1. ## c1 and c2

I need your help with particular solution of the following DE

$y'' + 2y'+5y=e^x , y(0)=0.125, y'(0)=0.125$

The solution I get is:
$y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x$

To get C1 and C2 I do this:
$y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)$
$C_1=0.125$

$y_1^{'}=-C_1 e^{-x} cos2x-2C_1 e^{-x} sin2x-C_2 e^{-x} sin2x+2C_2 e^{-x} cos2x$

$C_2=0.125$

I know that the particular solution should be: $y=0.125e^{x}$
Can you tell me what I'm doing wrong?

2. Originally Posted by hekt
I need your help with particular solution of the following DE

$y'' + 2y'+5y=e^x , y(0)=0.125, y'(0)=0.125$

The solution I get is:
$y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x$

To get C1 and C2 I do this:
$y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)$
$C_1=0.125$

$y_1^{'}=-C_1 e^{-x} cos2x-2C_1 e^{-x} sin2x-C_2 e^{-x} sin2x+2C_2 e^{-x} cos2x$

$C_2=0.125$

I know that the particular solution should be: $y=0.125e^{x}$
Can you tell me what I'm doing wrong?
I haven't checked your general solution but you should realise that to find C1 and C2 you substitute the given boundary conditions into $y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x$ not $y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)$.