Show that (2xy+x^2y+(y^3)/(3))dx+(x^2+y^2)dy=0 has an…

Show that $\left(2xy+x^2y+\dfrac{y^3}{3}\right)dx+(x^2+y^2)dy=0$ has an integrating factor of the form $\mu=\mu(x)$ or $\mu=\mu(y)$, and find its general integral.

Step 1: Set up $M,N$ and test exactness

With $M=2xy+x^2y+\dfrac{y^3}{3}$ and $N=x^2+y^2$:

\[M_y=2x+x^2+y^2,\qquad N_x=2x.\]

Since $M_y\ne N_x$, the equation is not exact.

Step 2: Find the integrating factor

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

a constant (in particular, a function of $x$ alone), so $\mu=\mu(x)$ exists with $\dfrac{\mu'}{\mu}=1$, giving $\mu(x)=e^{x}$.

Step 3: Multiply and verify exactness

Multiplying by $e^x$: $M_1=e^x\left(2xy+x^2y+\dfrac{y^3}{3}\right)$, $N_1=e^x(x^2+y^2)$. Then

\[(M_1)_y=e^x(2x+x^2+y^2)=(N_1)_x,\]

so the new equation is exact.

Step 4: Integrate to find the potential $F$

Integrate $F_y=N_1=e^x(x^2+y^2)$ in $y$: $F=e^x\left(x^2y+\dfrac{y^3}{3}\right)+h(x)$. Differentiating in $x$ and matching against $M_1$ forces $h'(x)=0$, so $h(x)$ is constant.