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anonymous
 4 years ago
Differential equations
anonymous
 4 years ago
Differential equations

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have: \[(2x^{2} + y) dx + (x^{2} y  x) dy = 0 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you get it @sirm3d ??

sirm3d
 4 years ago
Best 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\[\large 2x\frac{ y }{ x }+\frac{ y^2 }{ 2 }=\frac{ C }{ 2 }\]\[\large 4x^22y+2xy^2=Cx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got \[2x + \frac{ 1 }{ 2 } y^{2} \frac{ y }{ x } = C\]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.2that is also correct. the difference in our answer is the choice of the constant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0very nice..., Thank You for Checking :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354568766432:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank 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 \[2x + \frac{ 1 }{ 2 } y^{2}  \frac{ y }{ x } =C\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I thank all of you guys.., , @sirm3d and @mahmit2012 thank u so much :)
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