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Multiply the equation by $x$ to obtain $x{y}^{\prime \prime }-P(x)y^{\prime }-Q(x)y=0$. The leading terms after a Frobenius method trial $y=x^r$ arethe term involving $Q_0$ first appears at order $x^r$. The indicial equation is therefore

To obtain the first correction to a regular singular solution by reduction of order, start from the assumed solution $y_1=1+\beta x+O(x^2)$ and use the Abel identity:We work initially on $x>0$ near zero; since $P_0$ is a positive integer, the resulting expansion is an ordinary power series. Also $y_1^{-2}=1-2\beta x+O(x^2)$, so the reduction of order integral has integrandChoose its integration constant to be zero, eliminating an added multiple of $y_1$. Integrating and multiplying by $y_1$ givesWith $C=P_0+1$ the requested normalized answer isIts nonzero Wronskian makes it independent of $y_1$. The existence of the asserted $y_1$ also forces $P_0\beta +Q_0=0$; there is no need to introduce $Q_0$ into the normalized second coefficient.

For two solutions $y_1,y_2$, define their Wronskian byDifferentiation and substitution of the second-order linear differential equation giveThis proves the Abel identity and hence

On an interval where the known solution $y_1$ does not vanish,Integrating gives the reduction of order formulaChoosing the normalization $C=1$ makes the Wronskian nonzero, so this solution is linearly independent of $y_1$. To verify it really solves the equation, write $y_2=y_{1}h$. Its equation reduces tothe integrand gives $h^{\prime }=\exp (-\int p)/y_1^2$, which satisfies this equation on differentiation. Adding an integration constant only adds a multiple of $y_1$. The formula is local where $y_1\ne 0$; at a zero of $y_1$ with regular coefficients, the actual solution can be continued by initial-value existence and uniqueness, even if its integral representation has an apparent singularity.

Substitute the power series and compare coefficients. The constant coefficient gives $-a_1-a_0=0$. For $k\ge 1$, the coefficient of $x^k$ givesThus $a_1=-a_0$, while the $k=1$ equation places no restriction on $a_2$. For $k\ge 2$,The power series therefore sums toRenaming the two arbitrary constants gives $\boxed{y(x)=C(1-x)+D{e}^{-x}}$.

Both functions solve the equation by direct substitution. Their Wronskian, in the order $1-x,e^{-x}$, is $x{e}^{-x}$, which is nonzero for $x\ne 0$. Hence they give the full two-dimensional solution space on either interval away from the singular point. Both are entire, and the displayed formula also gives every twice continuously differentiable solution through zero: matching $y(0)$ and $y^{\prime \prime }(0)$ fixes the same $C,D$ on the two sides.

Set $v=u^{\prime \prime }$. The resulting second-order linear differential equation is $v^{\prime \prime }+2v=4t^2$. A polynomial particular solution $v_p=a{t}^2+b$ requires $2a=4$ and $2a+2b=0$, hence $v_p=2t^2-2$. The homogeneous characteristic roots of a constant-coefficient differential equation are $\pm i\sqrt{2}$, soIntegrating twice and renaming constants gives the general real solutionThe four independent homogeneous terms account for the four initial data of the fourth-order linear ordinary differential equation.

For the substitution $u(t)=y(t^2)$, repeated use of the chain rule, with $x=t^2$, givesThereforeFor $x>0$, the map $t=\sqrt{x}$ is invertible, and the transformed equation is exactly the equation just solved. Hence the general real solution on $x>0$ isThere is no requirement that $u$ extend as an even function through $t=0$: only the branch $t>0$ is needed, so the sine and square-root terms must be retained.

Let $D=d/dx$ and write the constant-coefficient linear ordinary differential equation as $p(D)y=0$. Since $D^k{e}^{\lambda x}={\lambda }^k{e}^{\lambda x}$,The exponential never vanishes, so $\boxed{e^{\lambda x}\text{ solves the equation }⟺p(\lambda )=0}$.

For the second claim, the product rule givesfor $k\ge 1$, with the $k=0$ term treated separately. Summing the coefficients yieldsA root of multiplicity at least two has $p(\mu )=p^{\prime }(\mu )=0$, proving $x{e}^{\mu x}$ is also a solution. This directly establishes the repeated-root contribution without assuming a solution formula. If the displayed coefficient list has leading zeros, the exponential calculation still applies to its actual polynomial degree.

Interpret the random choice as an independent selector $B$ with $P(B=1)=r$, and write $Y=BX+(1-B)X^{\prime }$. Its distribution is again uniform on $\{-1,1\}$, so $EX=EY=0$ and $\mathrm{Var}X=\mathrm{Var}Y=1$. By independence,Consequently $\boxed{\rho (X,Y)=r}$. The independent selector is the standard meaning of the stated mixture; a choice depending on the signs would describe a different model.

Put $U=X-EX$ and $V=Y-EY$. Their second moments are the finite positive variances of $X,Y$. The Cauchy-Schwarz inequality for square-integrable random variables givesDivide by the positive product of variances and take square roots. The correlation bound is $\boxed{-1\le \rho (X,Y)\le 1}$. As usual, non-constant here means not almost surely constant, so that the denominator is nonzero.

