Solve the wave equation $u_{tt} = c^2 u_{xx}$ for $u(x,0) = sin(pi x)$ and $u_t(x,0) = 0$ on the domain $[0, 1]$ with fixed endpoints.

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∻Solution by Steps∻ ‖ step 1 ⋮ We start by solving the wave equation $u_{tt} = c^2 u_{xx}$ with the given initial conditions $u(x,0) = \sin(\pi x)$ and $u_t(x,0) = 0$ on the domain $[0, 1]$ with fixed endpoints. ‖ ‖ step 2 ⋮ We use the method of separation of variables. Assume $u(x,t) = X(x)T(t)$. Substituting into the wave equation, we get $X(x)T''(t) = c^2 X''(x)T(t)$. ‖ ‖ step 3 ⋮ Dividing both sides by $X(x)T(t)$, we obtain $\frac{T''(t)}{c^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda$. This gives us two ordinary differential equations: $T''(t) + \lambda c^2 T(t) = 0$ and $X''(x) + \lambda X(x) = 0$. ‖ ‖ step 4 ⋮ For the spatial part, $X''(x) + \lambda X(x) = 0$, with boundary conditions $X(0) = 0$ and $X(1) = 0$. The solutions are $X_n(x) = \sin(n \pi x)$ for $\lambda = (n \pi)^2$. ‖ ‖ step 5 ⋮ For the temporal part, $T''(t) + (n \pi c)^2 T(t) = 0$. The solutions are $T_n(t) = A_n \cos(n \pi c t) + B_n \sin(n \pi c t)$. ‖ ‖ step 6 ⋮ Combining these, we get $u_n(x,t) = \left( A_n \cos(n \pi c t) + B_n \sin(n \pi c t) ight) \sin(n \pi x)$. ‖ ‖ step 7 ⋮ Using the initial condition $u(x,0) = \sin(\pi x)$, we find $A_1 = 1$ and $A_n = 0$ for $n eq 1$. ‖ ‖ step 8 ⋮ Using the initial condition $u_t(x,0) = 0$, we find $B_n = 0$ for all $n$. ‖ ‖ step 9 ⋮ Therefore, the solution is $u(x,t) = \cos(\pi c t) \sin(\pi x)$. ‖ ∻Answer∻ ⚹ $u(x,t) = \cos(\pi c t) \sin(\pi x)$ ⚹ ∻Key Concept∻ ⚹Wave Equation Solution⚹ ∻Explanation∻ ⚹The solution to the wave equation with given initial and boundary conditions is found using separation of variables, leading to a product of sine and cosine functions.⚹◊What method can be used to solve the wave equation $u_{tt} = c^2 u_{xx}$ with the given initial conditions on the domain $[0, 1]$?⍭ Generate me a similar question◊

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