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An integrating factor for a differential one-form is a nonzero function $\mu (x,y)$ for which $\mu (dy+fdx)$ is an exact differential, say $dF$. Then solutions lie on level sets $F=\text{constant}$. Locally, the compatibility condition is ${\partial }_x\mu ={\partial }_y(\mu f)$.

Here multiplication by $2y{e}^x$, which is nonzero for $y>0$, producesThe initial condition makes this first integral equal to $a^2$. Taking the positive branch yields the solutionon the connected interval containing zero where the radicand is positive. This domain restriction matters because the original differential equation is singular at $y=0$.

Completing the square gives $2x+x^2=(x+1{)}^2-1\ge -1$. As $y>0$, the differential equation consequently impliesAt $x=0$, $y^{\prime }=-a/2<0$. Thus, moving a little to the left from $y(0)=a\ge 1$, the curve rises above $a$. Throughout $y>1$ every segment of the direction field points downwards as $x$ increases, by the inequality above. Tracing backwards therefore makes $y$ rise further, so it cannot return to $y=1$. This graphical barrier argument proves $y>a\ge 1$ and $y^{\prime }<0$ for every $x<0$ on the solution. The explicit first integral also shows this branch extends throughout $x<0$: writing $t=-x>0$, the minimum of $e^t/t^2$ is $e^2/4>1$, so $a^2{e}^t-t^2>0$ for $a\ge 1$. Including the initial point, $y^{\prime }<0$ for all $x\le 0$.

For $a=1$ the curve is $y=\sqrt{e^{-x}-x^2}$ on $(-\infty ,b)$, where the strictly increasing function $x^2{e}^x$ has the unique positive root $x^2{e}^x=1$ at $b\simeq 0.703467$. The curve decreases through $(0,1)$ with slope $-1/2$ and reaches $(b,0)$ only as a limiting endpoint. Its derivative isAs $x\to -\infty$, $y\sim e^{-x/2}$ and $y^{\prime }\sim -\frac{1}{2}{e}^{-x/2}\to -\infty$. As $x\uparrow b$, the numerator tends to $-b^2-2b<0$ and the denominator to zero from above, so again $y^{\prime }\to -\infty$. More precisely, $y\sim \sqrt{b(b+2)(b-x)}$ at the endpoint. Both ends have the requested unbounded negative slope.

After $n+1$ independent tosses, an even count arises in exactly two disjoint ways: an even count after $n$ tosses followed by a tail, or an odd count followed by a head. Independence of the last toss from the preceding ones therefore givesSubtracting the fixed point $1/2$ turns this into the linear recurrenceAs zero is even, ${\pi }_0=1$, so the probability isFor $p=1/2$ this is $1/2$ for every $n\ge 1$; for $p=0$ it is always one; for $p=1$ it alternates between one and zero according to the parity of $n$.

The indicator function $I_A$ equals one when the event $A$ occurs and zero otherwise. Its expected value is $\mathbb{E}{I}_A=\mathbb{P}(A)$. By linearity of expectation, no independence assumption is needed to obtainSince $I_i{I}_j=I_{A_i\cap A_j}$, including $p_{ii}=p_i$, the variance isIn particular, correlations between the events cannot simply be discarded.

Write $\mu =\mathbb{E}N$. If $\mu >0$, then $N=0$ implies $|N-\mu |\ge \mu$. Applying Chebyshev's inequality therefore givesEquivalently, the contribution of $N=0$ to $\mathbb{E}[(N-\mu {)}^2]$ is ${\mu }^2\mathbb{P}(N=0)$. A positive mean is necessary for the displayed ratio to be defined. If $\mu =0$, nonnegativity instead gives $N=0$ almost surely, and the printed ratio would be $0/0$.

This is a separable differential equation. On the solution with $y<1$, integration from the initial condition givesThe right side lies in $[0,1)$, so taking its sine stays on the required branch. The solution isNear the origin, $1-\sqrt{1-x^2}=x^2/2+O(x^4)$, so $y=x^2/2+O(x^4)$ and the tangent line is horizontal. At the other end, $y\to \sin 1$, whileThus the curve has a vertical limiting tangent line as $x\uparrow 1$, even though $x=1$ is excluded from the domain.

For the direction field, the slope is nonnegative everywhere in the square. The segments are horizontal on $x=0$ and on $y=1$, become steeper as $x$ increases at fixed $y<1$, and become flatter as $y$ increases at fixed $x$. Near $x=1$ they are almost vertical unless $y$ is correspondingly close to one. The excluded corner $(1,1)$ has no single limiting slope: the ratio $(1-y^2)/(1-x^2)$ depends on the path of approach.

The characteristic equation of the homogeneous linear differential equation is $m^2+m-2=(m-1)(m+2)=0$, giving the homogeneous solution $A{e}^t+B{e}^{-2t}$. For the first forcing, a particular solution of the form $C{e}^{-t}$ gives $-2C=1$, soThe initial conditions impose $A+B=1/2$ and $A-2B=-1/2$, whence $A=1/6$, $B=1/3$. Thus the first solution isFor the second forcing, $e^t$ is already a homogeneous solution, so the usual exponential trial must be multiplied by $t$. Substitution of the particular solution $Ct{e}^t$ gives $3C{e}^t=e^t$, hence $C=1/3$. Now $y=A{e}^t+B{e}^{-2t}+t{e}^t/3$, and the initial conditions give $A+B=0$, $A-2B=-1/3$. Therefore the resonant solution isThe additional factor $t$ reflects resonance in a differential equation.