Expanding the determinant along the first row givesThe eigenvalues and normalized spatial eigenvectors, in increasing frequency order, areFor each mode both signs of ${\omega }_j$ give the same cosine displacement. An arbitrary overall mode amplitude remains free. These vectors are the $N=3$ instances of the sine eigenvectors of the fixed-end spring chain.

With $\mathbf{a}=(a_1,a_2,a_3{)}^T$, the eigenvalue equation isThe dimensionless matrix $D$ is the stiffness matrix divided by $k$; its eigenvalues determine squared angular frequencies.

Each proposed displacement has second derivative ${\ddot{x}}_i=-{\omega }^2{a}_i\cos \omega t$. Matching the coefficients of the common time factor in the three equations givesThese algebraic relations hold for the entire oscillation, including the isolated times when the common cosine vanishes.

An eigenvalue equation has the form $A\mathbf{a}=\lambda \mathbf{a}$, where $\mathbf{a}\ne 0$ is an eigenvector and $\lambda$ its eigenvalue. It means that the linear map preserves the vector's line and acts there by multiplication by $\lambda$. Nonzero solutions occur precisely when $\det (A-\lambda I)=0$; the zero vector is excluded because it would satisfy the equation for every $\lambda$.

Put $y=ux$, so $y^{\prime }=u+x{u}^{\prime }$. On an interval with $x\ne 0$, substitution and cancellation giveSince $u^2+2u+2=(u+1{)}^2+1>0$, no constant-$u$ solution is lost when separating. Integrateto obtainAfter restoring $u=y/x$, a local implicit general solution isThe inverse tangent is chosen continuously on a solution branch. Using the continuous polar angle of $(x,x+y)$ extends the expression to charts through $x=0$ when the original equation permits it. A differentiable graph cannot pass through $y=x\ne 0$, because there the original equation would require $5x=0$.

For a continuously differentiable differential one-form, exactness requires $\boxed{P_y=Q_x}$. It is locally sufficient, and globally sufficient on a simply connected domain; on a domain with holes, closedness alone need not imply exactness.

An integrating factor for a differential one-form depending only on $x$ must satisfyThus, where $Q\ne 0$,If the ratio depends only on $x$, this defines a nonzero factor and makes it a closed differential form, hence locally exact. For a factor depending only on $y$, the analogous calculation givesprovided the ratio is a function only of $y$ on the region where $P\ne 0$. A nonzero constant multiple of either factor works equally well.

The repeated factor in the differential operator suggests $y=e^{-x}u(x)$. Direct differentiation gives $y^{\prime \prime }+2y^{\prime }+y=e^{-x}{u}^{\prime \prime }$, so $u^{\prime \prime }=4$ andThe first value gives $C=1$. The second gives $u(1)=ey(1)=1$, hence $2+B+1=1$ and $B=-2$. Therefore

Under the same independence assumption, inclusion-exclusion givesOn $0<t<T$,There is one interior inflection point, where the curvature changes from positive to negative:At $t=T$ the graph reaches one with zero slope and then stays constant. There is no further change from one sign of curvature to the opposite sign there.

Without independence, $R=P_1+P_2-P$ still holds, but only $\boxed{s\le R(t)\le \min (1,s+s^2)}$ is forced by the marginals. Its inflection behaviour is then not determined. The shared-$U$ construction in (e), for example, gives $R=s$ before the plateau.

The product formula requires independent failure times. This assumption is not stated in the original PDF. Under that intended assumption, put $s=t/T$ and use the failure probabilities for two independent lifetimes:Its graph is a cubic increasing from $(0,0)$ to $(T,1)$, convex before reaching its plateau.

Without independence, the marginal distributions do not determine $P(t)$. The Fréchet inequalities give the sharp pointwise boundsFor example, taking $t_1=TU$ and $t_2=T\sqrt{U}$ with the same uniform random variable $U$ preserves both prescribed marginals but gives $P(t)=s^2$, rather than $s^3$.

The second cumulative distribution function is the integral of its density:Its graph is an upward-curving parabolic arc from $(0,0)$ to $(T_2,1)$, followed by a horizontal plateau. Both cumulative functions are continuous at their upper support endpoints.

Integrating the first probability density function gives the cumulative distribution functionIts graph is a straight ramp from $(0,0)$ to $(T_1,1)$, followed by a horizontal plateau.

Normalization requires $1={\int }_0^{T_2}Atdt=A{T}_2^2/2$. ThereforeThe graph rises linearly from zero to the left-hand limit $2/T_2$, then drops to zero beyond $T_2$. Its enclosed triangular area is one. The combined sketch in (f) uses dimensionless time and shows this density together with the cumulative probabilities.

A uniform distribution has constant density on its support, fixed by normalization. ThusThe density's assigned values at the endpoints do not affect probabilities.

Parametrize the diagonal by $\mathbf{x}(t)=(t,t,t)$ for $0\le t\le 1$. ThenThe second expression is $(t^3\cosh t{)}^{\prime }$, immediately giving $\int \mathbf{G}\cdot d\mathbf{x}=\cosh 1$. The first differs from it by $t^2\cosh t$. Two applications of integration by parts giveHenceThe changed value for $\mathbf{F}$ explicitly demonstrates path dependence, while $\mathbf{G}$ gives the same endpoint-dependent result.

On the first straight segment, $y=z=0$ and only $dx$ is nonzero, so both integrals vanish. On the second segment, $x=1,z=0$ and only $dy$ is nonzero; each integrand is one, contributing $1$. On the last segment, $x=y=1$ and only $dz$ is nonzero; each integrand is $\sinh z$, contributing $\cosh 1-1$. ThusThe second result also follows from the fundamental theorem for line integrals, using the potential found in (a). Agreement on this one path does not make $\mathbf{F}$ conservative.

The curl of the first vector field iswhich is not identically zero. Thus $\mathbf{F}$ is not conservative on ${\mathbb{R}}^3$ and cannot equal a negative gradient there.

For the second field, direct differentiation showsTherefore $\mathbf{G}$ is conservative, and the requested negative-gradient potential isIts curl vanishes, consistently with the curl of a gradient identity.