Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 2

The direction field has slope $1-y^2$, independent of $x$. Its line segments slope upward for $-1<y<1$, are horizontal at $y=\pm 1$, and slope downward for $|y|>1$. Thus $y=1$ is an attracting equilibrium in forward $x$, while $y=-1$ is repelling. The two constant solutions $y\equiv \pm 1$ must be included separately when separating variables.

The following original sketch shows the slope field and one solution in each requested region. The lower exterior solution ends at a finite pole; it cannot be continued across that pole as a real finite-valued solution.

Separate variables and integrate on $|y|<1$:Hence the general solution in this region isEvery such solution increases from values near $-1$ towards $1$, remains strictly between the two equilibria, and is defined for all real $x$.

For $|y|>1$, the separated integral isInverting givesThe upper branch has $x+C>0$, where $y>1$ decreases towards $1$ as $x\to \infty$. The lower branch has $x+C<0$, where $y<-1$ decreases to $-\infty$ as $x\uparrow -C$. These are separate maximal intervals: the pole is not part of a solution. For the domain $x\ge 0$, a lower-branch initial value below $-1$ necessarily reaches that pole in finite positive $x$. The direction field sketch above displays both exterior branches and a central hyperbolic tangent branch.

With $h=\delta t>0$, the central finite difference converts the decay equation toThe continuous ordinary differential equation has the general solution $x(t)=C{e}^{-Kt}$. The discrete linear recurrence is second order and requires two initial values, such as $x_0,x_1$. Trying $x_n={\rho }^n$ gives the characteristic equation of a linear recurrence ${\rho }^2+2Kh\rho -1=0$. Its roots aresoThere are no repeated roots, and the same formula extends to negative integer $n$ if required.

To demonstrate the continuum limit, put $a_h=\mathrm{arsinh}(Kh)$. Then ${\rho }_{+}=e^{-a_h}$ and ${\rho }_{-}=-e^{a_h}$, with $a_h/h\to K$. For $t_n=nh\in [0,T]$,uniformly, since $0\le t_n\le T$. More precisely $a_h/h=K-K^3{h}^2/6+O(h^4)$, giving an $O(h^2)$ error on a fixed interval. Thus $B=0$, $A=C$ selects the physical discrete mode and corresponds to $x_1={\rho }_{+}{x}_0$.

The other root is a parasitic amplification root:For fixed $B\ne 0$, even and odd mesh points approach values differing by $2B{e}^{Kt}$. Equivalently successive mesh values have a nonvanishing alternating jump despite their time separation tending to zero. This mode cannot converge uniformly to a continuous ODE solution; its limiting envelope also grows rather than decays.

For step-dependent initial data the precise result is the centered two-step discretization of exponential decay criterion:Sufficiency follows from the two uniformly bounded exponential envelopes on a fixed interval. Necessity follows from convergence at the adjacent points $0,h$: $x_0\to C$, $x_1-x_0\to 0$, and the coefficient formulas force $B_h\to 0$, $A_h\to C$. In particular a nonzero but vanishing starting error is not excluded. Starting with the exact value $x_1=C{e}^{-Kh}$ gives $B_h=C{K}^3{h}^3/12+O(h^4)$ and still converges on each fixed interval. The assertion about “remaining” solutions therefore concerns a fixed nonzero parasitic amplitude, not every nonzero step-dependent amplitude. The method is nevertheless not damping-stable for arbitrarily long times because $|{\rho }_{-}|>1$.

For a nonnegative integer-valued random variable $X$, its probability generating function isinitially defined at least for $|z|\le 1$. A general real-valued random variable need not have a probability generating function in this sense. For a binomial distribution, the binomial theorem givesDifferentiating the generating function produces factorial moments:Since $X^2=X(X-1)+X$, the mean and variance areThese formulas include the degenerate endpoint probabilities and $n=0$.

For independent random variables, their sum has the product of their probability generating functions:Comparing polynomial coefficients identifies a binomial distribution with parameters $n+m,p$:and the probability is zero at all other values. The common success probability is essential to this binomial form; independent binomial counts with different probabilities do not in general add to a binomial count.

The transformation is strictly increasing on $0\le x<1$, has range $[0,\infty )$, and its inverse is $x=y/(y+3)$. Therefore the cumulative distribution function isDifferentiation gives the probability density functionIts value at zero may be assigned arbitrarily without changing the distribution. There is no atom at zero or infinity. The expression is undefined at $X=1$, but that event has probability zero under the continuous uniform distribution, so any convention there leaves the law unchanged. Finally ${\int }_0^{\infty }3/(y+3{)}^{2}dy=1$, as required.

