Show that (xy^2-y^3) dx+(1-xy^2) dy=0 has an integrating…
Show that $(xy^2-y^3)\,dx+(1-xy^2)\,dy=0$ has an integrating factor of the form $\mu=\mu(x)$ or $\mu=\mu(y)$, and find its general integral.
Explanation of Method
With $M=xy^2-y^3,\ N=1-xy^2$, show $\mu(x)$ fails but $\dfrac{N_x-M_y}{M}=-\dfrac2y$ depends only on $y$, giving $\mu(y)=1/y^2$.
Step 1: Test exactness
$M_y=2xy-3y^2$, $N_x=-y^2$; since $M_y\ne N_x$, the equation is not exact.
Step 2: Rule out $\mu(x)$, find $\mu(y)$
\[\frac{M_y-N_x}{N}=\frac{2xy-2y^2}{1-xy^2}\]
depends on both variables. But
\[\frac{N_x-M_y}{M}=\frac{-y^2-2xy+3y^2}{xy^2-y^3}=\frac{2y(y-x)}{y^2(x-y)}=-\frac2y,\]
a function of $y$ alone, so $\mu(y)=y^{-2}$.
Step 3: Multiply and verify exactness
Multiplying by $1/y^2$: $M_1=x-y,\ N_1=\dfrac{1}{y^2}-x$, and $(M_1)_y=-1=(N_1)_x$: exact.
Step 4: Integrate to find $F$
$F=\displaystyle\int(x-y)\,dx=\dfrac{x^2}{2}-xy+g(y)$; matching $F_y=-x+g'(y)$ to $N_1$ gives $g'(y)=\dfrac1{y^2}$, so $g(y)=-\dfrac1y$.
Result
Final answer:
\[\boxed{\,\frac{x^2}{2}-xy-\frac1y=C\,}\qquad\text{equivalently}\qquad\boxed{\,x^2y-2xy^2-2=Cy\,}\]
(verified with sympy: after multiplying by $\mu=1/y^2$ the equation is exact, and $F_x-\mu M\equiv0$, $F_y-\mu N\equiv0$ identically.)