1. **Problem statement:** Determine the solution of the initial-value problem for the wave equation on a semi-infinite string with given initial and boundary conditions. 2. **Given for part (b):** $$u_{tt} = 16 u_{xx}, \quad 0 < x < \infty, t > 0$$ $$u(x,0) = 1 + x, \quad u_t(x,0) = x^3, \quad 0 \leq x < \infty$$ $$u_x(0,t) = \cos x, \quad 0 \leq t < \infty$$ 3. **Given for part (c):** $$u_{tt} - c^2 u_{xx} = 0, \quad 0 < x < \infty, t > 0$$ $$u(x,0) = \log(1 + x^2), \quad u_t(x,0) = 2, \quad 0 \leq x < \infty$$ 4. **Method:** For semi-infinite domain wave equations, the method of reflection or d'Alembert's formula adapted with boundary conditions is used. 5. **Part (b) solution:** Wave speed $c = 4$ since $16 = c^2$. The general d'Alembert solution for infinite domain is: $$u(x,t) = \frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds$$ Since domain is semi-infinite with Neumann boundary condition $u_x(0,t) = \cos x$, we use even extension of initial data to satisfy $u_x(0,t)$ condition. Define even extensions: $$f_e(x) = \begin{cases} f(x), & x \geq 0 \\ f(-x), & x < 0 \end{cases}$$ $$g_e(x) = \begin{cases} g(x), & x \geq 0 \\ g(-x), & x < 0 \end{cases}$$ Given $f(x) = 1 + x$, $g(x) = x^3$. Then solution: $$u(x,t) = \frac{f_e(x-ct) + f_e(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g_e(s) ds$$ Substitute $c=4$: $$u(x,t) = \frac{f_e(x-4t) + f_e(x+4t)}{2} + \frac{1}{8} \int_{x-4t}^{x+4t} g_e(s) ds$$ 6. **Part (c) solution:** General wave equation with speed $c$ and initial conditions: $$u(x,0) = \log(1 + x^2), \quad u_t(x,0) = 2$$ Boundary condition $u(0,t) = x$ is non-homogeneous and depends on $x$, which is unusual; assuming a typo and boundary condition is $u(0,t) = 0$ or similar for well-posedness. Assuming $u(0,t) = 0$, use odd extension for $f$ and even for $g$ to satisfy Dirichlet boundary condition. Define odd extension: $$f_o(x) = \begin{cases} f(x), & x \geq 0 \\ -f(-x), & x < 0 \end{cases}$$ Even extension for $g$: $$g_e(x) = \begin{cases} g(x), & x \geq 0 \\ g(-x), & x < 0 \end{cases}$$ Then solution: $$u(x,t) = \frac{f_o(x-ct) + f_o(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g_e(s) ds$$ 7. **Summary:** - For (b), use even extensions of $f$ and $g$ with $c=4$. - For (c), assuming $u(0,t)=0$, use odd extension of $f$ and even extension of $g$. This approach ensures boundary conditions are satisfied and initial conditions are met. **Final answers:** (b) $$u(x,t) = \frac{f_e(x-4t) + f_e(x+4t)}{2} + \frac{1}{8} \int_{x-4t}^{x+4t} g_e(s) ds$$ where $$f_e(x) = \begin{cases} 1 + x, & x \geq 0 \\ 1 - x, & x < 0 \end{cases}, \quad g_e(x) = \begin{cases} x^3, & x \geq 0 \\ (-x)^3 = -x^3, & x < 0 \end{cases}$$ (c) $$u(x,t) = \frac{f_o(x-ct) + f_o(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g_e(s) ds$$ where $$f_o(x) = \begin{cases} \log(1 + x^2), & x \geq 0 \\ -\log(1 + x^2), & x < 0 \end{cases}, \quad g_e(x) = \begin{cases} 2, & x \geq 0 \\ 2, & x < 0 \end{cases}$$