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The origin is a regular singular point of this Kummer differential equation. Apply the Frobenius method, writing $y={\sum }_{n=0}^{\infty }{a}_n{x}^{n+r}$ with $a_0\ne 0$. The lowest power gives the indicial equationFor $n\ge 1$, comparison of the coefficient of $x^{n+r-1}$ givesSince $0<c<1$, both roots give valid independent branches. Normalize $a_0=1$:Here $(c{)}_n$ is the rising factorial, and the first series is the confluent hypergeometric function of the first kind $M(1,c,x)$. The two leading powers are distinct, proving linear independence. We describe real solutions on $x>0$, with their one-sided behavior at zero.

For the inhomogeneous equation a particular solution is $-(x+c)/2$, verified by direct substitution. Thus $y=A{y}_1+B{y}_2-(x+c)/2$. Since $y_2^{\prime }\sim (1-c)x^{-c}$, a finite derivative at zero forces $B=0$. Then $y(0)=0$ forces $A=c/2$, and the required solution isIn particular $y^{\prime }(0)=0$. The singular homogeneous branch cannot be added without violating the derivative condition.

Assume $\omega >0$. The characteristic roots of the damped harmonic oscillator are $-k\pm \sqrt{k^2-{\omega }^2}$. Its general homogeneous solution isThese are respectively the underdamped, critically damped and overdamped regimes.

For the velocity-switched driven oscillator, the initial acceleration is $-{\omega }^2{x}_1<0$. Its velocity is therefore negative on the first half-cycle, and the forcing is zero:Indeed $\dot{x}=-({\omega }^2{x}_1/p)e^{-kt}\sin pt<0$ in that open interval. Set $r=e^{-k\pi /p}$; at the first turning point, $x_2=-r{x}_1$ and $\dot{x}=0$.

Put $\tau =t-\pi /p$. On the second half-cycle the velocity is positive, so the particular solution is $a/{\omega }^2$. Matching position and velocity givesIts derivative is positive for $0<\tau <\pi /p$. At the next turning point,The turning-point return map therefore repeats the initial state exactly whenThis value exceeds $a/{\omega }^2$, so the subsequent acceleration initiates the required negative-velocity half-cycle again. The switched solution is continuous with continuous velocity, and is piecewise twice differentiable; acceleration has one-sided jumps where the forcing switches.

For $k>\omega$, the specified positive initial value givesThere is no next turning point: the forcing stays zero and $x$ approaches zero monotonically. The motion from the specified initial state cannot be periodic. More generally, a solution started at any positive maximum has this same obstruction; a nonconstant periodic orbit would require such a maximum. The zero equilibrium is a separate constant solution.

On an interval where $y>0$, the Bernoulli differential equation is transformed by $u=y^{1-p}$:For $f_1=1$, $f_2=x$, put $s=1-p\ne 0$. An integrating factor $e^{-sx}$ solves $u^{\prime }-su=sx$, giving $u=C{e}^{sx}-x-s^{-1}$. Thus the positive branches areon intervals where the expression in parentheses is positive. The transformation omits the separate solution $y\equiv 0$. For $0<p<1$, the right-hand side need not be locally Lipschitz continuous at zero, so nonnegative solutions can also join zero and positive branches where the original equation permits; the formula describes every strictly positive segment.

For the autonomous case, $y^{\prime }=y-{\alpha }^2{y}^p$. Its two nonnegative equilibrium points areAt the positive equilibrium point, $f^{\prime }(y_{\ast })=1-p$. If $p>1$, $f(y)>0$ below $y_{\ast }$ and $f(y)<0$ above it; hence $0$ is unstable and $y_{\ast }$ is asymptotically stable. Every strictly positive solution tends to $y_{\ast }$ forward in time.

