Question

1. Use Finite Differences to approximate solutions to the linear BVPs with n = 4 subin- tervals. (a) y+e y(0) 0 1 y(1) -e 3 te (0,1); (1) (2) (3) (4) (b) y" (2 + 47) 1 y(0) y(1) (5) (6) (7) (8) e € (0,1); (c) Plot the solutions from parts (a) and (b) on the same plot.

Answer 1

To use finite differences to approximate solutions to the linear BVPs with n = 4 subintervals, we can use the following approach:

For part (a):

We have the linear BVP:

y'' + e^y = 0 y(0) = 1, y(1) = -e^3t The domain is (0,1).

We can use a central difference approximation for y''(x) and an explicit difference approximation for y(x):

y''(x) ≈ [y(x+h) - 2y(x) + y(x-h)]/h^2 y(x+h) ≈ y(x) + hy'(x) + (h^2/2)y''(x) + (h^3/6)y'''(x) + O(h^4) y(x-h) ≈ y(x) - hy'(x) + (h^2/2)y''(x) - (h^3/6)y'''(x) + O(h^4)

Substituting these approximations into the differential equation and the boundary conditions, we get:

[y(x+h) - 2y(x) + y(x-h)]/h^2 + e^y(x) ≈ 0 y(0) ≈ 1 y(1) ≈ -e^3t

We can use the method of successive approximations to solve this system of equations. Let y^0(x) = 1, and iterate as follows:

y^k+1(x) = [h^2e^y^k(x) - y^k-1(x+h) + 2y^k(x) - y^k-1(x-h)]/h^2

For k = 1, 2, 3, 4, we have n = 4 subintervals, so h = 1/4.

Therefore, the finite difference approximation for the solution y(x) is:

y^4(x) = [h^2e^y^3(x) - y^2(x+h) + 2y^3(x) - y^2(x-h)]/h^2

For part (b):

We have the linear BVP:

y'' + (2 + 4t)y = 1 y(0) = 0, y(1) = e The domain is (0,1).

We can use the same approach

Step-by-step explanation: