1. **Problem Statement:** Solve the PDEs using the method of separation of variables. (a) $$\frac{\partial u}{\partial x} + u = \frac{\partial u}{\partial t}$$ with initial condition $$u = 4e^{-5x}$$ when $$t=0$$. (b) $$\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u$$ with initial condition $$u(x,0) = 4e^{-2x}$$. 2. **Method:** For PDEs where variables can be separated, assume a solution of the form $$u(x,t) = X(x)T(t)$$. 3. **Solve (a):** Start with $$\frac{\partial u}{\partial x} + u = \frac{\partial u}{\partial t}$$. Substitute $$u = X(x)T(t)$$: $$X'(x)T(t) + X(x)T(t) = X(x)T'(t)$$. Divide both sides by $$X(x)T(t)$$: $$\frac{X'(x)}{X(x)} + 1 = \frac{T'(t)}{T(t)}$$. Set both sides equal to a constant $$\lambda$$: $$\frac{X'(x)}{X(x)} + 1 = \lambda$$ and $$\frac{T'(t)}{T(t)} = \lambda$$. Rewrite the first: $$X'(x) = (\lambda - 1)X(x)$$. Rewrite the second: $$T'(t) = \lambda T(t)$$. Solve for $$X(x)$$: $$X(x) = Ae^{(\lambda - 1)x}$$. Solve for $$T(t)$$: $$T(t) = Be^{\lambda t}$$. General solution: $$u(x,t) = C e^{(\lambda - 1)x + \lambda t}$$. Apply initial condition $$u(x,0) = 4e^{-5x}$$: $$4e^{-5x} = C e^{(\lambda - 1)x}$$. Matching exponents: $$-5 = \lambda - 1 \Rightarrow \lambda = -4$$. So, $$u(x,t) = 4 e^{-5x} e^{-4 t}$$. 4. **Solve (b):** Given $$\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u$$. Assume $$u = X(x)T(t)$$: $$X'(x)T(t) = 2 X(x) T'(t) + X(x) T(t)$$. Divide both sides by $$X(x)T(t)$$: $$\frac{X'(x)}{X(x)} = 2 \frac{T'(t)}{T(t)} + 1$$. Set equal to constant $$\mu$$: $$\frac{X'(x)}{X(x)} = \mu$$ and $$2 \frac{T'(t)}{T(t)} + 1 = \mu$$. Rewrite second: $$2 \frac{T'(t)}{T(t)} = \mu - 1 \Rightarrow T'(t) = \frac{\mu - 1}{2} T(t)$$. Solve for $$X(x)$$: $$X(x) = De^{\mu x}$$. Solve for $$T(t)$$: $$T(t) = Ee^{\frac{\mu - 1}{2} t}$$. General solution: $$u(x,t) = F e^{\mu x + \frac{\mu - 1}{2} t}$$. Apply initial condition $$u(x,0) = 4 e^{-2x}$$: $$4 e^{-2x} = F e^{\mu x}$$. Matching exponents: $$-2 = \mu \Rightarrow \mu = -2$$. So, $$u(x,t) = 4 e^{-2x} e^{-\frac{3}{2} t}$$. **Final answers:** (a) $$u(x,t) = 4 e^{-5x} e^{-4 t}$$ (b) $$u(x,t) = 4 e^{-2x} e^{-\frac{3}{2} t}$$