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The sine addition formula gives $(\sin x+\cos x)/\sqrt{2}=\sin (x+\pi /4)$. On the requested interval the shifted angle runs from $0$ to $\pi /2$, so the graph rises from $(-\pi /4,0)$ through $(0,1/\sqrt{2})$ to $(\pi /4,1)$.

Its derivative $\cos (x+\pi /4)$ is nonnegative and reaches zero only at the right endpoint. Its second derivative is $-\sin (x+\pi /4)\le 0$, so the rising arc is concave downward and ends with a horizontal tangent.

The domain of a function excludes $x=3$. Both branches are positive, with a vertical asymptote at $x=3$ approached from both sides with value $+\infty$, and horizontal asymptote $y=0$ at either infinity. The graph is symmetric about $x=3$ because it depends only on $|x-3|$.

For $x<3$, the formula is $2/(3-x)$ and the derivative is positive; for $x>3$, it is $2/(x-3)$ and the derivative is negative. It passes through $(0,2/3)$ and has no zeros.

There are $N+1$ terms, namely $1,2,\dots ,N+1$. The arithmetic series formula givesPairing the first and last terms gives pair sum $N+2$, explaining both factors.

This geometric series has first term $1$ and ratio $0.999$, which lies strictly inside the unit interval. ThereforeThe decimal is treated as the exact value $999/1000$.

The Gaussian function $e^{-x^2}$ is even and $\sin x$ is odd, so their product is an odd function. Moreover $|e^{-x^2}\sin x|\le e^{-x^2}$, whose integral on the real line is finite. Thus the improper integral is absolutely convergent, not merely a symmetric principal value, and the contributions of its two half-lines cancel:

For $x>0$, use integration by parts with $u=\ln x$ and $dv=dx$. ThenDifferentiating the answer checks cancellation of the extra constant term from $(x\ln x{)}^{\prime }$.

For absolute convergence, examine ${\sum }_{n\ge 1}|x-3{|}^n/n$. When $0<|x-3|<1$, the ratio of successive terms tends to $|x-3|<1$, so the ratio test proves convergence; at $x=3$ every term is zero. For $|x-3|>1$, the terms do not tend to zero.

At either boundary $|x-3|=1$, the absolute-value series is the divergent harmonic series. Thus for real $x$ the complete answer isThe original series at $x=2$ is conditionally convergent by the alternating-series criterion, but it is not absolutely convergent and is correctly excluded here.

The derivative of the hyperbolic sine is the hyperbolic cosine. Hence the chain rule gives

Apply the chain rule to the exponential, differentiating its exponent $-x^2$:

Seek a separated nonzero solution $u=X(x)Y(y)$. Division by $XY$ on intervals where the factors are nonzero gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-25$. Write $X^{\prime \prime }+\lambda X=0$ and $Y^{\prime \prime }+(25-\lambda )Y=0$, with both factors zero at their endpoints $0,\pi$.

For $X$, a nonpositive $\lambda$ gives only the zero solution under the two endpoint conditions: linear functions vanish identically when $\lambda =0$, and the hyperbolic solutions do likewise for $\lambda <0$. For $\lambda >0$, the endpoints force $X=A\sin (mx)$ with positive integer $m$ and $\lambda =m^2$. Applying the same argument to $Y$ gives $Y=B\sin (ny)$ and $25-m^2=n^2$, with positive integer $n$.

The only positive integer pairs satisfying $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By superposition,To establish that this is the most general sufficiently regular solution, expand $u$ in the complete product Fourier sine series on the square. Substitution, or integration by parts against each sine mode, gives $(25-m^2-n^2)c_{mn}=0$ for every coefficient. Hence only the two displayed modes remain; all boundary terms vanish because of the given zero values. This is the eigenvalue-$25$ case of the Dirichlet eigensystem of a rectangle. The zero solution is included when $A=B=0$.

The partial derivatives are $f_x=e^{xy}(1+xy)$ and $f_y=x^2{e}^{xy}$. At $(1,1)$ they have values $2e$ and $e$. The first-order Multivariate Taylor polynomial therefore gives the changeThus the requested linear approximation is $e(2\delta x+\delta y)$; it is the change from $f(1,1)=e$, not the new function value itself.

If $Pdx+Qdy=dF$ with continuous second partial derivatives of $F$, then $P=F_x$, $Q=F_y$. The symmetry of second derivatives requiresThis is a necessary condition for an exact differential. With continuously differentiable coefficients it is also locally sufficient, and is globally sufficient on a simply connected open domain; a domain with holes can require additional global information.

Differentiate the Airy ordinary differential equation $y^{\prime \prime }=sy$ to obtain $y^{(3)}=y+s{y}^{\prime }$. Combining part (a)'s derivatives yieldsTo make this vanish for an arbitrary Airy profile, the time powers must agree: $a-1=4a$, giving $a=-1/3$. Their common coefficient must then vanish, giving $-1/3+b^{-1}=0$, hence $b=3$. ThereforeFor $t>0$ the two terms cancel exactly:This is an Airy similarity solution of the linear dispersive equation. The Airy equation here is the linear partial differential equation $u_t+u_{xxx}=0$, the reduced equation specified in the question; it is distinct from the profile's Airy ordinary differential equation. The construction works for every linear combination of its two independent profiles, including zero. No condition at $t=0$ is supplied, and this singular time is excluded. With real cube roots the final formula can also be used separately on $t<0$.

Homogeneous equation. Its characteristic equation is $r^2-r=r(r-1)=0$. The two independent solutions are $1$ and $e^x$, soForced equation. A particular solution is $-x^2/2-x$, since its second derivative minus its first derivative equals $x$. Thus $y=C_1+C_2{e}^x-x^2/2-x$ and $y^{\prime }=C_2{e}^x-x-1$. The derivative condition at one gives $C_{2}e-2=2$, so $C_2=4/e$. The value condition then gives $C_1+4-1/2-1=1$, so $C_1=-3/2$. HenceBoth initial conditions and the forced equation follow directly on substitution.