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

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

Step 1: Check exactness

With $M=xy^2$, $N=x(xy-1)$: $M_y=2xy$, $N_x=2xy-1$; since $M_y\ne N_x$, the equation is not exact.

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

Exactness requires $\dfrac1\mu\dfrac{d\mu}{dp}=\dfrac{M_y-N_x}{Ny-Mx}$. Here $M_y-N_x=1$ and $Ny-Mx=xy(xy-1)-x^2y^2=-xy=-p$, so

\[\frac{1}{\mu}\frac{d\mu}{dp}=\frac{1}{-p}=-\frac1p,\]

a function of $p=xy$ alone.

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

$\ln|\mu|=-\ln|p|$, so $\mu(p)=\dfrac1p=\dfrac{1}{xy}$.

Step 4: Multiply and verify exactness

$M_1=y$, $N_1=x-\dfrac1y$. Then $\partial_yM_1=1=\partial_xN_1$, so the equation is exact.

Step 5: Find the potential function

Integrating $F_x=M_1$ gives $F=xy+g(y)$. Matching $F_y=x+g'(y)$ to $N_1$ gives $g'(y)=-\dfrac1y$, so $g(y)=-\ln|y|$.