For a fixed horse $i$, independence of the race results gives probability $p_i^m$ of winning every race. The events that different horses win every race are disjoint, soThis is a homogeneous polynomial of degree $m$. Apply part (a) to the nonnegative numbers $p_i$, whose sum is one, to obtain $\boxed{Q\ge N^{1-m}}$. For $m>1$ the smallest probability occurs when every horse has winning probability $1/N$; for a single race $Q=1$ regardless of these probabilities.

For $m\ge 2$, $f^{\prime \prime }(x)=m(m-1)x^{m-2}\ge 0$ on $[0,\infty )$, so $f$ is a convex function; for $m=1$ it is linear and also convex. The equally weighted random variable has expected value $EX=N^{-1}{\sum }_i{x}_i=1/N$. Applying Jensen's inequality givesMultiplying by $N$ provesFor $m>1$ equality requires all $x_i=1/N$, by strict convexity; for $m=1$ equality holds for every admissible vector.

The superposition principle applies because the recurrence operator is linear. Add the particular solutions from parts (ii) and (iii), then add the general homogeneous linear recurrence relation solution from part (i). This gives

The forcing base $-2$ is a root of the characteristic equation of a linear recurrence, so a constant multiple of $(-2{)}^n$ cannot be a particular solution. For $P(r)=r^3-3r+2$, direct substitution givesFor the resonant case of exponential forcing in a linear recurrence, at $r=-2$, $P(-2)=0$ and $(-2)P^{\prime }(-2)=-18$. Hence $-(n/18)(-2{)}^n$ is a particular solution of the inhomogeneous linear recurrence. The general solution is

Let $L[y{]}_n=y_{n+3}-3y_{n+1}+2y_n$. For the exponential trial sequence $D{2}^n$,Thus $D=1/4$ gives a particular solution. The difference of any two solutions of the inhomogeneous linear recurrence satisfies the homogeneous linear recurrence relation in part (i). Therefore

Substitution of $y_n=r^n$ gives the characteristic equation of a linear recurrenceThe repeated characteristic root of a linear recurrence $1$ supplies the solutions $1$ and $n$, while the simple root $-2$ supplies $(-2{)}^n$. They are linearly independent: their values at three consecutive indices have nonzero determinant. A third-order linear recurrence relation is uniquely determined by three consecutive initial values, so these three solutions span its solution space. The general solution is

Now complete the square in the other order:Divide the joint probability density function by the positive marginal densityThe factor $e^{-b^2/2}$ cancels, giving the conditional probability densitySincethis is a normalized normal distribution density. HenceConditioning on an exact observation is interpreted through this conditional probability density, rather than conditioning by a ratio of event probabilities.

The other marginal is $X_2\sim N({\mu }_2,{\sigma }_2^2)$ by the same square-completion argument with the indices exchanged. Its moment-generating function is therefore

The nonsingular bivariate normal distribution has ${\sigma }_1,{\sigma }_2>0$ and $|\rho |<1$. Put $a=(x_1-{\mu }_1)/{\sigma }_1$ and $b=(x_2-{\mu }_2)/{\sigma }_2$. Completing the square givesConsequently the joint probability density function factors asFor fixed $x_1$, the second factor is a normalized normal distribution density in $x_2$, with mean ${\mu }_2+\rho {\sigma }_{2}a$ and standard deviation ${\sigma }_2\sqrt{1-{\rho }^2}$. Its integral is one; explicitly substitute $w=(b-\rho a)/\sqrt{1-{\rho }^2}$ in the Gaussian integral. Thus the marginal density is

Summing the geometric series for the offspring probability generating function givesThe fixed-point equation $s=1/(2-s)$ is $(s-1{)}^2=0$, whose only root in $[0,1]$ is one. The Galton-Watson extinction fixed point result therefore proves extinction is certain.

For the explicit iterates, the proposed formula at $n=1$ is $1/(2-s)$, so the base case holds. If it holds at $n$, part (a) givesThis is exactly the same formula with $n$ replaced by $n+1$. By induction,Evaluating the probability generating function at zero selects the mass of an empty generation:These probabilities increase to one, also confirming the extinction conclusion directly.

The eventual extinction probability is the smallest solution $\boxed{q\in [0,1]\text{ of }{F}_1(q)=q}$. This is the Galton-Watson extinction fixed point result for one initial individual. Equivalently,since having an empty generation is an absorbing event and the extinction-by-generation events increase. In particular extinction is certain when the smallest fixed point is one. With $p_0>0$, an offspring mean at most one implies this, whereas a mean greater than one gives a fixed point below one. The fixed-point statement itself requires no finite-mean assumption.

Conditional on $X_n=m$, the next generation is the sum of $m$ independent offspring counts, each with probability generating function $F_1(s)$. HenceThis also holds for $m=0$, when the empty sum is zero and the empty product is one. The law of total expectation givesFor complex $|s|\le 1$, the expectations are absolutely bounded because $|F_1(s)|\le 1$. This proves the branching-process generating-function iteration, with $F_0(s)=s$.