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## gerryliyana 2 years ago Differential equations

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1. gerryliyana

I have: $(2x^{2} + y) dx + (x^{2} y - x) dy = 0$

2. gerryliyana

did you get it @sirm3d ??

3. 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$

4. gerryliyana

I got $2x + \frac{ 1 }{ 2 } y^{2} -\frac{ y }{ x } = C$

5. sirm3d

that is also correct. the difference in our answer is the choice of the constant.

6. gerryliyana

very nice..., Thank You for Checking :)

7. mahmit2012

|dw:1354568766432:dw|

8. gerryliyana

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 $2x + \frac{ 1 }{ 2 } y^{2} - \frac{ y }{ x } =C$

9. gerryliyana

I thank all of you guys.., , @sirm3d and @mahmit2012 thank u so much :)

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