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The product rule and the derivative of the exponential function give

The quotient rule and the derivative of the exponential function give

Multiply the two displayed equations by $e^{-x}$ and subtract the one with indices reversed. The product terms combine into the Lagrange identity for a Sturm-Liouville operator:Integrating from zero to infinity givesAll factors $L_j,L_j^{\prime }$ are polynomials. The endpoint term at zero vanishes because of the factor $x$, and the one at infinity vanishes because the exponential function dominates every polynomial. Therefore, if $m\ne n$,The integral is absolutely convergent. This is orthogonality in the weighted inner product with weight $e^{-x}$; the boundary cancellation is the self-adjoint mechanism behind the Laguerre differential operator on polynomials.

The product rule gives the two partial derivativesThey both vanish at the origin. The exponential function is never zero, so any further stationary point must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm \sqrt{2}$; if $x=0$, they are $y=\pm \sqrt{2}$. Hence the full list is

Apply the product rule and the chain rule to the exponential function:

The partial derivatives are $f_{xx}=4f$ and $f_t=\omega f$. Since the exponential function never vanishes, substituting into the partial differential equation gives $4=6\omega$. Hence

The defining exponential function formulas for the hyperbolic functions give

The outer exponential function differentiates to itself, and the inner sine has derivative cosine. The chain rule gives

This exponential function is positive everywhere, with $y(0)=1$, $y\to \infty$ as $x\to -\infty$ and $y\to 0$ as $x\to \infty$. Its derivatives areIt is strictly decreasing overall, with a horizontal stationary inflection point at $(0,1)$. A second, nonstationary inflection point occurs at $x=(2/3{)}^{1/3}$, where $y=e^{-2/3}$. The only horizontal asymptote is $y=0$ toward positive infinity; the graph is not symmetric about the vertical axis.

At a single point of contact the line is a tangent line, so its slope equals the derivative of the exponential function. At the contact coordinate $x_0$,Since $m>0$, dividing gives $x_0=1$, hence $m=e$. To verify uniqueness, the difference $e^x-ex$ has positive second derivative everywhere and its unique minimum is zero at $x=1$. Thus this line touches the curve once.

Using the contact coordinate $x_0=1$ and substituting into the exponential function givesBoth the curve and the line with the slope found in part (a) pass through this point.

In the homogeneous equation, substitute the exponential function $e^{kx}$. The characteristic equation isIts distinct roots are $0,1,-1,i,-i$. Combining the complex-conjugate pair into real sine and cosine gives the complementary functionFive linearly independent terms supply the required five homogeneous degrees of freedom.

The base $2e$ is constant. Write the exponential function as $\exp [\cos x\ln (2e)]$ and apply the chain rule:

Because the exponential function is strictly increasing, the level sets $f=L>0$ are exactly those of $G=\ln L$. For $a=1$ they satisfyThere are no contours for $L<1$, the level $L=1$ is the origin, and $L>1$ gives nested closed ovals symmetric about both axes. Their stronger fourth-power growth in $x$ makes them relatively narrower horizontally as the level rises.

For $a=-1$, the contour equation becomesThe central local minimum is surrounded by closed ovals for $1<L<e^{1/4}$, accompanied by outer open branches. At $L=e^{1/4}$, the level passes through the two saddle points $(\pm 1,0)$ and changes its connectivity. Levels $L<1$ have only outer branches, while levels above the saddle value have an upper and a lower branch. The sketch is restricted to the requested square; labels give $G=\ln f$, so they identify the corresponding contours of $f$ without exponentially large labels.

Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.

Rewrite the exponential function as $2^x=e^{x\ln 2}$. The chain rule gives

The exponential function is strictly positive, so the zero level set is determined entirely by its prefactor:This is a rectangular hyperbola with branches in the first and third quadrants.

The product rule gives $y^{\prime }=e^x(\cos x-\sin x)$. The exponential function never vanishes, so a stationary point requires $\tan x=1$. In the given interval this means $x=\pi /4$. Substitution givesThe derivative changes from positive to negative there, making it a strict local maximum.

Expand hyperbolic sine in exponential functions:Integrating each term gives the indefinite integral

Substitute $x^3$ into the Taylor series for the exponential function:The general nonzero term is $x^{3n}/n!$; the power series coefficients of powers not divisible by three are zero. The radius of convergence is infinite.