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.

Away from $x=-1$ the separable differential equation givesOn the interval containing $x=1$, one has $x>-1$, so the positive initial value selects $y=K(1+x^3{)}^{1/3}$. Substitution of $y(1)=2$ gives $K=2^{2/3}$. ThusThe zero solution omitted by division by $y$ cannot satisfy the initial value. Although the selected solution tends to zero as $x\downarrow -1$, its derivative diverges there; it does not extend through that point as a classical differentiable solution.

Use the product substitution for a first-order equation $v=xy$. The product rule gives $v^{\prime }=x{y}^{\prime }+y$, turning the equation into $v^{\prime }=e^v$. Consequently $(e^{-v}{)}^{\prime }=-1$, so $e^{-v}=C-x$.

At $x=1$ the initial value gives $v=1$, hence $C=1+e^{-1}$. Recovering $y=v/x$ givesThis is the maximal real graph interval containing the initial point: the logarithm requires $x<1+e^{-1}$, and the numerator has a nonzero limit at zero, so this particular solution cannot extend through zero. Differentiating $xy=-\ln (C-x)$ verifies $x{y}^{\prime }+y=1/(C-x)=e^{xy}$.

Multiply the two defining Gamma integrals. For $p,q>0$ the nonnegative product integrand allows Tonelli's theorem:Use $r=x+y$, $u=x/(x+y)$, so $x=ru$, $y=r(1-u)$ and the positive quadrant becomes $r>0$, $0<u<1$. The Jacobian determinant has absolute valueAfter including this factor, the transformed integral separates:This proves the beta-gamma identity,All changes of variables are made inside the open positive domains; their boundaries do not contribute to the integrals.

Write $B(n,n)={\int }_0^1[x(1-x){]}^{n-1}dx$ and set $v=2x-1$. Since $x(1-x)=(1-v^2)/4$ and $dx=dv/2$,In the last integral put $u=v^2$, giving $dv=du/(2\sqrt{u})$. This yields $B(n,n)=2^{1-2n}B(1/2,n)$, and part (a) interchanges the arguments. HenceThe equal-argument beta integral identity holds for every positive real $n$, not only positive integers.

The increasing substitution $x={\sin }^2\theta$ maps $0\le \theta \le \pi /2$ onto $0\le x\le 1$, with $dx=2\sin \theta \cos \theta d\theta$. Since sine and cosine are nonnegative there, powers combine without a sign ambiguity:This is the trigonometric form of the beta function. The powers at the endpoints are integrable because $p,q>0$.

In the integral for $B(q,p)$, put $u=1-x$, so $dx=-du$ and the endpoints are interchanged. ThenThus exchanging the positive parameters leaves the Euler beta function unchanged. Positivity ensures convergence at both endpoints of this improper integral.

Let $d=b\cdot x$. Taking a dot product with $b$ reduces the vector equation to $(1-b\cdot c)d=b\cdot a$. If $b\cdot c\ne 1$, this determines a unique scalar and hence a unique vector:Direct substitution verifies it, so necessity has also produced sufficiency. The formula includes $c=0$, when $x=a$.

If $b\cdot c=1$, the reduced scalar equation becomes $0=b\cdot a$. When $b\cdot a\ne 0$, there is no solution. When $b\cdot a=0$, all solutions are $x=a+dc$ with arbitrary $d\in \mathbb{R}$: their dot product with $b$ is indeed $d$. Since $b\cdot c=1$ forces $c\ne 0$, this is an entire affine line, not a single point. These alternatives are the solvability of an affine rank-one vector equation.

Put $d=b\cdot x$ and $B=|b{|}^2$. Taking the dot product of the original equation with $b$ eliminates $b\times x$ and gives $d=a\cdot b-Bd$. ThusNext take its cross product with $b$. The term $db$ disappears, and the vector triple product identity givesInsert this expression back into $x=a-db-b\times x$. The two longitudinal terms cancel, leaving $(1+B)x=a-b\times a$. ConsequentlyThe denominator is strictly positive for every real $b$, including $b=0$. Substitution using the same triple-product identity verifies the solution and establishes uniqueness. This is the cross-product operator with a longitudinal rank-one term.

Rearrange to $z^4=1+i\sqrt{3}=2e^{i\pi /3}$. A fourth root has modulus $2^{1/4}$ and an argument one quarter of $\pi /3+2\pi k$. Therefore all four solutions areTheir arguments differ by $\pi /2$, so they are distinct. Raising any of them to the fourth power returns $1+i\sqrt{3}$.