An integrating factor for a differential one-form makes the multiplied form exact. If $dF=\mu gdx+\mu fdy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed partial derivatives yieldsThis is a local compatibility condition under the usual smoothness assumptions.

For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y^{\prime }=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The direction field below uses normalized line segments, so steep directions remain visible.

Multiplication by $\sin x\cos y$ givesThus the multiplying function is an integrating factor, and the implicit general solution isIn the larger square, $\cos x\ge 0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation iswith endpoints tending to $y=\mathrm{sgn}(C)\pi /2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the integrating factor, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.

After multiplication by ${\sec }^{2}x$, recognize both sides as derivatives:The nontrivial first integral is thereforeIt necessarily depends on the derivative, correcting the printed notation $h(x,y)$. A second integration giveson the specified interval. Since $y(0)=0$, $C_2=0$. A finite derivative at zero also implies $y{y}^{\prime }\to 0$ there, forcing $C_1=0$. A positive particular solution is consequentlyThe negative of this function is the other sign choice. Near zero, $-2\log (\cos x)=x^2+x^4/6+O(x^6)$, so the positive branch has $y=x+x^3/12+O(x^5)$ and $y^{\prime }(0)=1$; the negative branch has derivative $-1$.

For the positive branch, $y^{\prime }=\tan x/y>0$. Its second derivative has numerator $-2\log \cos x-{\sin }^{2}x$, multiplied by a positive factor. This numerator is zero at zero and has derivative $2{\sin }^{3}x/\cos x>0$, so the curve is convex for positive $x$. It starts tangent to $y=x$ and rises without bound as $x\uparrow \pi /2$, with $y\sim \sqrt{-2\log (\pi /2-x)}$. The third panel of the sketch above shows this branch.

For two solutions $y_1,y_2$, define the WronskianDifferentiate and substitute the second-order linear differential equation:Integration proves the Abel identity, including the identically zero case:For $p=-2/x$, this is $W=C{x}^2$ on either interval excluding zero, since the integral uses $\log |x|$. The coefficient singularity at zero must not be ignored.

Multiply the particular equation by $x^2$ and substitute the Frobenius solution $y=x^{\lambda }{\sum }_{n\ge 0}{a}_n{x}^n$. With $a_{-1}=a_{-2}=0$, equating coefficients givesThe leading coefficient is nonzero, so the indicial equation is $\lambda (\lambda -3)=0$. Thus only $\lambda =0,3$ are possible, and the recurrences below construct both.

For $\lambda =0$, $a_1=0$, $a_2=-a_0/2$, and the $n=3$ recurrence reads $0=a_1=0$, leaving $a_3$ free. For all later indices,This is an undetermined coefficient at Frobenius resonance; choosing $a_3=0$ selects a convenient even basis solution but is not forced by the differential equation.

For $\lambda =3$, all odd-indexed coefficients vanish andThe factorial denominators show convergence for every finite $x$. Expanding the given hyperbolic functions directly yieldsConsequently the exponent-zero family is $a_0(\cosh x-x\sinh x)-3a_3(\sinh x-x\cosh x)$, and the normalized exponent-three solution is $-3a_0(\sinh x-x\cosh x)$. This verifies the coefficient formulas and supplies two independent basis functions.

Finally $y_1^{\prime }=-x\cosh x$ and $y_2^{\prime }=-x\sinh x$, soThis agrees with the Abel calculation. Its vanishing at zero does not contradict independence on regular intervals: the original equation is singular there.

For the homogeneous solution, expand in the independent eigenvectors of $A$. Each coefficient obeys $c_j^{\prime }={\lambda }_j{c}_j$, so the complementary function isIf ${\mathbf{x}}_p$ is any particular integral, subtracting it from any other solution gives a homogeneous solution. Hence $\mathbf{x}={\mathbf{x}}_p+{\mathbf{x}}_c$ is the general solution. Complex-conjugate modes can be combined to produce real solutions when the forcing is real.

For a real two-by-two matrix, the characteristic polynomial is ${\lambda }^2-(\mathrm{tr}A)\lambda +\det A$. Under the stated distinct-eigenvalue assumption, all nonzero homogeneous modes are purely oscillatory exactly when the eigenvalues are $\pm i\omega$ with $\omega >0$. Their sum and product imply $\mathrm{tr}A=0$, $\det A={\omega }^2>0$. Conversely these trace and determinant conditions force those eigenvalues, so there is no exponential growth, decay or Jordan-block secular growth.

