Show that y(1+xy) dx-x dy=0 has an integrating factor of the…

Show that $y(1+xy)\,dx-x\,dy=0$ has an integrating factor of the form $\mu=\mu(x)$ or $\mu=\mu(y)$, and find its general integral.

Step 1: Expand and test exactness

$M=y+xy^2,\ N=-x$. Then $M_y=1+2xy$, $N_x=-1$, so $M_y\ne N_x$: not exact.

Step 2: Rule out $\mu(x)$, find $\mu(y)$

\[\frac{M_y-N_x}{N}=\frac{(1+2xy)+1}{-x}=-\frac{2(1+xy)}{x}\]

depends on both variables, so no $\mu(x)$ exists. But

\[\frac{N_x-M_y}{M}=\frac{-1-(1+2xy)}{y+xy^2}=\frac{-2(1+xy)}{y(1+xy)}=-\frac2y,\]

a function of $y$ alone, so $\mu(y)$ exists with $\mu'/\mu=-2/y$, giving $\mu(y)=y^{-2}$.

Step 3: Multiply and verify exactness

Multiplying by $1/y^2$: $M_1=\dfrac1y+x,\ N_1=-\dfrac{x}{y^2}$, and $(M_1)_y=-\dfrac1{y^2}=(N_1)_x$: exact.

Step 4: Integrate to find $F$

$F=\displaystyle\int\left(\frac1y+x\right)dx=\frac{x}{y}+\frac{x^2}{2}+g(y)$; matching $F_y=N_1$ forces $g'(y)=0$.