From (X')/(X) = k, integrating with respect to x gives X(x) = C_1 e^(kx). From (2T' + T)/(T) = k, we have 2(T')/(T) + 1 = k, which simplifies to (T')/(T) = (k - 1)/(2). Integrating with respect to t gives T(t) = C_2 e^((k - 1)/(2)t).

From (X')/(X) = k, integrating with respect to x gives X(x) = C_1 e^(kx). From (2T' + T)/(T) = k, we have 2(T')/(T) + 1 = k, which simplifies to (T')/(T) = (k - 1)/(2). Integrating with respect to t gives T(t) = C_2 e^((k - 1)/(2)t).

Multiplying the solutions gives u(x, t) = C e^(kx + (k - 1)/(2)t), where C = C_1 C_2. Applying the given condition u(x, 0) = 6e^(-3x), we match terms to find C = 6 and k = -3.

Substitute k = -3 and C = 6 into the general solution. The exponent in t evaluates to (-3 - 1)/(2)t = -2t. Thus, the unique solution is u(x, t) = 6e^(-3x - 2t).

Method of Separation of variables 1. Use the method of separation of variables to solve the equation 𝜕𝑢𝜕𝑥 = 2𝜕𝑢𝜕𝑡+ u; 𝑢(𝑥, 0) = 6𝑒−3𝑥.

Answer: $u(x, t) = 6e^{-3x - 2t}$

Step-by-step solution

Step 1: Assume a separable product solution

We assume a trial solution in product form $u(x, t) = X(x)T(t)$ , where $X$ is a function solely of $x$ and $T$ is a function solely of $t$ . Computing the partial derivatives gives $\frac{\partial u}{\partial x} = X'T$ and $\frac{\partial u}{\partial t} = XT'$ .

Step 2: Separate the variables

Substitute the derivatives into the PDE $\frac{\partial u}{\partial x} = 2\frac{\partial u}{\partial t} + u$ to get $X'T = 2XT' + XT$ . Dividing both sides by $XT = u(x, t)$ separates the variables into an expression in $x$ alone on the left and an expression in $t$ alone on the right. Since they depend on independent variables, both sides must equal a constant $k$ .

Step 3: Solve the ordinary differential equations

From $\frac{X'}{X} = k$ , integrating with respect to $x$ gives $X(x) = C_1 e^{kx}$ . From $\frac{2T' + T}{T} = k$ , we have $2\frac{T'}{T} + 1 = k$ , which simplifies to $\frac{T'}{T} = \frac{k - 1}{2}$ . Integrating with respect to $t$ gives $T(t) = C_2 e^{\frac{k - 1}{2}t}$ .

Step 4: Combine and apply the initial condition

Multiplying the solutions gives $u(x, t) = C e^{kx + \frac{k - 1}{2}t}$ , where $C = C_1 C_2$ . Applying the given condition $u(x, 0) = 6e^{-3x}$ , we match terms to find $C = 6$ and $k = -3$ .

Step 5: Write the final particular solution

Substitute $k = -3$ and $C = 6$ into the general solution. The exponent in $t$ evaluates to $\frac{-3 - 1}{2}t = -2t$ . Thus, the unique solution is $u(x, t) = 6e^{-3x - 2t}$ .

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