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The first field has vanishing curl and is the gradient of a monomial:For the second field, direct component differentiation givesIt is not identically zero. Thus ${\mathbf{F}}_1$ is conservative and ${\mathbf{F}}_2$ is not on ${\mathbb{R}}^3$. The conditional path-integral requests therefore apply to ${\mathbf{F}}_1$.

Substitution first makes the square-root branch transparent:On the given interval this is exactly $\sin x$, so direct differentiation givesin agreement with the chain-rule calculation. Outside this interval one must retain the absolute value; there are cusps at integer multiples of $\pi$.

Differentiate the polar form of a complex number using the product rule and chain rule:A second differentiation givesThe two contributions to $2\dot{r}\dot{\theta }$ arise separately from differentiating $r\dot{\theta }$ and from differentiating the moving exponential factor.

Write $\mathbf{a}=(a_x,a_y,a_z)$. The cross product gives ${\mathbf{v}}_1=(a_{y}z-a_{z}y,a_{z}x-a_{x}z,a_{x}y-a_{y}x)$, whose three diagonal partial derivatives vanish. For the second vector field, direct differentiation gives ${\partial }_x(2xy)=2y$, ${\partial }_y(x^2+3y^2+z^2)=6y$, and ${\partial }_z(2yz)=2y$. Thus

Separate variables near the nonzero initial condition: $y^{-2}dy=dx$. Integration gives $-1/y=x+C$, and $y(1)=1$ fixes $C=-2$. ThereforeThis is the maximal connected interval containing the initial point. Direct differentiation gives $y^{\prime }=1/(2-x{)}^2=y^2$; the pole at $x=2$ prevents continuation as a finite classical solution of a differential equation through that point.

The positive-domain graph starts with limiting value $f(0^{+})=1$ and limiting slope $-1$, then decreases towards the horizontal asymptote $f=0$, with $f(q)\sim 1/(2q)$ for large $q$. In fact direct differentiation givesThe first inequality follows from $e^s>1+s$ for $s>0$. Thus the graph is strictly decreasing and convex upward, with $0<f(q)<1$ and no stationary points. The integral representation $f(q)={\int }_0^1(1-e^{-s/q})ds$ also shows its positivity.

These properties belong to the reusable exponential spherical mass factor.

Put $D={\sigma }^2>0$ and $\xi =x-x_0$. Logarithmic differentiation of the positive Gaussian function givesDifferentiating once more in space givesMultiplying by $D$ provesThe assumption $\sigma \ne 0$ ensures both the normalization and exponential denominator are defined. This is the heat kernel, with diffusion coefficient $D$ and centre $x_0$.

On the negative half-line, differentiation gives $d\ln (-u)/du=1/u$. The integration path from $-1$ to $x<0$ stays away from the singularity at zero. ThereforeThis choice satisfies $F(-1)=0$.

Let $H(u)={\sin }^{2}u+\ln u$. The chain rule gives $H^{\prime }(u)=\sin 2u+1/u$, so the two given summands combine into its logarithmic derivative, $f(u)=H^{\prime }(u)/H(u)$. For $u>1$, $H(u)>0$, so the real logarithm is defined throughout the integration path. Since $H(\pi )=\ln \pi$,This satisfies $F(\pi )=0$, and differentiation recovers both terms of the integrand.

The scaling relation says that $z$ is a homogeneous function of degree $n$. Fix $(x,y)$ and differentiate the identity with respect to the positive parameter $\lambda$. The chain rule givesSetting $\lambda =1$ proves the Euler theorem for homogeneous functions identityDifferentiate once more with respect to $\lambda$, using equality of the mixed partial derivatives for the assumed twice continuously differentiable function:At $\lambda =1$ this is the second-order Euler identity for homogeneous functions:The repeated parameter differentiation accounts for the factor $n(n-1)$, rather than $n^2$.

Write $F(x)=\ln (e^{ax}-b)$. Direct differentiation, using the chain rule, givesThe stated parameter restrictions ensure that the logarithm is defined near $x=1$, and that $F^{\prime }(1)>0$ and $F^{\prime \prime }(1)<0$. With $h=x-1$, the Taylor series beginsIf $e^a-b\ne 1$, the constant term is nonzero, and the first two nonzero terms areThis covers both $a>\ln (1+b)$ and $0<a<\ln (1+b)$; the constant term is respectively positive or negative.

If $a=\ln (1+b)$, which is possible precisely with $0<a<\ln 2$ and $b=e^a-1$, the constant term vanishes. The first two nonzero terms are thenNo further case is needed, since the linear and quadratic coefficients cannot vanish under the given restrictions. All remainders use Big O notation.

