1. ## Initial condition problem

y''-2y'+2y=x+1, y(0) = 3, y'(0)=0

I know I need to find the homogenous solution and add it to a particular solution.

Homogenous:

$\displaystyle r^2-2r+2 = 0$

So $\displaystyle r = \{ 1+i , 1-i \}$, then the homogenous solution is $\displaystyle y_{h} = c_{1}e^{x}cosx+c_{2}e^xsinx$

Particular:

Let y = Ax + B, y'=A, y''=0

0 -2 (A) + 2(Ax+B) = x+1

A=1/2, B=1, $\displaystyle y_{p}= \frac {1}{2} x + 1$

yh + yp =$\displaystyle c_{1}e^{x}cosx+c_{2}e^xsinx + \frac {1}{2} x + 1$

y''-2y'+2y=x+1, y(0) = 3, y'(0)=0

I know I need to find the homogenous solution and add it to a particular solution.

Homogenous:

$\displaystyle r^2-2r+2 = 0$

So $\displaystyle r = \{ 1+i , 1-i \}$
You can do it the same way you do with real solutions:
$\displaystyle y(x) = Ae^{(1 + i)x} + Be^{(1 - i)x}$

We typically would want a real expression for this, so we may do the following:
$\displaystyle y(x) = e^x(Ae^{ix} + Be^{-ix})$

$\displaystyle = e^x(Acos(x) + Ai~sin(x) + Bcos(x) - Bi~sin(x))$

$\displaystyle = e^x((A + B)cos(x) + (A - B)i~sin(x))$
and we may replace the two arbitrary constants A and B with two new arbitrary constants:
$\displaystyle y(x) = e^x(a~cos(x) + b~sin(x))$

(What would happen here if you solved the problem using the original exponent form is that A and B would end up being complex constants.)

So
$\displaystyle y(0) = 3 = a$

$\displaystyle y^{\prime}(0) = a + b = 0$

So a = 3 and b = -3, giving
$\displaystyle y(x) = 3e^x(cos(x) - sin(x))$

I wasn't sure... Are you asking for help with the particular solution as well? Or do you have a handle on that?

-Dan