For this first-order linear differential equation, an integrating factor on an interval avoiding zeros of $\cos x$ is $\cos x$. The product rule then givesThe initial condition gives $3/\sqrt{2}=1/\sqrt{2}+C$, hence $C=\sqrt{2}$. ThusThe displayed interval is the maximal one containing the specified initial point, since the coefficient $\tan x$ is undefined at its endpoints.

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.

Set $w=e^z\ne 0$. The hyperbolic cosine identity gives $w+w^{-1}=2i$, hence the quadratic equationLet $L=\ln (1+\sqrt{2})>0$. The first root has modulus $1+\sqrt{2}$ and complex argument $\pi /2$; the second has modulus $\sqrt{2}-1=e^{-L}$ and complex argument $-\pi /2$. Taking all complex logarithm values givesThe two families are distinct because their real parts are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible complex arguments have been included.

For real $x$, $\cosh x>0$, so multiplication by the hyperbolic cosine is reversible. The identities $\cosh x-\sinh x=e^{-x}$ and $\mathrm{sech}x=1/\cosh x$ giveThus $e^{-2x}=3$. The exponential function is injective, so the unique real solution is

For $t>0$, differentiate the printed ansatz:Matching the independent coefficients in $h_t=(h{h}_x{)}_x-1$ givesThere are two parameter pairs, including the degenerate spatially constant case:Their corresponding solutions areThe request for “the general form” cannot mean the complete solution set of this nonlinear partial differential equation: substitution of one ansatz cannot provide arbitrary initial data, and the superposition principle does not apply.

A useful wider family follows by allowing an independent time-dependent constant term and quadratic coefficient, $h=A(t)+B(t)x^2$. Coefficient comparison gives $B^{\prime }=6B^2$ and $A^{\prime }=2AB-1$. For $B\ne 0$, write $\tau =t-t_0>0$ to obtain $B=-1/(6\tau )$ and $A=C{\tau }^{-1/3}-3\tau /4$. Spatial translation, also a symmetry of the equation, therefore givesHere $C,x_0,t_0$ are arbitrary constants; the displayed nontrivial ansatz corresponds to $C=1/6$ and $x_0=t_0=0$. There are also affine solutions $h=px+(p^2-1)t+q$, including the spatially constant branch. These are valid solution families, rather than a claim of a full general solution.

For a separated nonzero solution $T=f(x)g(y)$, substitution and division by $fg$ givesThe homogeneous Dirichlet boundary conditions impose $g(0)=g(1)=0$. For a nonzero spatial factor, integration by parts gives ${\lambda }^2{\int }_0^1{g}^{2}dy={\int }_0^1(g^{\prime }{)}^{2}dy$, so the separation value is nonnegative. The zero value gives only a trivial linear $g$ under these boundary conditions. The nontrivial separated factors, for $a\ne 0$, are thereforeFor the intended forward diffusion equation, assume $a>0$. The superposition principle gives a Fourier sine series, whose coefficients must expand the initial ramp in the interior:Consequently the standard heat-flow solution isIt satisfies both lateral boundary conditions for $x>0$, and tends to $y$ as $x\downarrow 0$ for $0<y<1$, also converging to the ramp in the L2 norm. The exponential coefficient decay justifies differentiating the series for positive $x$.

There is a genuine incompatibility in the printed data: at $(0,1)$ the lateral condition demands zero while the initial condition demands one. Thus no function satisfies every stated value at that corner. The displayed series solves the usual consistent interpretation with initial data on $0<y<1$ and lateral data for $x>0$; it cannot be continuous at the upper initial corner. This discrepancy is present in the original PDF.

The sign of $a$ is also unstated. If $a=0$, the spatial equation with two zero endpoints forces $T=0$, incompatible with the ramp. If $a<0$, the separated Fourier coefficients grow like $e^{n^2{\pi }^{2}x/|a|}/n$ and fail to be square summable for every $x>0$; these data therefore have no forward solution in the standard L2 space class. The positive-$a$ assumption is essential to the stated heat-flow interpretation.

The repeated index sums the diagonal entries of $A$. This is its trace:

Scalar entries commute, so $B_{jk}{A}_{ij}=A_{ij}{B}_{jk}$ is exactly the summed entry formula for matrix multiplication. Hence

Introduce the second constraint $q(x,y,z)=y-h(x)=0$ and a second Lagrange multiplier $\mu$. At points where the two constraint gradients are linearly independent, solveThese are five scalar equations for $x,y,z,\lambda ,\mu$. The two constraint normal vectors span the normal plane to the feasible curve. Points where those gradients become dependent must be examined separately; the regular two-multiplier condition need not cover them.