Factor the right side as $(x-1{)}^2(x-3)$. On $-1<x<1$, division by $x^2-1$ gives a first-order linear differential equation with integrating factorMultiplication makes the equation $(\mu y{)}^{\prime }=3-x$, so $\mu y=3x-x^2/2+C$. At zero, $\mu (0)y(0)=1$, hence $C=1$. Simplifying givesIts derivative isThe derivation solves the initial-value problem on $(-1,1)$. The simplified expression also extends smoothly through $x=1$ and satisfies the original undivided equation there, giving a solution on all $x>-1$; its pole prevents continuation through $x=-1$.

For the forcing $8e^{-2x}$, the method of undetermined coefficients gives the particular term $2e^{-2x}$. The term $e^{-x}$ is resonant with a homogeneous solution, so use $ax{e}^{-x}$ for its forcing; substitution gives $a=1$.

Thus $y=C_1{e}^{2x}+C_2{e}^{-x}+2e^{-2x}+x{e}^{-x}$. The two initial conditions reduce to $C_1+C_2=3$ and $2C_1-C_2=0$, so $C_1=1$, $C_2=2$. The solution of the initial value problem isThe extra factor $x$ in the resonant particular solution is essential.

The characteristic polynomial of this constant-coefficient differential equation is $r^2-r-2=(r-2)(r+1)$. Its distinct roots give

For $h=1/N$, the composite trapezoidal rule isWith $f(x)=1/(4-x^2)$ this becomesWrite the requested right Riemann sum as $S_N$. The trapezoidal endpoint correction for a right Riemann sum givesThe integrand is continuous on $[0,1]$, so the trapezoidal values converge to its definite integral; the endpoint correction tends to zero. Therefore

A horizontal strip at height $y$ has width $4-2\sin (\pi y/2)$. The sine term is odd, so the area isReflection symmetry in the $y$-axis gives $\bar{x}=0$. For uniform density, the constant density cancels from the center of mass ratio. Its first moment in the vertical coordinate isConsequentlyThe wider part below the horizontal axis accounts for the negative vertical coordinate.

For positive bounds, integration by parts with $u=\log x$ and $dv=x^{2}dx$ givesFor negative bounds, specify the complex logarithm on the negative real axis. With argument $\pi$, namely the upper-bank boundary value of the principal logarithm, $\log x=\log (-x)+i\pi$. Substitute $t=-x$ in the real part:More generally, a continuous branch taking argument $(2k+1)\pi$ replaces $i\pi$ by $(2k+1)i\pi$. The imaginary part depends on that choice. The analytic principal branch excludes the negative axis itself, so the boundary-value convention is part of this interpretation.

Use partial fraction decomposition:Integrating givesThis real antiderivative is valid on each interval avoiding $x=\pm 2$, with its own integration constant. In particular, $I(1)-I(0)=\frac{1}{4}\log 3$.

Complete the square:The minimum locus of a squared function occurs where its residual vanishes. Hence the minimum is $-6$ on $y=-x^2/2$, provided these points lie in the ellipse. Setting $t=x^2$ in the constraint gives $t+t^2/2\le 9$, or $0\le t\le \sqrt{19}-1$. ThusFor the maximum, let $q=x^2+2y$. The ellipse givesAlso $q\ge 2y\ge -3\sqrt{2}$, whose absolute value is less than $19/2$. Therefore $q^2\le (19/2{)}^2$. Equality is attained at $y=1/2$, $x^2=17/2$, givingThese attained bounds are global, not merely stationary-point tests. The minimizing locus is a parabola segment.

Write $\dot{x}=dx/du$, $\ddot{x}=d^{2}x/d{u}^2$, $A_v=dA/dv$, $A_{vv}=d^{2}A/d{v}^2$, and $y_w=dy/dw$, $y_{ww}=d^{2}y/d{w}^2$. Put $S=A+A^T$. The product rule and chain rule giveHere $w^{\prime },w^{\prime \prime }$ are derivatives with respect to $v$. In the first derivative, $x^{T}A\dot{x}={\dot{x}}^T{A}^{T}x$ because the expression is scalar. Likewise ${\dot{x}}^{T}S\dot{x}=2{\dot{x}}^{T}A\dot{x}$ if desired. This differentiation of a parameter-dependent quadratic form does not assume that $A$ is symmetric. The positive linear term $y^{T}b$ contributes the positive chain-rule terms in the $v$ derivatives; it is independent of $u$.

The gradient equations areThus $x=3$ gives $y=2$, while $x=-3$ gives $y=8$. The Hessian matrix isAt both critical points the determinant is $-36<0$, so the Hessian is indefinite and the second-derivative test classifies both as saddle points. Direct substitution givesThe factorization $f+30=(x^2-9)(x+y-5)$ also shows the crossing level-set branches through each saddle.