For the given differential form, set $P=x(y^4+1)$ and $Q=y(2x^2{y}^2+1)$. Their partial derivatives satisfyIn fact an explicit potential issince $F_x=P$ and $F_y=Q$. Thus the differential form is an exact differential form on all of ${\mathbb{R}}^2$.

Replacing $y^2$ by $u$ suggests the potentialDirect differentiation verifies that $dH/dx$ is exactly the left-hand side of the required ordinary differential equation. This direct verification remains valid even for negative $u$; it does not require a real square root $y=\sqrt{u}$. ThereforeThe initial condition gives $C=1$. Solving this quadratic equation in $u$ and selecting the branch that has $u(1)=0$ givesThe second expression also defines the removable value $u(0)=1$. The maximal open interval containing $1$ on which this is a differentiable real solution isAt either endpoint the square root vanishes and $2x^{2}u+1=0$, while $x(u^2+1)\ne 0$, so the original ordinary differential equation cannot hold there with a finite derivative.

A rational lower bound for the natural logarithm proves positivity for every $n$, including $n=1$. DefineThen $h(0)=0$, while direct differentiation givesThus $h(t)>0$ for $t>0$. Setting $t=1/n$ in the expression from part (b) givesTherefore $\boxed{u_n>0\text{ for every }n\ge 1}$, and $(a_n)$ is a strictly increasing sequence.

An antiderivative of the initial velocity is $-2c{e}^{-s^2/2}$. Thus D'Alembert's formula givesThis is a positive Gaussian function moving right and a negative Gaussian function moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct differentiation gives ${\psi }_t(x,0)=2cx{e}^{-x^2/2}$.

For every $t>0$, the profile is an odd function of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct{)}^2-(x-ct{)}^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height Gaussian functions centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive global maximum occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.

Choose the left homogeneous solution $v(x)=\sinh x$ and the right one $w(x)=\sinh (1-x)$. Their Wronskian is $v{w}^{\prime }-v^{\prime }w=-\sinh 1$. The Dirichlet modified Helmholtz Green function on an interval, with operator sign $D_x^2-1$, is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative jumps by $+1$, giving $(D_x^2-1)G=\delta (x-\xi )$. The Green-function representation then givesThe two integrals are $(e^{2x}-1)/4-x/2$ and $e(1-x)/2-(e-e^{2x-1})/4$, respectively. Substituting them and simplifying givesDirect differentiation verifies $u^{\prime \prime }-u=e^x$ and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.

In monic form the Chebyshev differential equation has $P(x)=-x/(1-x^2)$ and $Q(x)=p^2/(1-x^2)$. Both are analytic near zero, so it is an ordinary point. At $x=1$, $(x-1)P=x/(x+1)$ and $(x-1{)}^{2}Q=-p^2(x-1)/(x+1)$ are analytic; at $x=-1$, $(x+1)P=x/(x-1)$ and $(x+1{)}^{2}Q=-p^2(x+1)/(x-1)$ are analytic. Thus both endpoints are regular singular points.

Insert $y_q={\sum }_{n\ge 0}{a}_n{x}^{n+q}$ and equate coefficients. The recurrence relation isExplicitly the factors are $(n^2-p^2)/[(n+2)(n+1)]$ for $q=0$ and $[(n+1{)}^2-p^2]/[(n+3)(n+2)]$ for $q=1$. The lowest equation for $q=1$ forces $a_1=0$. For $q=0$, one may choose $a_1=0$ to select an even solution, independent of the odd $q=1$ solution. The leading coefficient $a_0$ is free in either chosen parity branch.

For a nonterminating branch the consecutive parity-term ratio tends to $x^2$. Hence the ratio test gives convergence for $|x|<1$, and radius of convergence one. Terminating branches are polynomials with infinite radius. For $p=m\ge 0$ an integer, choose $q$ of the same parity as $m$: the recurrence stops once $n+q=m$, producing a degree-$m$ Chebyshev polynomial. Normalization at $1$ gives all nonzero coefficients through degree three:

Set $Y(\theta )=y(\cos \theta )$. Direct differentiation gives $Y^{\prime \prime }={\sin }^2\theta y^{\prime \prime }-\cos \theta y^{\prime }$, so the equation becomes $Y^{\prime \prime }+m^{2}Y=0$. For $m>0$, $Y=A\cos (m\theta )+B\sin (m\theta )$. A polynomial $y$ has finite derivative at $x=1$, and therefore $Y^{\prime }(0)=-\sin 0y^{\prime }(1)=0$, forcing $B=0$. Also $Y(0)=T_m(1)=1$, so $A=1$. For $m=0$, the transformed solution is affine in $\theta$ and the same derivative condition removes its linear term. ThusThe endpoint values follow by continuity. The finite-derivative condition is what excludes the independent nonpolynomial branch, rather than normalization alone.

