- A semi-infinite solid ($x=0 \sim \infty$), with an initial temperature of $T_{0}$. At time $\theta=0$, a heat flux $q_{0}$ (input) is applied through the surface at $x=0$, and this flux remains constant. Assume the relevant material properties are constants. Starting from the energy equation in a Cartesian coordinate system (6-26a), simplify it into a special form for this specific case, and explain the basis for the simplification process. Differentiate both sides of the above equation with respect to $x$ to obtain an equation expressed in terms of the heat flux $q$, and state the corresponding boundary conditions. Solve this equation to obtain an expression for the temperature distribution.

Options:
A. $$
t=\frac{k q_{0}} {\sqrt{4 \beta\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\xi} \operatorname{e x p} (-\xi^{2} ) \mathrm{d} \xi} {\int_{0}^{\infty} \operatorname{e x p} (-\xi^{2} ) \mathrm{d} \xi} \right] \mathrm{d} \eta+t_{0}
$$
B. $$
t=\frac{k q_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} {\int_{0}^{\infty} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} \right] \mathrm{d} \eta+t_{0}
$$
C. $$
t=\frac{l q_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} {\int_{0}^{\infty} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} \right] \mathrm{d} \eta+t_{0}
$$
D. $$
t=\frac{k q_{0}} {\sqrt{5 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\beta} \operatorname{e x p} (-\beta^{2} ) \mathrm{d} \beta} {\int_{0}^{\infty} \operatorname{e x p} (-\beta^{2} ) \mathrm{d} \beta} \right] \mathrm{d} \eta+t_{0}
$$
E. $$
t=\frac{c q_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{e x p} (-\xi^{2} ) \mathrm{d} \xi} {\int_{0}^{\infty} \operatorname{e x p} (-\xi^{2} ) \mathrm{d} \xi} \right] \mathrm{d} \eta+t_{0}
$$
F. $$
t=\frac{k q_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\gamma} \operatorname{e x p} (-\gamma^{2} ) \mathrm{d} \gamma} {\int_{0}^{\infty} \operatorname{e x p} (-\gamma^{2} ) \mathrm{d} \gamma} \right] \mathrm{d} \eta+t_{0}
$$
G. $$
t=\frac{k h_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} {\int_{0}^{\infty} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} \right] \mathrm{d} \eta+t_{0}
$$
H. $$
t=\frac{k q_{0}} {\sqrt{4 \alpha\theta}} \int_{0}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{e x p} (-\eta^{2} ) \mathrm{d} \eta} {\int_{0}^{\infty} \operatorname{e x p} (-\lambda^{2} ) \mathrm{d} \lambda} \right] \mathrm{d} \eta+t_{0}
$$
I. $$
t=\frac{k q_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{s i n} (-\eta^{2} ) \mathrm{d} \eta} {\int_{0}^{\infty} \operatorname{s i n} (-\eta^{2} ) \mathrm{d} \eta} \right] \mathrm{d} \eta+t_{0}
$$
J. $$
t=\frac{k q_{0}} {\sqrt{4 \alpha\theta}} \int_{\infty}^{t} \left[ 1-\frac{\int_{0}^{\eta} \operatorname{e x p} (-\eta^{3} ) \mathrm{d} \eta} {\int_{0}^{\infty} \operatorname{e x p} (-\eta^{3} ) \mathrm{d} \eta} \right] \mathrm{d} \eta+t_{0}
$$
Answer: B