I started by trying to find the auxillary equation, as this is a homogenous second order DE.

I get to this:

k^2 - 2k + 1 + E^2

What can I do from here? (k-1)(k-1) + E^2 ???

Printable View

- Nov 29th 2010, 08:54 AMchr91Second order homogenous DE - Tricky
I started by trying to find the auxillary equation, as this is a homogenous second order DE.

I get to this:

k^2 - 2k + 1 + E^2

What can I do from here? (k-1)(k-1) + E^2 ??? - Nov 29th 2010, 09:11 AMTheEmptySet
For part a you are on the right path

$\displaystyle \displaystyle (k+1)^2+\epsilon^2=0 \implies k+1=\pm i\epsilon \iff k = -1\pm i\epsilon$

Now write down the solution and see what happens as $\displaystyle \epsilon \to 0$

For part b put the equation in standard form and solve. Then take the limits (Wink) - Nov 29th 2010, 09:23 AMchr91
Thanks, so for a, the complementary function is y = e^x (A cos E x + B sin E x)

But I'm finding it hard to find the values for A and B to find the general solution.

For example y = e^x (A cos E x + B sin E x) and x=0, y=1

1 = 1(AcosE)

So A = 1/cosE ?

EDIT : it's k = 1+or- E