second order ODE. I have this second order ODE:
$$-au''+bu'=x^2$$
where $u=u(x)$, $a=\text{constant}>0$, $b=\text{constant}\ge0 $
and  $$u(0)=1 ,u(1)=0$$
I tried to solve it and I get:
$$u(x)=D+Ce^{bx/a}$$
where $D,C$ are constant.

Is my answer correct?
How I can find $D$ and $C$?
I realy need help because there are 3 other quistions depending on this one.
Thanks.

Hint
I'd guess particular solution as
\begin{align}
u_p' &= Ax^2+Bx+C \\
u_p'' &= 2Ax + B
\end{align}
Now, substitute it to the ODE given
$$
-2Aax - Ba + Abx^2 + Bbx + Cb = Abx^2 + (Bb - 2Aa)x + Cb - Ba = x^2
$$
from which you can find
\begin{align}
Ab &= 1 \\
Bb - 2Aa &= 0 \\
Cb - Ba &= 0
\end{align}
$$
A = \frac 1b; B = \frac {2a}{b^2}; C = \frac {2a^2}{b^3}
$$
So, particular solution
$$
u_p = \frac A3 x^3 + \frac B2 x + Cx = \frac 1{3b}x^3 + \frac a{b^2}x^2 + \frac {2a^2}{b^3} x
$$
Therefore, general solution
$$
u = C_1 + C_2 e^{\frac ba x} + \frac 1{3b}x^3 + \frac a{b^2}x^2 + \frac {2a^2}{b^3} x
$$
If you substitute BCs you get
\begin{align}
u(0) &= C_1 + C_2 = 1 \\
u(1) &= C_1 + C_2 e^{\frac ba} + \frac 1{3b} + \frac a{b^2} + \frac {2a^2}{b^3} = 0
\end{align}
After solving which you get your constants $C_1$ and $C_2$.
Update
As for the case with $b = 0$, that's different story. You need to solve
$$
-au'' = x^2
$$
which has a solution
$$
u = -\frac 1{12a}x^4 + C_1 x + C_2
$$
and you need to substitute BC to this one rather than the solution I provided above.
