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For the first-order ODE: $$(2y^2-3xy)dx + (4xy-3x^2)dy = 0,$$

find the integrating factor (of form below) that makes the ODE exact... $$\mu(x,y)=x^my^n$$

My attempt:

$$M_y = 4y - 3x; N_x = 4y - 6x$$

I'm pretty much stuck at this point...any help?

Thanks

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3 Answers 3

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$$\frac{\partial}{\partial y}\left(\left(2y^2-3xy\right)x^my^n\right)=x^{m-1}y^{n-1}\left((4+2n)xy^2-(3+3n)x^2y\right) $$ $$\frac{\partial}{\partial x}\left(\left(4xy-3x^2\right)x^my^n\right)=x^{m-1}y^{n-1}\left((4+4m)xy^2-(3m+6)x^2y\right)$$ Since we want $M_y=N_x$, we have to equate the corresponding coefficients: $$\begin{cases} 4+4m=4+2n \\ 3m+6=3+3n \end{cases} $$ The solution is $m=1, n=2$, so the integrating factor is $\mu(x,y)=xy^2$. After you multiply by it, the differential equation is exact, and can be solved in a standard way. The answer is $$x^2y^4-x^3y^3=C $$

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You could group the terms by total degree, $$ 2[y^2\,dx+2xy\,dy]-3[xy\,dx+x^2\,dy]=0, $$ which makes it very easy to see what the integrating factors for each term are $$ 2\,d(xy^2)-3x\,d(xy)=0. $$ Use the identified $u=xy^2$ and $v=xy$ as new variables, see that $x=v^2/u$, so that multiplication with $u=xy^2$ renders the full expression integrable, $$ 2u\,du-3v^2\,dv=0\implies u^2-v^3=C. $$

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$$(2y^2-3xy)dx + (4xy-3x^2)dy = 0,$$ Rearrange terms: $$(-3xydx-3x^2dy)+2(y^2dx+2xydy)=0$$ $$-3xdxy+2(y^2dx+xdy^2)=0$$ $$-3xd(xy)+2d(xy^2)=0$$ Integrating factor is $$\mu (x,y)=xy^2$$ $$-3(xy)^2d(xy)+2xy^2d(xy^2)=0$$ Integrate $$(xy)^3-(xy^2)^2=K$$ $$\boxed {x^3y^3-x^2y^4=K}$$

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