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Let $u=x^{-5}$, which tends to zero as $|x|\to \infty$. Factor the denominator and use the cosecant expansion near zero:Substitution gives the first two nonzero terms and the requested omitted order:More precisely the next term is $7/(360x^{18})$, followed by $O(x^{-28})$. The outside factor $x^2$ is essential when converting the remainder order from powers of $u$ to powers of $x$.

Choose the upward orientation of the paraboloid cap. Its boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise as viewed from above. By Stokes theorem,On the boundary, $dz=0$ and $A\cdot dr=ydx-3xydy$. Parametrize by $x=2\cos \theta$, $y=2\sin \theta$, $0\le \theta \le 2\pi$. ThenThe second term integrates to zero over a full period, while the first contributes $-4\pi$. Equivalently, replace the cap by the upward disk at $z=1$: there $(\nabla \times A{)}_z=-3y-1$, whose integral is $-4\pi$ by symmetry.

An orientation is not stated in the original question. The displayed result uses the upward normal, also the outward normal on the curved top of the enclosed cap; the opposite orientation gives $4\pi$.

Take the sphere's outward normal. Since $F=k(x,y,z)$ with constant $k$, its divergence is $3k$. The divergence theorem givesThe translated centre does not enter the answer, because the divergence is constant. Equivalently the constant field produced by translating coordinates has zero net flux through a closed sphere. The inward orientation would negate the result.

Choose the plane normal to be $+\hat{y}$. Only the second field component contributes, and on $y=0$ it is $e^{-(x^2+z^2)}$. Thus the surface integral factors into two Gaussian integrals:The original question does not specify an orientation; choosing $-\hat{y}$ instead gives $-\pi$.

The printed first component $\sin y/y$ is undefined on the entire integration plane. Its removable continuous extension there is $1$, since ${\lim }_{y\to 0}\sin y/y=1$. The stated flux uses this natural extension; in fact the normal component alone is already well-defined and integrable on the plane, so the first component does not affect either oriented value.

The divergence isFor any closed piecewise smooth surface bounding a volume, use the divergence theorem with the outward orientation:Reversing the orientation still gives zero. The field is smooth everywhere, so there is no hidden singular source inside the surface.

View $a=a(b,c)$ with $b=b(d,e)$ and $c=c(d,e)$. Hold $e$ fixed and apply the multivariable chain rule:Dividing the changes along the fixed-$e$ variation by $dd$ gives exactly the two terms in the question. Thus the simpler expression isThe subscripts specify which coordinate is held fixed and must be retained.

Away from zeros of $\sin x$, an integrating factor for a differential one-form is $\mu (x)=1/\sin x$. Multiplication converts the form toTherefore the implicit solution family is $xy+\ln |\sin x|=C$. On each graph interval between consecutive multiples of $\pi$, this becomesNo such open interval contains zero. At $x=k\pi$, the undivided graph equation would read $\cos (k\pi )=0$, which is impossible for a finite derivative; hence no differentiable graph extends across one of these points.

If solutions are instead understood as arbitrary parametrized integral curves of the differential form, the vertical lines $x=k\pi$ are additional solutions: along them $dx=0$ and $x\sin xdy=0$. They were lost on division by $\sin x$ and are not graphs $y(x)$ over an interval.

If the form is $dF$ for a scalar function with continuous second partial derivatives, then $p=F_x$ and $q=F_y$. Equality of the mixed partial derivatives therefore requiresThis is necessary for an exact differential. For continuously differentiable coefficients on a simply connected open domain it is also sufficient; on a domain with holes, the equality alone need not give a single-valued global potential.

Set $w=z^3$. Then $w^2+w+1=0$ gives $w=e^{2\pi i/3}$ or $e^{-2\pi i/3}$. Taking all three cube roots of each yields six distinct solutions.

Equivalently, multiply the original polynomial by $z^3-1$ to obtain $z^9-1$. The original roots are exactly the ninth roots of unity that are not cube roots of unity: a cube root would give $1+1+1=3$, not zero. ThusAll six have modulus one, and their distinct arguments show that no roots have been repeated or omitted from the degree-six polynomial.

The complementary function is $C\cos 2x+D\sin 2x$. A particular solution is $(\sin x+\cos x)/3$, since applying $d^2/d{x}^2+4$ to either sine or cosine multiplies it by $3$. ThusAt zero, $C+1/3=3$, giving $C=8/3$. At $x=\pi /4$, the cosine term vanishes and the particular term is $\sqrt{2}/3$, so $D=-\sqrt{2}/3$. ConsequentlyThe two boundary conditions determine the two constants uniquely because their evaluation matrix on $\cos 2x,\sin 2x$ is the identity.

Construct the scalar potential $F(x,y)=y{e}^{x^2}-x^2$. Its differential iswhich proves exactness directly. Multiplying the differential equation by $e^{x^2}$ shows that a solution satisfies $dF/dx=0$, so $y{e}^{x^2}-x^2=C$. The initial value at zero gives $C=-1$. ThereforeThe factor $e^{x^2}$ is also an integrating factor for the associated first-order linear equation.

Set $v=y^{\prime }$ to reduce the second-order linear differential equation to $x{v}^{\prime }+v=0$. Equivalently, $(xv{)}^{\prime }=0$, so $v=C_1/x$ on any interval excluding zero. The derivative condition at $x=1$ gives $C_1=2$.

Integrating on the positive interval containing the initial point yields $y=2\ln x+C_2$. The value $y(1)=-1$ then fixes $C_2=-1$, givingIts derivative is $2/x$, and direct substitution verifies the equation. The selected solution diverges as $x\downarrow 0$ and cannot extend through zero as a finite-valued function.