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gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
I have: \[(2x^{2} + y) dx + (x^{2} y  x) dy = 0 \]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
did you get it @sirm3d ??
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
\[\large 2x^2dx+ydx+x^2ydyxdy=0\]\[\large 2x^2dx(xdyydx)+x^2ydy=0\]divide everything by x^2 \[\large2dx\frac{ xdyydx }{ x^2 }+ydy=0\]thesecond term is the differential d(y/x)\[\large2dxd\left(\frac{y}{x}\right)+ydy=0\]integrate\[\large 2x\frac{ y }{ x }+\frac{ y^2 }{ 2 }=\frac{ C }{ 2 }\]\[\large 4x^22y+2xy^2=Cx\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
I got \[2x + \frac{ 1 }{ 2 } y^{2} \frac{ y }{ x } = C\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
that is also correct. the difference in our answer is the choice of the constant.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
very nice..., Thank You for Checking :)
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354568766432:dw
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
thank u @mahmit2012 i'll post my answer \[(2x^{2} + y) dx + (x^{2} y x) dy = 0\] where \((2x^{2} + y) = M\) and \(x^{2}y  x = N\) \[\frac{ dM }{ dy } = 1\] \[\frac{ dN }{ dy } = 2xy  1\] So this isn't exact differential equations for integration factor: \(P (x) = \frac{ 1 }{ N } \left( \frac{ dM }{ dy }  \frac{ dN }{ dx } \right)\) because it's only f(x) \[P (x) = \frac{ 1 }{ x^{2}y  x } \left[ 1 (2xy 1) \right] = \frac{ 2(1xy) }{ x^{2}yx}= \frac{ 2 }{ x }\] \[\mu(x) = \exp (2 \int\limits_{0}^{x} \frac{ dx }{ x }) = \exp(2 \ln x) = e^{\ln x^{2}} = x^{2} = \frac{ 1 }{ x^{2}}\] \[\mu(x) = \exp(\int\limits_{0}^{x} P(x) dx)\] \[(2+ \frac{ y }{ x^{2} } ) dx + (y  \frac{ 1 }{ x }) dy = 0\] Look! its exact differential equations..., multiply all (1) terms by integration factor, becomes: \[x^{2} (2x^{2} + y) dx + x^{2} (x^{2}y  x) dy = 0\] Then, using grouping method: \[2 dx + \frac{ y }{ x^{2} } dx + y dy  \frac{ dy }{ x } = 0\] \[2dx + y dy +\frac{ y }{ x^{2} } dx  \frac{ dy }{ x } = 0\] \[d(2x) + d(\frac{ 1 }{ 2 }y^{2}) + d(\frac{ y }{ x }) = dC\] \[\int\limits d(2x) + \int\limits d(\frac{ 1 }{ 2 }y^{2}) + \int\limits d(\frac{ y }{ x }) = \int\limits dC\] becomes \[2x + \frac{ 1 }{ 2 } y^{2}  \frac{ y }{ x } =C\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
I thank all of you guys.., , @sirm3d and @mahmit2012 thank u so much :)
 one year ago
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