If $0<p<1$, those signs reverse: $0$ is stable from the nonnegative side and $y_{\ast }$ is unstable. The stability at zero follows from the phase-line signs, not a finite derivative there. For initial value $0<y_0<y_{\ast }$, the transformed equation is $u^{\prime }=s(u-{\alpha }^2)$ with $s>0$. The solution reaches zero atand remains zero thereafter as a nonnegative forward solution. This is finite-time extinction in a sublinear Bernoulli equation. Initial values above $y_{\ast }$ increase without bound; the sublinear negative term prevents neither that growth nor the instability of $y_{\ast }$.

The events $B\cap A_i$ are pairwise disjoint and their union is $B$. Finite additivity and the definition of conditional probability therefore give the law of total probability:For the birthday problem, let $D$ mean that all birthdays differ. For $n\le 365$, sequential conditioning on previous distinct birthdays and independence of random variables giveHere each factor uses the stated bound $1-x\le e^{-x}$. For $n\ge 29$, $n(n-1)\ge 812>730\log 3$, soFor $n>365$ a match is certain by the pigeonhole principle. The threshold from the exponential bound is $(1+\sqrt{1+2920\log 3})/2\approx 28.8$, so $29$ suffices.

The characteristic roots of the homogeneous linear ordinary differential equation are $2$ and $3$. The forcing is resonant with the $e^{3x}$ solution, so the method of undetermined coefficients uses the particular solution $x{e}^{3x}$. Thus $y=A{e}^{2x}+B{e}^{3x}+x{e}^{3x}$. The two initial conditions are $A+B=0$ and $2A+3B+1=0$, giving

This is a logistic equation, with positive equilibrium point $N=\alpha$ and initial population above that value. On the positive branch, put $v=1/N$. It satisfies the linear ordinary differential equation $v^{\prime }+\alpha v=1$, so $v={\alpha }^{-1}+C{e}^{-\alpha t}$. The initial condition gives $C=-1/(2\alpha )$, henceThe population decreases monotonically toward the stable equilibrium point; it never reaches it at a finite time.

The law of total probability and the definition of conditional probability give Bayes' theorem:The denominator is positive because $P(E)>0$.

For the observed cargo, set $p_A=0.05$, $p_B=0.03$, $p_C=0.01$. The binomial likelihood under origin $i$ isEqual prior probabilities and Bayes' theorem make the posterior proportional to these likelihoods. The binomial coefficient cancels in the likelihood ratios:Thus $L_A/L_B\approx 4.99\times 10^{-45}$ and $L_C/L_B\approx 2.74\times 10^{-9}$, givingEven the permitted approximation $\log (1-p)\approx -p$ gives log ratios about $-93.83$ and $-23.72$, which already show overwhelming posterior preference for B. The exact logarithms are preferable because multiplication by $9800$ magnifies the approximation error. An observed contamination fraction halfway between two hypothesized rates does not give equal likelihoods: the binomial distribution assigns asymmetric probabilities to those deviations, especially at this large sample size.

Integrating the uniform density gives $\mathbb{E}{X}^k={\int }_0^1{x}^{k}dx=1/(k+1)$. By independence of random variables, $\mathbb{E}(XY{)}^k=\mathbb{E}{X}^k\mathbb{E}{Y}^k=1/(k+1{)}^2$. For the remaining expected value, integrate first in $x$:The value at $y=0$ is supplied by continuity. Therefore, with the harmonic number $H_m={\sum }_{j=1}^{m}1/j$,

Conditional on one sample point being $(x,y)$, it is a maximal external point precisely when none of the other $n-1$ points falls in the upper-right rectangle $(x,1]\times (y,1]$. Its area is $(1-x)(1-y)$, so its conditional probability of being maximal is $[1-(1-x)(1-y){]}^{n-1}$. Both $1-X$ and $1-Y$ are independent uniform random variables, hence a given point is maximal with probability $H_n/n$. Summing their indicator random variables and applying linearity of expectation gives the expected number of coordinatewise maxima:This grows only logarithmically with the sample size, even though all $n$ points are potential candidates.