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Write the linear system of differential equations asA homogeneous power $t^{\lambda }\mathbf{v}$ solves this Cauchy-Euler differential system precisely when $A\mathbf{v}=\lambda \mathbf{v}$. The eigenvalues are $3$ and $6$, with eigenvectors $(2,1{)}^T$ and $(1,-1{)}^T$. The homogeneous solution is therefore $C{t}^3(2,1{)}^T+D{t}^6(1,-1{)}^T$.

For a particular solution, put ${\mathbf{z}}_p=t\mathbf{c}$. Then $(I-A)\mathbf{c}=\mathbf{b}$, which gives $\mathbf{c}=(3,0{)}^T$. At $t=1$, the two initial conditions become $2C+D+3=0$ and $C-D=0$, so $C=D=-1$. ThusThe two powers correspond to the two eigenvalues; the lower-degree particular term accounts for the constant forcing. Substitution gives both original right sides and the zero initial data.

Completing squares reveals the geometry of the sheared quartic double-well gradient flow:The critical points satisfy $2(x+y)=0$ and $4x^3-2x+2y=0$, so they are $(0,0)$, $(1,-1)$ and $(-1,1)$. The Hessian matrix isAt the origin its determinant is $-8$, so $(0,0)$ is a saddle point, with $V=0$. At either other critical point, the determinant is $16$ and the upper-left entry is $10$, so it is a positive-definite matrix. They are strict local minima; the completed squares show that both are global minima, with $V=-1$.

A contour at level $c$ obeysThere are no contours for $c<-1$, two isolated minimum points for $c=-1$, and two separate closed ovals for $-1<c<0$. At $c=0$ the two lobes meet at the saddle point; near the origin their tangents are $y=(-1\pm \sqrt{2})x$. For $c>0$ one closed contour surrounds both minima.

Along the gradient flow, the gradient-flow dissipation identity isAt the stated initial point, $V(1,-1/2)=-3/4$. Its trajectory stays in the invariant sublevel set $V\le -3/4$. This set is compact because $V$ is a coercive function. It also excludes every point with $x=0$, where $V=y^2\ge 0$; continuity therefore keeps the trajectory in its positive-$x$ component.

To justify its limit rather than merely infer it from the sketch, integrate the dissipation identity: ${\int }_0^{\infty }\Vert \nabla V{\Vert }^{2}dt$ is finite. On this compact set the gradient and its time derivative are bounded, so $\Vert \nabla V{\Vert }^2$ is uniformly continuous. A nonnegative uniformly continuous function with finite integral tends to zero: otherwise separated intervals of a fixed positive height and width would force an infinite integral. Every accumulation point is consequently a critical point. In the positive-$x$ component below this energy level the only one is $(1,-1)$. Compactness then implies convergence, andThis is a compact gradient-flow trapping criterion; the drawn trajectory illustrates, rather than replaces, the convergence proof.

Use a power-series solution of a differential equation, $y={\sum }_{n\ge 0}{c}_n{x}^n$. Comparing coefficients givesIn particular $c_0=0$, while $c_1$ and $c_2$ are free. The two prescribed derivative pairs select $(c_1,c_2)=(a,0)$ and $(0,b)$. The next coefficients are $c_3=c_1/2$, $c_4=c_2/6$, $c_5=c_1/24$, $c_6=c_2/120$. HenceThese are the first three nonzero terms when the respective parameter is nonzero; if $a=0$ or $b=0$, the corresponding solution vanishes identically.

For closed forms, set $y=xu$. Cancellation of the first-derivative terms reduces the equation, for $x\ne 0$, to $u^{\prime \prime }-u=0$. The analytic solutions extend across zero, and their hyperbolic cosine and hyperbolic sine expansions identifyThis is the hyperbolic reduction of a regular-singular differential equation: although the original leading coefficient vanishes at zero, both selected power-series solutions of a differential equation are entire.

For a nonnegative integer-valued random variable, the probability generating function iswith $0\le s\le 1$ always allowed. Its derivative at $1$ from below gives the expectation when finite.

The waiting time $N$ has the positive-integer geometric distribution $P(N=j)=p{q}^{j-1}$. For the capped geometric waiting time, $X=j<k$ exactly when $N=j$, while $X=k$ occurs whenever $N\ge k$. Thus the terminal mass is $q^{k-1}$, givingThe finite sum is a polynomial; any apparent singularity of the rational expression is removable. This includes $k=1$, where $G_X(s)=s$.

Differentiating the rational form at $s=1$ yieldsAs a second interpretation, the expectation is the sum of the tail probabilities $P(X\ge j)=q^{j-1}$ for $1\le j\le k$. Their finite geometric sum gives the same answer.

For events with $P(B)>0$, the conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives Bayes' theorem:Conditionally on $N=n$, independent fair tosses give a binomial distribution for the head count, so $P(H=1\mid N=n)=n{2}^{-n}$. The joint probability mass function is consequentlyUse the law of total probability and the differentiated geometric series, ${\sum }_{n\ge 1}n{r}^n=r/(1-r{)}^2$, to obtain $P(H=1)=4/9$. The posterior coin count after exactly one head is thereforeThese probabilities sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.

A constant orbit is a fixed point of $F(u)=u(1+u)/2$. Solving $F(u)=u$ gives $u(u-1)=0$, henceFor fixed point stability for an iteration, perturb a fixed point $u_{\ast }$ by $v_n$. Its local evolution is $v_{n+1}=F^{\prime }(u_{\ast })v_n+O(v_n^2)$, where $F^{\prime }(u)=u+1/2$. At zero, $F^{\prime }(0)=1/2$, so perturbations contract: zero is locally asymptotically stable. For example, if $|u|\le r<1$, then $|F(u)|\le (1+r)|u|/2$; this gives an invariant neighborhood and geometric convergence to zero.

At one, $F^{\prime }(1)=3/2$, so perturbations expand: one is unstable. More explicitly, putting $u_n=1+v_n$ gives $v_{n+1}=3v_n/2+v_n^2/2$, whose magnitude increases while a nonzero perturbation remains sufficiently small. The derivative test concerns local stability; it does not assert attraction from every real initial value.

The repeated-root constant-coefficient differential equation has characteristic polynomial $(r+2{)}^2$. Its two fundamental solutions areThey are linearly independent: their Wronskian is $e^{-4x}$, which never vanishes. For the forced equation, exploit the repeated factor by writing $y=e^{-2x}u$. Direct differentiation gives $(D+2{)}^{2}y=e^{-2x}{u}^{\prime \prime }$, so the forcing reduces to $u^{\prime \prime }=1$. The initial data imply $u(0)=0$ and $u^{\prime }(0)=0$, hence $u=x^2/2$. ThereforeThe extra factor $x^2$ reflects resonance of the forcing with the repeated characteristic root; the exponential substitution obtains it without guessing a particular solution.

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.