Show that (x^m+y) dx-x dy=0 has an integrating factor of the…
Show that $(x^m+y)\,dx-x\,dy=0$ has an integrating factor of the form $\mu=\mu(x)$ or $\mu=\mu(y)$, and find its general integral.
Explanation of Method
With $M=x^m+y,\ N=-x$, show $\dfrac{M_y-N_x}{N}=-\dfrac2x$ depends only on $x$, giving $\mu(x)=1/x^2$; integrate, splitting into the cases $m\ne1$ and $m=1$.
Step 1: Test exactness and find $\mu(x)$
$M_y=1$, $N_x=-1$, so $M_y\ne N_x$: not exact. But $\dfrac{M_y-N_x}{N}=\dfrac{1-(-1)}{-x}=-\dfrac2x$, a function of $x$ alone, so $\mu(x)=x^{-2}$.
Step 2: Multiply and verify exactness
Multiplying by $1/x^2$: $M_1=x^{m-2}+\dfrac{y}{x^2},\ N_1=-\dfrac1x$, and $(M_1)_y=\dfrac1{x^2}=(N_1)_x$: exact.
Step 3: Integrate to find $F$ (case $m\ne1$)
$F=\displaystyle\int-\dfrac1x\,dy=-\dfrac{y}{x}+h(x)$; matching $F_x=\dfrac{y}{x^2}+h'(x)$ to $M_1$ gives $h'(x)=x^{m-2}$, so $h(x)=\dfrac{x^{m-1}}{m-1}$.
Step 4: The special case $m=1$
If $m=1$, $h'(x)=x^{-1}$ instead, so $h(x)=\ln|x|$.
Result
Final answer:
For $m\ne1$: \[\boxed{\,\frac{x^{m-1}}{m-1}-\frac{y}{x}=C\,}\qquad\text{equivalently}\qquad\boxed{\,x^m-(m-1)y=C(m-1)x\,}\]
For $m=1$: \[\boxed{\,\ln|x|-\frac{y}{x}=C\,}\qquad\text{equivalently}\qquad\boxed{\,x\ln|x|-y=Cx\,}\]
(verified with sympy for general symbolic $m$: after multiplying by $\mu=1/x^2$ the equation is exact, and $F_x-\mu M\equiv0$, $F_y-\mu N\equiv0$ identically in both cases.)