Show that (2x^3y^2-y) dx+(2x^2y^3-x) dy=0 has an integrating…

Show that $(2x^3y^2-y)\,dx+(2x^2y^3-x)\,dy=0$ has an integrating factor of the form $\mu=\mu(xy)$, and solve it.

Step 1: Check exactness

$M_y=4x^3y-1$, $N_x=4xy^3-1$; since $M_y\ne N_x$, the equation is not exact.

Step 2: Impose the condition for $\mu(p)$, $p=xy$

For $\mu=\mu(p)$, exactness requires $\dfrac{1}{\mu}\dfrac{d\mu}{dp}=\dfrac{M_y-N_x}{Ny-Mx}$. Here $M_y-N_x=4xy(x^2-y^2)$ and

\[Ny-Mx=y(2x^2y^3-x)-x(2x^3y^2-y)=-2x^2y^2(x^2-y^2),\]

so $\dfrac{1}{\mu}\dfrac{d\mu}{dp}=\dfrac{4xy(x^2-y^2)}{-2x^2y^2(x^2-y^2)}=-\dfrac{2}{xy}=-\dfrac2p$, a function of $p=xy$ alone.

Step 3: Solve for $\mu(p)$

$\ln|\mu|=-2\ln|p|$, so $\mu(p)=\dfrac{1}{p^2}=\dfrac{1}{x^2y^2}$.

Step 4: Multiply and verify exactness

$M_1=2x-\dfrac{1}{x^2y}$, $N_1=2y-\dfrac{1}{xy^2}$. Then $\partial_yM_1=\dfrac{1}{x^2y^2}=\partial_xN_1$, so the equation is exact.

Step 5: Find the potential function

Integrating $F_x=M_1$ gives $F=x^2+\dfrac{1}{xy}+g(y)$. Matching $F_y=-\dfrac{1}{xy^2}+g'(y)$ to $N_1$ gives $g'(y)=2y$, so $g(y)=y^2$.