Yes: $y=2-x$. Its derivative is $-1$, while substitution into the right-hand side gives $-2/(6-4)=-1$. This is the equilibrium $v=2$ of the transformed autonomous differential equation; it must be restored separately after dividing by $v-2$ in the separated solution.

It is the only straight-line solution. Substituting $y=mx+d$ into the original equation shows that $m\ne -1$ would require $m=-1/3$ from the $x$ coefficients, but the constant coefficients would then give $4/3=0$. For $m=-1$, the constant equation gives $d=2$.

With $v=x+y$, the chain rule givesFor $v\ne 2$, this separable differential equation givesRestoring $v$ and absorbing constants, the nonconstant branches satisfyThe constant branch lost in division is $v=2$, giving the additional solution $\boxed{y=2-x}$. All branches must avoid $v=4/3$, where the original differential equation is undefined.

On the loop, $y=tx$, so $dy=tdx+xdt$ and $xdy-ydx=x^{2}dt$. The traversal is anticlockwise. The area formula from Green's theorem becomesSet $u=1+t^3$ to obtainAlthough the parameter interval is unbounded, the return to the origin closes the loop and the integral converges.

For $a>0$, both coordinates are nonnegative exactly on the loop branch $t\ge 0$. It leaves the origin at $t=0$ and returns there in the limit $t\to +\infty$. Thus the loop is traced once bywith the limiting endpoint identified with the starting point. The same parameter range traces the rotated loop when $a<0$.

The orientation assumption gives $\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})>0$. The same open half-space is therefore characterized byEquality would put the point in the separating plane, rather than strictly on the same side.

The independent vectors $\mathbf{a},\mathbf{b}$ span the plane through the origin, and their cross product is a nonzero normal vector. Thus the required nonparametric equation is

The integrand field is exactly the curl computed in part (a), so use its polynomial vector potential $\mathbf{G}=(z^3,x^3,y^3)/3$. The upper cap meets the plane $z=0$ in an ellipse. Since the parameters are only assumed nonzero, set $a_0=|a|$ and $b_0=|b|$ to use positive semiaxes. Its correctly oriented boundary isOn that boundary, $G_x=0$ and $dz=0$, so only $G_{y}dy=x^{3}dy/3$ contributes. The Stokes theorem therefore givesThe last integral follows by expanding ${\cos }^{4}t=(3+4\cos 2t+\cos 4t)/8$. ThusThe result is independent of $c$. A sign check uses the divergence theorem: closing the cap with its base disk adds outward normal $-\hat{\mathbf{z}}$ there. Since $\nabla \cdot \mathbf{F}=0$, the cap flux equals ${\int }_{\mathrm{disk}}{x}^{2}dxdy>0$, consistent with the boxed expression and its absolute values.

For a smooth oriented surface $S$ with piecewise smooth boundary $C=\partial S$ and a continuously differentiable vector field $\mathbf{G}$, the Stokes theorem statesThe boundary orientation is fixed by the right-hand rule: curling the fingers along the positive boundary direction makes the thumb point along the surface normal. For an outward upper cap, its boundary ellipse is traversed counterclockwise when viewed from the positive $z$ direction. The figure shows two possible caps with the same oriented boundary, anticipating the last part.

Take the curl component by component:Its divergence is zero, as expected because the divergence of a curl is zero.

Yes. For any sufficiently smooth scalar $\chi$, the gauge transformationleaves its curl unchanged because $\nabla \times \nabla \chi =0$. This freedom is much larger than adding a constant vector: for example, $\chi =xy$ adds the nonconstant vector $(y,x,0)$. Two vector potentials for the same field differ by a curl-free field; on a simply connected domain that difference is a gradient, while nontrivial topology may allow additional global freedom.

On a connected domain, a vector field is a conservative vector field if and only if it is the gradient of a single-valued scalar potential, $\boxed{\mathbf{H}=\nabla \varphi }$. Equivalently, its line integral is path independent, or its integral around every closed curve is zero. For a continuously differentiable field on a simply connected domain, this is equivalent to $\nabla \times \mathbf{H}=0$. Without the domain hypothesis, vanishing curl alone is only a local test and is not the necessary-and-sufficient global statement.

Apply the recurrence repeatedly down to $m=2$:Including the base factor $1/(n-1)$ and rewriting the products as factorials givesThe formula also includes $m=2$ because $0!=1$. It is symmetric in $m,n$, consistent with the integration by substitution $x=1/t$. Equivalently it is the Euler beta function $B(m-1,n-1)$, although that identification is not needed for the recurrence derivation.

For $m=2$, evaluate the improper integral directly:For $m\ge 3$, use integration by parts with $u=x^{m-2}$ and $dv=(1+x{)}^{-(m+n-2)}dx$. Then $v=-(1+x{)}^{-(m+n-3)}/(m+n-3)$ andThe boundary term vanishes: at zero $m-2>0$, while at infinity its power is $1-n<0$. The restriction $m\ge 3$ belongs to this recurrence step; the separate $m=2$ base case must not be replaced by a formal zero times a divergent $I_{1,n}$.

The binomial series gives $(1+ϵ{)}^{-\alpha }=1-\alpha ϵ+O({ϵ}^2)$. ThereforeTo justify the limiting approximation over the infinite interval, also use the right-endpoint sum. It is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$, and monotonicity gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds have the same limit. The geometric partition bounds for a power-law improper integral and the squeeze theorem thus proveThis bounding argument matters because the geometric partition has unbounded widths far along the half-line; a finite-interval maximum-mesh argument cannot simply be applied to the entire infinite partition at fixed $ϵ$.

Index the stated geometric progression by $x_i=(1+ϵ{)}^i$, $i=0,1,\dots$. Then $\Delta x_i=ϵ(1+ϵ{)}^i$, and its left-endpoint contribution isHence the approximation is a geometric series withSince $ϵ>0$ and $\alpha >0$, the ratio satisfies $0<q<1$ and the series converges. The integrand is decreasing, so the left-endpoint sum is an upper bound for the improper integral.

Partition the finite interval as $a=x_0<x_1<\cdots <x_N=b$, with $\Delta x_i=x_{i+1}-x_i$. For a Riemann integrable function, the left-endpoint Riemann sums satisfyThe maximum subinterval width, rather than merely the number of subintervals, must tend to zero. Continuity of $f$ on the closed interval is a sufficient integrability hypothesis. More generally, any chosen sample point in each subinterval gives the same definite integral under these assumptions.

Impose the five initial conditions at zero on the preceding general solution. They giveThe second and fourth equations give $C_1=C_2$ and $C_4=0$. The third and fifth then yield $C_1=C_2=1/4$ and $C_3=-1/2$, while the first gives $C_0=1$. ThusFor a direct check, the difference of its fifth and first derivatives is $x$. At zero its value is one; its first, second, third and fourth derivatives vanish. In particular the quadratic term cancels the otherwise nonzero second derivative of the hyperbolic-cosine and cosine combination.

Add the particular solution to the complementary function. The real general solution isThe five constants remain arbitrary until initial data are specified.

A quadratic polynomial is sufficient: set $y_p=A{x}^2$. Its fifth derivative vanishes, so $y_p^{(5)}-y_p^{\prime }=-2Ax$. Matching the forcing gives $A=-1/2$, hence a particular integral is

In the homogeneous equation, substitute the exponential function $e^{kx}$. The characteristic equation isIts distinct roots are $0,1,-1,i,-i$. Combining the complex-conjugate pair into real sine and cosine gives the complementary functionFive linearly independent terms supply the required five homogeneous degrees of freedom.