Differential equations

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Differential equations

Mathematics
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I have: \[(2x^{2} + y) dx + (x^{2} y - x) dy = 0 \]
did you get it @sirm3d ??
\[\large 2x^2dx+ydx+x^2ydy-xdy=0\]\[\large 2x^2dx-(xdy-ydx)+x^2ydy=0\]divide everything by x^2 \[\large2dx-\frac{ xdy-ydx }{ x^2 }+ydy=0\]thesecond term is the differential d(y/x)\[\large2dx-d\left(\frac{y}{x}\right)+ydy=0\]integrate\[\large 2x-\frac{ y }{ x }+\frac{ y^2 }{ 2 }=\frac{ C }{ 2 }\]\[\large 4x^2-2y+2xy^2=Cx\]

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I got \[2x + \frac{ 1 }{ 2 } y^{2} -\frac{ y }{ x } = C\]
that is also correct. the difference in our answer is the choice of the constant.
very nice..., Thank You for Checking :)
|dw:1354568766432:dw|
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) \[P (x) = \frac{ 1 }{ x^{2}y - x } \left[ 1- (2xy -1) \right] = \frac{ 2(1-xy) }{ x^{2}y-x}= -\frac{ 2 }{ x }\] \[\mu(x) = \exp (-2 \int\limits_{0}^{x} \frac{ dx }{ x }) = \exp(-2 \ln x) = e^{\ln x^{-2}} = x^{-2} = \frac{ 1 }{ x^{2}}\] \[\mu(x) = \exp(\int\limits_{0}^{x} P(x) dx)\] \[(2+ \frac{ y }{ x^{2} } ) dx + (y - \frac{ 1 }{ x }) dy = 0\] 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 \[2x + \frac{ 1 }{ 2 } y^{2} - \frac{ y }{ x } =C\]
I thank all of you guys.., , @sirm3d and @mahmit2012 thank u so much :)

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