- #1

bhavik22

- 5

- 0

## Homework Statement

Use a Green's function to solve:

u" + 2u' + u = e

^{-x}

with u(0) = 0 and u(1) = 1 on 0[tex]\leq[/tex]x[tex]\leq[/tex]1

## 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

From the characteristic equation,

[tex]\lambda[/tex]

^{2}+ 2[tex]\lambda[/tex] + 1 = 0

[tex]\lambda[/tex] = -1 (repeated root)

Characteristic solution:

u

_{1}(x) = c

_{1}e

^{-x}and u

_{2}= c

_{2}e

^{-x}x

To satisfy boundary conditions,

u(0) = c

_{1}e

^{0}+ c

_{2}e

^{0}(0)

c

_{1}= 0

Thus take f(x) = e

^{-x}x

and

u(1) = c

_{1}e

^{-1}+ c

_{2}e

^{-1}(1)

get c

_{1}= [tex]\frac{e}{2}[/tex] and c

_{2}= [tex]\frac{e}{2}[/tex]

Thus take g(x) = [tex]\frac{1}{2}[/tex]e

^{-x+1}(x+1)

evaluating the wronskian, W = [tex]\frac{1}{2}[/tex]e

^{-2x+1}

I contruct green's function as per the formula provided above and carry out the integral and get final answer of,

u(x) = xe

^{-x}(x

^{2}- 3)

which obviously doesn't satisfy the second boundary condition of u(1) = 1.

I found out the final solution from Wolfram alpha:

u = [tex]\frac{1}{2}[/tex]e

^{-x}x(x+2e-1)

I've also tried many different combinations of f(x) and g(x) but none seen to work

Any help appreciated !