Second Order Diffrential Equation $y'' - 4y' + 5y = 25x - 3e^2x$
Where $y(0)=0$ and $y'(0)=0$
This is what I have done so far:-
$r^2 - r + 5 = 0$ 
$\sqrt{b^2-4ac} < 0$ ...... I get $2\pm i$
Thus $y(x)=e^2x(A\cos x + D\sin x)$
How do I find $A$ and $D$?
Substitute $y(0)=0$ into the equation?
What about the right hand side equation?
This is what I did for that...
$g = 25x-3e^2x$
$g' = 25-6e^2x$
$g'' = -12e^2x$
Then I sub the values, giving me
$-12e^2x - 4(25-6e^2x) + 5(25x-3e^2x) = 25x - 3e^2x$
which will give me $x = 1$
Am I doing any of this right?
 A: Solving an imhomogenous ODE is done in two steps. 


*

*Determine the general solution to the corresponding homogenous ODE.

*Find a particular solution to the inhomogenous ODE.


It looks like youve done the first part correctly, but for completeness I'll repeat it here. Consider 
$$ y'' - 4y' + 5y = 0.$$
This is a linear ODE, so we know that our solutions are going to be exponentials. Substitute $y = \exp[rx]$ into the equation and find that:
$$ r^2 - 4r + 5 = 0.$$
We solve using the quadratic formula and find that $r = 2 \pm i$. Since $r$ is complex, the solution is a combination of exponential growth and oscillation:
$$ y(x) = Ae^{2x}\cos(x) + Be^{2x}\sin(x).$$
This is the general solution. Up to this point your answer is good. It's only in the second step that you've gone a little off track. There are a couple of ways to find particular solutions; I like the method of undetermined coefficients (this is the technical term for making an educated guess).
If we look at the inhomogenous equation $$ y'' - 4y' + 5y = 25x - 3e^{2x}$$
we notice that the right hand side is the sum of a polynomial and an exponential. Our educated guess is that the particular solution will also be the sum of a polynomial and an exponential, $$y_{p} = Ce^{2x} + Dx + E.$$
Now we substitute this into the inhomogenous expression and solve for $C$, $D$ and $E$.
$$ 4Ce^{2x} -4(2Ce^{2x} + D) + 5(Ce^{2x} + Dx +E) = 25x - 3e^{2x},$$
$$5Dx +(5E - 4D) + (4C - 8C + 5C)e^{2x} = 25x -0 - 3e^{2x},$$
comparing the coefficients of the $x$ terms shows that $D = 5$, comparing the constants shows that $E = 4$ and comparing the coefficients in front of the exponential shows that $C = -3$. The solution we want is just the sum of the general and particular solutions,
$$ y = Ae^{2x}\cos(x) + Be^{2x}\sin(x) -3e^{2x} +5x +4.$$
To calculate the values of $A$ and $B$ you use the given initial values, $y(0) = 0$ and $y'(0)=0$. 
$$ y(0) = A -3 + 4 \Rightarrow A = -1.$$
$$ y'(0) = 2B - 6 + 5 \Rightarrow B = \frac{1}{2}.$$
Substitute these back into the expression for the solution to get the final answer.
$$ y = -e^{2x}\cos(x) + \frac{1}{2}e^{2x}\sin(x) -3e^{2x} +5x +4.$$
