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

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

Step 1: Rewrite and test exactness

$(x^2y^3+y)\,dx+(x^3y^2-x)\,dy=0$, with $M=x^2y^3+y$, $N=x^3y^2-x$: $M_y=3x^2y^2+1$, $N_x=3x^2y^2-1$, so $M_y\ne N_x$: not exact.

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

For $\mu=\mu(p)$, exactness of $\mu M\,dx+\mu N\,dy=0$ requires $\mu(M_y-N_x)=(Ny-Mx)\mu'(p)$ (writing $u=xy$, $u_x=y,\,u_y=x$). Here $M_y-N_x=2$ and

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

so $\dfrac1\mu\dfrac{d\mu}{dp}=\dfrac{2}{-2p}=-\dfrac1p$, a function of $p=xy$ alone.

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

$\ln|\mu|=-\ln|p|$, so $\mu(p)=\dfrac1p=\dfrac1{xy}$.

Step 4: Multiply and verify exactness

$M_1=xy^2+\dfrac1x$, $N_1=x^2y-\dfrac1y$. Then $(M_1)_y=2xy=(N_1)_x$: exact.

Step 5: Find the potential function

$F=\displaystyle\int\left(xy^2+\frac1x\right)dx=\frac{x^2y^2}{2}+\ln|x|+g(y)$; matching $F_y=x^2y+g'(y)$ to $N_1$ gives $g'(y)=-\dfrac1y$, so $g(y)=-\ln|y|$.