Recent content by bhavik22

It works! Using the change of variables the final result is u(x) = (1/2)xe-x(x + 2e - 1) which satisfies the BCs Thanks a ton :smile:

In all the examples we're given you 'swap' the green's functions over to the other limits on the integral (so the green's function for x<=z<=1 is evaluated in the integral from 0 to x and the green's function for 0<=z<=x is evaluated in the integral from x to 1)? So I went over the integral...

Ah I used the correct f(x) in the integral, just wrote it incorrectly in my second post

Apologies for the formatting, I couldn't get the proper characters to work for some reason. Ok I used g(x)=c_1(1-x)e^{-x} with c1 = 1 so g(x)=(1-x)e^{-x} and f(x) = e-x get W = -e-2x G(x,z) = -e-x x e-z(z-1)/-e-2z for 0<=x<=z G(x,z) = -ze-z e-x(x-1)/-e-2z for z<=x<=1 Thus...

Homework Statement Use a Green's function to solve: u" + 2u' + u = e-x with u(0) = 0 and u(1) = 1 on 0\leqx\leq1 Homework Equations This from the lecture notes in my course: The Attempt at a Solution Solving for the homogeneous equation first: u" + 2u' + u = 0...