For the final specified initial-value problem, direct multiplication gives $A^2=-9I$. The forcing vector $\mathbf{b}=(2,3i-1{)}^T$ satisfies $A\mathbf{b}=3i\mathbf{b}$, so the forcing is resonant and $t\mathbf{b}{e}^{3it}$ is a particular integral. The matrix exponential isSince the resonant particular integral vanishes initially, the initial-value solution isThis complex solution is appropriate to the complex forcing as written. Taking real parts gives the corresponding solution for the real part of that forcing.

Try a particular integral along each forcing eigenvector. Since $A{\mathbf{a}}_j={\lambda }_j{\mathbf{a}}_j$,With both denominators nonzero, add the two responses:Only the forcing component's own eigenvalue enters its denominator; a frequency matching a different eigenmode does not create resonance in this already diagonal forcing component.

The second component is resonant. Differentiating $t{e}^{{\lambda }_{2}t}$ supplies the term missing from the exponential ansatz:Thus a particular integral isThe linearly growing oscillatory envelope is forced resonance in a differential equation; it does not contradict bounded oscillation of the complementary function. The worked initial-value solution in the question's root Solution uses this same resonant mechanism.

Differentiate the proposed first integral along the vector field:At $\alpha =0$, $K$ is conserved and the system is an inverted quartic oscillator phase portrait. Its equilibria are $O=(0,0)$ and $S_{\pm }=(0,\pm 1/\sqrt{2})$. The Jacobian matrix isAt the origin its eigenvalues are $\pm i$. Moreover $K$ has a strict local minimum there, and its nearby positive contours are closed, so the origin is a genuine center equilibrium, not merely an inconclusive linear centre. The arrows circulate clockwise: on the positive $x$-axis, $\dot{y}<0$. At either $S_{\pm }$, the eigenvalues are $\pm \sqrt{2}$, giving saddle equilibria. The stable eigendirection is $\delta x=\sqrt{2}\delta y$, and the unstable one is $\delta x=-\sqrt{2}\delta y$.

The special energy level factors asBetween the two saddles, its left arc has $x<0$ and travels upward from $S_{-}$ to $S_{+}$; its right arc has $x>0$ and travels downward from $S_{+}$ to $S_{-}$. These two heteroclinic orbits bound the central eye. Trajectories inside the eye are closed periodic orbits and do not tend to the centre as $t\to \infty$.

More explicitly, a level $K=C$ hasFor $0<C<1/4$, its bounded central component is periodic, while its exterior components turn at $|y|=\sqrt{(1+\sqrt{1-4C})/2}$ and then escape. For $C>1/4$, $x$ never vanishes and the two branches run from one vertical infinity to the other. Levels $C\le 0$ have only exterior nonconstant components, apart from the origin at $C=0$.

On the upper outer separatrix, the $x>0$ branch arrives from positive infinity and approaches $S_{+}$ in forward infinite time; the $x<0$ branch leaves $S_{+}$ towards positive infinity. On the lower outer separatrix, the $x<0$ branch arrives from negative infinity and approaches $S_{-}$, while the $x>0$ branch leaves towards negative infinity. Approaching a saddle along its stable manifold is exponential at rate $\sqrt{2}$. A trajectory close to but off a separatrix can spend a long time near the saddle before completing a periodic orbit or escaping.

The escape is finite-time blowup, not an asymptote at $t=\infty$. On any escaping level, $|x|=\sqrt{y^4-y^2+C}\sim y^2$, so the remaining travel time is bounded by a convergent integral proportional to ${\int }^{\infty }dy/y^2$. Thus an outward upper branch has $y\sim (T-t{)}^{-1}$, $x\sim -(T-t{)}^{-2}$; an outward lower branch has $y\sim -(T-t{)}^{-1}$, $x\sim (T-t{)}^{-2}$. The statement about approaching from infinity likewise refers to a finite past endpoint. Only equilibria, periodic trajectories and trajectories tending to saddles have a future defined for all time in this conservative portrait.

For small positive $\alpha$, the equilibria remain in place, but the origin's eigenvalues become $\alpha /4\pm i\sqrt{1-{\alpha }^2/16}$, so it becomes a repelling focus. The saddles retain one eigenvalue of each sign, $\alpha /4\pm \sqrt{2+{\alpha }^2/16}$. Since $\dot{K}\ge 0$, and its time integral is strictly positive on every nonconstant trajectory, trajectories cross $K$-contours towards larger values. Near the centre this means outward spiralling. No nonconstant periodic orbit can remain, and the former saddle-to-saddle connections at equal $K=1/4$ are broken. Stable saddle trajectories approach from $K<1/4$; unstable ones depart into $K>1/4$. Generic central trajectories eventually escape, while the exceptional stable saddle manifolds still approach their saddles.

The inclusion-exclusion principle givesFor the coat problem, all $n!$ permutations are equally likely. If a specified set of $j$ people gets its own coats, the other coats can be assigned in $(n-j)!$ ways, so that intersection has probability $(n-j)!/n!$. Taking the complement of the union of correct-coat events gives the derangement probabilityTo have exactly $m$ fixed coats, choose their recipients and derange all remaining coats. Thus the fixed point count of a uniform random permutation hasand zero probability outside this range. The empty permutation has $D_0=1$, so the formula includes $m=n$; it also gives zero for $m=n-1$, since a lone remaining coat cannot be misplaced.

For fixed $m$, $n-m\to \infty$, and the exponential series givesThese are the probabilities of a Poisson distribution of mean one. The limit is for fixed $m$, not for $m$ increasing with $n$.

The probabilities satisfy $a,b,c,d\ge 0$ and $a+b+c+d=1$. Since the variables are indicators,Their covariance isThus the necessary and sufficient condition for uncorrelated random variables isThis uses zero covariance as the definition. If one variable is constant, a normalized correlation coefficient is undefined, but zero covariance and this criterion remain meaningful.

Independence requires each joint probability to equal the product of the relevant marginals. In particular it implies $d=(c+d)(b+d)$, equivalent to $ad=bc$.

Conversely let $p=c+d$ and $q=b+d$, and assume $ad=bc$. The covariance calculation gives $d=pq$. ThenAll four joint probabilities therefore factor, proving independence. HenceThis is Bernoulli independence from zero covariance, including degenerate marginals. It is special to two-point variables and is not a general equivalence for arbitrary random variables.

An uneventful day leaves one aphid, a one-offspring day leaves the parent plus its offspring, and a two-offspring day leaves three aphids. Death leaves none. Thus the daily descendant counts are $0,1,2,3$ with probabilities $(1-q)t$, $q$, $(1-q)r$, $(1-q)s$, respectively. The mean and probability generating function areIndependence across individuals and days makes the daily population a Galton-Watson process, with each surviving parent counted as one of the next day's descendants.

The standard Galton-Watson extinction fixed point result states that the eventual extinction probability of a branching process $e$ from one ancestor is the smallest root in $[0,1]$ of $G(e)=e$. Moreover a nondegenerate offspring law of mean at most one has extinction probability of a branching process one. Here, since $q<1$, the fixed-point equation reduces towhich contains no $q$. This proves laziness invariance of branching extinction: uneventful days change the timescale, not eventual extinction. The excluded case $q=1$ would keep one aphid forever.

If $2r+3s\le 1$, the offspring mean is at most one. In this parameter range $t>0$: if $t=0$, then $r+s=1$ would imply $2r+3s\ge 2$. Thus the offspring law is not deterministically one, and the extinction criterion applies. Extinction is certain.

For the specified probabilities, the fixed-point equation factors asIts roots in $[0,1]$ are $1/2$ and $1$, and the smaller root is the extinction probability of a branching process:

By rotational invariance, fix the first landing point at the north pole of a sphere of radius $R$. A band of central angle $\theta$ and width $d\theta$ has area $2\pi R^2\sin \theta d\theta$. Dividing by total area gives the central angle between independent uniform sphere points densityand zero outside this range. The radius cancels.

Given the first pair's central angle $\gamma$, the third point can communicate with each of them exactly when it lies in that point's positive hemisphere. Choose their intersection line as a polar axis. Each hemisphere permits an azimuth interval of length $\pi$, and the intervals' centres are separated by $\gamma$. Their overlap is a spherical lune of opening $\pi -\gamma$, with area $2R^2(\pi -\gamma )$. ThereforeThe union probability follows from inclusion-exclusion for two hemispheres:Thus the requested conditional answers are“Either” here is inclusive: a third ship linked to both is also linked to at least one.

For network connectivity, if $\gamma <\pi /2$ the first pair is already linked, and the third must link to at least one of them. If $\gamma >\pi /2$, the first pair is not linked, and both must link through the third. Consequently the connectivity of three acute-angle sphere points isUsing ${\int }_0^{\pi /2}\gamma \sin \gamma d\gamma =1$ and ${\int }_{\pi /2}^{\pi }\gamma \sin \gamma d\gamma =\pi -1$ givesAngles exactly equal to the threshold have probability zero. Connectivity allows relaying and is weaker than requiring direct links between every pair.