The solution of U_x = 4U_y : U(0, y) = 8 e^−3y is : 

To solve the partial differential equation U_x = 4U_y with the initial condition U(0, y) = 8 e^−3y, we will use the method of characteristics. This method involves finding curves along which the partial differential equation reduces to an ordinary differential equation.

Step-by-Step Solution:

The given partial differential equation is:

U_x = 4U_y

We introduce new variables along the characteristic curves. Let:

dx/dt = 1

dy/t = 4

dU/dt = 0

From the first equation, we have:

dx/dt = 1 implies x = t + C₁

From the second equation, we have:

dy/dt = 4 implies y = 4t + C₂

Since C₁ and C₂ are constants of integration, we can combine them as follows:

y = 4x + C, where C is a new constant.

From the third equation:

dU/dt = 0 implies U is constant along the characteristics, meaning U is a function of the combination of variables x and y, specifically in the form U(x, y) = f(4x + y).

Given the initial condition U(0, y) = 8 e^−3y, we can find the specific form of the solution:

U(0, y) = f(y) = 8 e^−3y

Hence, the function f must be:

f(4x + y) = 8 e^{−3(4x + y)}

Therefore, the general solution to the partial differential equation with the given initial condition is:

U(x, y) = 8 e^{−3(4x + y)}