1. **State the problem:** Solve the partial differential equation (PDE) $$\frac{\partial u}{\partial t} = \frac{1}{4} \frac{\partial^2 u}{\partial x^2}$$ with boundary conditions $$u(0,t) = 0$$ and $$u(2,t) = 0$$ for $$t > 0$$, and initial condition $$u(x,0) = 2 \sin\left(\frac{\pi x}{2}\right) - \sin(\pi x)$$ for $$0 \leq x \leq 2$$. 2. **Use separation of variables:** Assume $$u(x,t) = X(x)T(t)$$. 3. Substitute into the PDE: $$X(x) T'(t) = \frac{1}{4} X''(x) T(t)$$ Divide both sides by $$X(x) T(t)$$: $$\frac{T'(t)}{T(t)} = \frac{1}{4} \frac{X''(x)}{X(x)} = k$$ where $$k$$ is a separation constant. 4. Set $$k = -a^2$$ (to satisfy boundary conditions and get nontrivial solutions): $$\frac{X''(x)}{X(x)} = -a^2 \implies X''(x) + a^2 X(x) = 0$$ 5. Solve the spatial ODE: General solution: $$X(x) = A \cos(ax) + B \sin(ax)$$ Apply boundary conditions: - At $$x=0$$: $$u(0,t) = X(0)T(t) = 0 \implies X(0) = 0 \implies A = 0$$ - At $$x=2$$: $$X(2) = B \sin(2a) = 0$$ For nontrivial $$B \neq 0$$, require: $$\sin(2a) = 0 \implies 2a = n\pi, n=1,2,3,...$$ Thus, $$a_n = \frac{n\pi}{2}$$ 6. So eigenfunctions are: $$X_n(x) = \sin\left(\frac{n\pi x}{2}\right)$$ 7. Solve the temporal ODE: $$T'(t) = -\frac{1}{4} a_n^2 T(t) = -\frac{1}{4} \left(\frac{n\pi}{2}\right)^2 T(t) = -\frac{n^2 \pi^2}{16} T(t)$$ General solution: $$T_n(t) = C_n e^{-\frac{n^2 \pi^2}{16} t}$$ 8. The general solution is a sum over all $$n$$: $$u(x,t) = \sum_{n=1}^\infty C_n e^{-\frac{n^2 \pi^2}{16} t} \sin\left(\frac{n\pi x}{2}\right)$$ 9. Use initial condition to find coefficients $$C_n$$: $$u(x,0) = 2 \sin\left(\frac{\pi x}{2}\right) - \sin(\pi x) = \sum_{n=1}^\infty C_n \sin\left(\frac{n\pi x}{2}\right)$$ Comparing terms, only $$n=1$$ and $$n=2$$ appear: $$C_1 = 2, \quad C_2 = -1$$ 10. **Final solution:** $$\boxed{u(x,t) = 2 e^{-\frac{\pi^2}{16} t} \sin\left(\frac{\pi x}{2}\right) - e^{-\frac{\pi^2}{4} t} \sin(\pi x)}$$