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Work with $t>0$ and write $u=t^{-1/2}{e}^{-x^2/(4t)}$. The chain rule givesThus $u_{xx}=u_t$: it is a constant multiple of the one-dimensional Gaussian heat kernel.

For the modified equation, put $f=g(t)u$. Substitution gives $g{u}_{xx}=g^{\prime }u+g{u}_t-gu/t$. The heat-equation terms cancel, and $u>0$ permits division to obtain $\boxed{g^{\prime }=g/t}$. Solving this separable differential equation givesThe zero solution is included by $C=0$. This is the scalar integrating factor for a reaction-diffusion equation with reaction coefficient $1/t$, because the modified equation is $f_t=f_{xx}+f/t$. Its sign produces growth. Indeed ${\int }_{\mathbb{R}}udx=2\sqrt{\pi }$ is constant, while ${\int }_{\mathbb{R}}fdx=2C\sqrt{\pi }t$ increases linearly for $C>0$; neither profile is being assigned unit normalization here.

Differentiation and factorization give $f^{\prime }(x)=e^{-x^2}(1+x^2-2x^4)=e^{-x^2}(1-x^2)(1+2x^2)$. The exponential and $1+2x^2$ are positive, so the only stationary points are at $x=\pm 1$. The derivative is positive for $|x|<1$ and negative for $|x|>1$, givingThe function is odd, crosses the origin with slope one, and tends to zero from below as $x\to -\infty$ and from above as $x\to \infty$. These facts determine the sketch.

For the negative-sign equation, take a nonzero separated solution $\varphi =X(x)Y(y)$. Separation of variables gives $X^{\prime \prime }=-\mu X$ and $Y^{\prime \prime }=-(25-\mu )Y$. The zero Dirichlet boundary conditions on $X$ force $\mu =m^2$ with a positive integer $m$ and $X=\sin mx$, up to a constant. Zero boundary values on $Y$ similarly require $25-m^2=n^2$ with a positive integer $n$. A zero or negative separation eigenvalue would give linear or hyperbolic factors, whose two zero endpoint values force the trivial solution.

The only positive integer pairs with $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By the complete Dirichlet eigensystem of a rectangle, these span the whole eigenspace, soFor the positive-sign equation there is only the zero solution. One can use the same strictly positive eigenvalues of $-\Delta$, or directly apply the zero Dirichlet solution of the modified Helmholtz equation. Multiplying by $\varphi$ and using Green's first identity giveswhere $\Omega =(0,\pi {)}^2$. Both nonnegative integrals must vanish, proving $\boxed{\varphi =0}$.

For orthogonal matrices $Q_1,\dots ,Q_m$ of the same size, transpose reverses the order of multiplication. Thereforeby canceling the adjacent pairs $Q_j^T{Q}_j=I$ successively. Hence every finite product is an orthogonal matrix.