$y'' - 4y' + 5y = 25x - 3e^2x$
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?