Second order ODE question.

I don't know what's wrong with me. I'm in my second year of uni and I feel embarrassed to ask this question, because I SHOULD know it.

i have to solve this second order ODE:

$\frac{d^2y}{dx^2} +\frac{dy}{dc} -2y = -2$

$y(0) = 1$ $y(1)=2$ find $y(0.2)$


the thing that's annoying me is that 2.
What do we do with it?
Here's what I've done so far:

associated homogeneous equation: $y'' + y'-2y = 0$
so $m=-2, 1$

complementary solution: $Y_c = C_1e^x + C_2e^{-2x}$

now for the particular solution: $Y_p=A$ so $Y_p' = y_p''=0$
and substituting that in i get A =1 so $Y_p =1$

to satisfy the initial conditions we want,
$C_1 + C_2 = 1$
$C_1e + C_2e^{-2x} = 2$

so $C_2 = \frac{e^2}{1-e^3}$ and $1-\frac{e^2}{1-e^3}$

and now $y = C_1e^x + C_2e^{-2x} + 1$

$y(0.2) = 3.1446$ but the answer is 1.49588


Any help will be greatly appreciated!

before working out the simultaneuous equation, you need to say general=CF +PI and THEN work out the value of the constants - you have forgotten the PI which you worked out earlier.

$C_1 + C_2 = 1$
$C_1e + C_2e^{-2} = 2$

Thanks for replying.
ok I did that, but i get 0.4789 which is still wrong.

you have misunderstood me- because you didn't take into account the extra 1
y=Ae^x+Be^2x+1

NOW use the intial conditions. You should find that you get:
A+B=0 and Ae^3+B=e^2

Edit: by the way I subbed y(0.2) and I got 1.21335...... I checked it on wolframalpha