying until I got the answer. Thanks for your help!!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
@sirm3d You're right about the F and Fx thing! Sorry for my earlier mistakes!!! When I tried w=y/x, I got something quite ugly :S
 one year ago
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