For a quasilinear partial differential equation, the characteristic equations for a quasilinear equation are the ordinary differential equationsA solution surface is locally a differentiable graph $z=u(x,y)$ whose tangent plane contains the vector $(P,Q,R)$ at every point. Its normal is $(-u_x,-u_y,1)$, so the tangency condition is precisely $P{u}_x+Q{u}_y=R$. Thus every characteristic lying on that graph follows the vector field without leaving the graph.

Parameterize the given initial curve by $\gamma (t)=(x_0(t),y_0(t),z_0(t))$. For each $t$, solve the characteristic equations for a quasilinear equation with initial value $\gamma (t)$ at $s=0$. The resulting map $(s,t)\mapsto (x(s,t),y(s,t),z(s,t))$ sweeps out a candidate solution surface. If the coefficients and initial data are continuously differentiable, the Picard-Lindelöf theorem supplies this local family. To make it a graph we also need the projection $(s,t)\mapsto (x,y)$ to be locally invertible. At $s=0$ its Jacobian determinant isThe Noncharacteristic Cauchy data condition is $J(0,t)\ne 0$. The inverse function theorem then gives $s=s(x,y)$ and $t=t(x,y)$, and hence $u(x,y)=z(s(x,y),t(x,y))$. Its derivatives obey the PDE by the chain rule. The same characteristic initial-value uniqueness gives local uniqueness of this solution wherever the projection remains invertible and the relevant characteristic meets the prescribed data.

The same variation of parameters construction works on $[0,\pi ]$. All solutions satisfying the boundary condition at zero have the formAt $x=\pi$, the homogeneous term vanishes for every $C$; it can no longer be chosen to correct the final boundary value. Since $\sin (\pi -t)=\sin t$, that boundary value is zero exactly whenThis condition is necessary by the displayed general solution, and sufficient by the same formula. When it holds, the full answer isFor complex-valued forcing, allow $C\in \mathbb{C}$ instead. The freedom is addition of the nonzero homogeneous Dirichlet solution $e^{-2x}\sin x$. At this endpoint separation, the homogeneous boundary problem has a nonzero solution. The resulting Fredholm solvability condition restricts the forcing to the compatible functions described above. In the language of the Fredholm alternative, the condition is orthogonality to the adjoint null solution $e^{2x}\sin x$ of $D^2-4D+5$ with zero endpoint values.

For this forcing the integral in part (a) simplifies toConsequently the unique boundary value problem solution isIt vanishes at zero and at $\pi /2$. To check the differential equation directly, set $y=e^{-2x}w$. Then $Ly=e^{-2x}(w^{\prime \prime }+w)$. For $w=1-\sin x-\cos x$, one has $w^{\prime \prime }+w=1$, as required.