Partial differential equation

[Jncik]

Homework Statement

solve

u_x = 2*u_y when u(0,y) = e^(-y)

The Attempt at a Solution

this my attempt

using separation of variables I get

X'(x) - 2 λ X(x) = 0
Y'(y) - 2 λ Y(y) = 0

now I have to figure out for different λs what's happening

so

for λ = 0 I get X'(x) = 0 hence X(x) = A where A is a constant

also Y'(y) = 0 hence Y(y) = B where B is a constant

this gives us a solution C = BA where u(x,y) = C

now from the boundary conditions we have u(0,y) = e^(-y) => C = e^(-y)

hence u(x,y) = e^(-y)

for λ>0 I get a solution X(x) = A e^(λ x) and Y(y) = B e^(λx)

from the boundary conditions I get AB = e^(-y) hence

u(x,t) = e^(λx - y)

for λ<0 I have the same thing as for λ>0 but a solution u(x,t) = e^(-y) * e^(-3 λ x)

is this correct?

 

[mighty2000]

Your attempt at solving the given problem is mostly correct. However, there are a few errors and areas of improvement that can be made. Let's go through them one by one:

1. Separation of variables: In your attempt, you correctly separate the given equation into two separate equations with respect to x and y. However, the notation you use is a bit confusing. Instead of using X'(x) and Y'(y), it would be better to use u'(x) and v'(y) to represent the derivatives of the functions X(x) and Y(y).

2. Boundary conditions: You correctly apply the given boundary condition u(0,y) = e^(-y) to find the value of C. However, in your solution, you have written u(x,y) = e^(-y), which is not correct. The correct solution should be u(x,y) = e^(-y) * X(x) * Y(y).

3. Solutions for different λ values: For λ = 0, your solution is correct. However, for λ > 0, the solution should be u(x,y) = e^(λx - y). Also, for λ < 0, the solution should be u(x,y) = e^(-3λx - y). This can be easily verified by substituting these solutions back into the original equation.

4. General solution: Your solution for the general case is close, but not entirely correct. The correct general solution should be u(x,y) = e^(λx - y) + e^(-3λx - y), where λ can take on any value (positive or negative).

Overall, your attempt at solving the given problem is good, but there are a few areas that can be improved upon. Keep practicing and you'll get better at solving these types of problems. Good luck!