Show that (x^2+y^2)(x dy-y dx)=(a+x)x^4 dx (with parameter…

Show that $(x^2+y^2)(x\,dy-y\,dx)=(a+x)x^4\,dx$ (with parameter $a$) has an integrating factor of the form $\mu=\mu(x)$ or $\mu=\mu(y)$, and find its general integral.

Step 1: Rearrange into standard form

$(x^2+y^2)(x\,dy-y\,dx)-(a+x)x^4\,dx=0 \;\Rightarrow\; \big[x^2y+y^3+ax^4+x^5\big]dx - x(x^2+y^2)\,dy=0$, so $M=x^2y+y^3+ax^4+x^5,\ N=-x^3-xy^2$.

Step 2: Test exactness

$M_y=x^2+3y^2,\ N_x=-3x^2-y^2$, so $M_y\ne N_x$: not exact for any $a$.

Step 3: Find the integrating factor

\[M_y-N_x=(x^2+3y^2)-(-3x^2-y^2)=4x^2+4y^2=4(x^2+y^2),\]

\[\frac{M_y-N_x}{N}=\frac{4(x^2+y^2)}{-x(x^2+y^2)}=-\frac4x,\]

a function of $x$ alone (the parameter $a$ cancels entirely, since $a$ only entered $M$ through the term $ax^4$, which does not affect $M_y$). So $\mu(x)=x^{-4}$.

Step 4: Multiply and verify exactness

Multiplying by $x^{-4}$: $M_1=\dfrac{y}{x^2}+\dfrac{y^3}{x^4}+a+x,\ N_1=-\dfrac1x-\dfrac{y^2}{x^3}$, and $(M_1)_y=\dfrac1{x^2}+\dfrac{3y^2}{x^4}=(N_1)_x$: exact.

Step 5: Integrate to find $F$

$F=\displaystyle\int\Big(-\frac1x-\frac{y^2}{x^3}\Big)dy=-\frac{y}{x}-\frac{y^3}{3x^3}+h(x)$; matching $F_x=M_1$ gives $h'(x)=a+x$, so $h(x)=ax+\dfrac{x^2}{2}$.