The substitution separates the uniform equilibrium from the decaying perturbation. Direct differentiation givesConsequently the reaction-diffusion equation reduces to the heat equation $v_t=v_{xx}$. Its data become

When the three constants coincide, the Dirichlet boundary conditions for $v$ are homogeneous. Separation of variables gives $X^{\prime \prime }+n^{2}X=0$ and $T^{\prime }=-n^{2}T$, selecting the normal modes $\sin (nx)e^{-n^{2}t}$ for $n\ge 1$. ThusTwo integrations by parts give ${\int }_0^{\pi }x(\pi -x)\sin (nx)dx=2[1-(-1{)}^n]/n^3$, while the constant term uses ${\int }_0^{\pi }\sin (nx)dx=[1-(-1{)}^n]/n$. Therefore the Fourier sine series isFor $u_0\ne 2$, its first coefficient is nonzero. Multiplying by $e^{-\beta t}$ and retaining the slowest-decaying mode givesThe remainder estimate holds uniformly in $x$ for sufficiently large positive time. If $u_0=2$, the fundamental vanishes and the next possible mode has decay rate $\beta +9$, explaining the exclusion.

There is an initial endpoint compatibility issue unless $u_0=0$: the given polynomial profile vanishes at the endpoints, whereas the boundary values equal $u_0$. The series realizes the initial data in L2 space and pointwise in the open interval, imposing the prescribed boundary values for $t>0$. For nonzero $u_0$ it cannot also be continuous at both initial endpoint corners.

The homogeneous equation is a Cauchy-Euler equation. Trying $y=x^m$ gives the repeated indicial rootHence, for $x>0$,Both independent solutions diverge at the origin, so no nonzero homogeneous combination satisfies the zero-origin conditions by continuous extension.

For a source point $\xi >0$, finiteness on $0<x<\xi$ forces the Green function to vanish on that side. On $x>\xi$, continuity and the unit first-derivative jump requireUsing the homogeneous solutions gives the causal Green function for a repeated-root Euler operatorIndeed, the Wronskian of $x^{-1}$ and $x^{-1}\log x$ is $x^{-3}$, and direct differentiation gives $G_x({\xi }^{+},\xi )=1$. Since $G$ is continuous, its second derivative produces precisely a unit Dirac delta, with no derivative of a delta. For each positive source it is identically zero near the origin, so both required origin values vanish. The differential operator itself is interpreted on $x>0$; no value of its singular coefficients at zero is needed.

Integrating against the forcing yieldsThe last integral equals $1/16$, by integration by parts or differentiating ${\int }_0^1{s}^{p}ds=1/(p+1)$. ConsequentlySubstitution gives $y^{\prime \prime }+3y^{\prime }/x+y/x^2=(6+9+1)x/16=x$, while $y(0)=y^{\prime }(0)=0$. The divergent homogeneous solutions prove uniqueness in the regular class. No boundary condition at infinity was imposed or required.

Direct differentiation givesOn the unit disk $D$, polar coordinates give ${\int }_D{y}^{2}dA=\pi /4$, ${\int }_{D}xdA=0$ and ${\int }_{D}dA=\pi$. Consequently the volume integral isFor an independent surface integral calculation, the relevant cross product isThe bottom face contributes ${\int }_{D}xydA=0$ and the top contributes ${\int }_D(-h{y}^2-xy)dA=-\pi h/4$. On the cylindrical side, put $x=\cos \theta$, $y=\sin \theta$, $\hat{\mathbf{n}}=(x,y,0)$ and $dS=d\theta dz$. The flux density is $x^3+2xy{z}^2+2y^{2}z-x{y}^2$. All terms except $2y^{2}z$ integrate to zero because they are odd functions of $x$ or $y$, givingThus minus the total outward flux is $-(\pi h^2-\pi h/4)$, in agreement with the volume integral.

The source factor cancels the singular denominator of the Green function. Splitting the integral at $\xi =x$ givesUsing $\int (1+\xi )e^{-\xi }d\xi =-(\xi +2)e^{-\xi }$, these two integrals are $e-(x+2)e^{-x}$ and $5/2-x-e(x+2)e^{-x}/2$, respectively. Simplification givesThe exponential and linear terms are homogeneous solutions, while direct differentiation gives $L(-x^2)=x$ for $x\ne 0$. Also $y(-1)=y(1)=0$, and the expression is analytic through zero. Uniqueness in this regular class follows because any difference is a global linear combination $A{e}^x+B(1+x)$: the left boundary forces $A=0$, and the right then forces $B=0$.