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Find a multiplicative factor $\beta(x,y)=x^n y^m$ such that $(2y^2-6xy) dx+(3xy-4x^2) dy=0$ is exact.

So I know an exact equation is in the form of $M(x,y)dx+N(x,y)dy = 0$ and that: $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$ They should be equal for the given equation to be exact. I also know a multiplicative factor can be multiplied to the equation to make the equation exact.

Up until now, I have found a multiplicative factor either in terms of $x$ only or $y$, by using the formulas.

To solve this problem, if I were to multiply the given multiplicative factor and equate the partial derivatives, it still wouldn't give me the respective values of $m$ and $n$ as there is one equation with $2$ unknown variables. Should I perhaps first try to find the multiplicative factor in terms of $x$, then in terms of $y$ and the product of do would be the answer?

Is that the correct way to go about solving this?

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Hint: $x^ny^m(2y^2-6xy)dx+x^ny^m(3xy-4x^2)dy=0$

$(2x^ny^{m+2}-6x^{n+1}y^{m+1})dx+(3x^{n+1}y^{m+1}-4x^{n+2}y^{m})dy=0$

$$\frac{\partial}{\partial y}(2x^ny^{m+2}-6x^{n+1}y^{m+1})=2(m+2)x^ny^{m+1}-6(m+1)x^{n+1}y^{m}=x^ny^m(2(m+2)y-6(m+1)x)$$ $$\frac{\partial}{\partial x}(3x^{n+1}y^{m+1}-4x^{n+2}y^{m})=3(n+1)x^ny^{m+1}-4(n+2)x^{n+1}y^{m}=x^ny^m(3(n+1)y-4(n+2)x)$$ \begin{cases} -6(m+1)=-4(n+2),\\ 2(m+2)=3(n+1). \end{cases} gives us $m=1=n$.

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  • $\begingroup$ Your welcome. . $\endgroup$ – Nosrati Mar 25 '17 at 8:51

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