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

Explanation:

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)}

Conclusion:

The correct solution is U(x, y) = 8 e^{−3(4x + y)}. This corresponds to Option 1 in the provided choices.

Additional Information:

To further understand the analysis, let’s evaluate the other options:

Option 2: 8 e^{−3(4x + y)}

This option is incorrect because it does not match the correct form of the solution. The factor of 4 in the exponent is misplaced, and it does not correctly reflect the relationship between x and y in the characteristic solution.

Option 3: 8 e^{3(4x + y)}

This option is incorrect due to the positive exponent. The solution must have a negative exponent to satisfy the initial condition.

Option 4: 8 e^{3(3x + 4y)}

This option is incorrect because both the exponent and the coefficients of x and y inside the exponent do not match the derived solution.

Understanding the method of characteristics and the process of solving partial differential equations with initial conditions is crucial for correctly identifying the solution. The given partial differential equation U_x = 4U_y with the initial condition U(0, y) = 8 e^−3y has the solution U(x, y) = 8 e^{−3(4x + y)}, which corresponds to Option 1.