Solve the equation y"+529y= e^2x where y(0)=y'(0)=0

First, I got the homogeneous solution:
y_c= A*cos(529x)+B*sin(529x)

Then by assuming
y_p=C*e^2x
and y"=4C*e^2x, plugging it into y"+529y= e^2x,
I find that C =1/533
Therefore, y_p= (1/533)*e^2x right?

Then, using the initial conditions I find that A=B=0
So the final solution should be y=(y_c)+(y_p)
making it y= (0)cos(529x) +(0)sin(529x) +(1/533)*e^2x

Which would just be
y=(1/533)*e^2x right? But I'm told it's incorrect...did I miss something or am I just outright doing it wrong.

first find the complementary solution.
y"+529y=0
r^2+529=0
r^2=-529
r=(+/-)23i
y_c=C1*cos(23x)+C2*sin(23x)
now you need to find the particular solution.
you have e^2x
Assumption: Ae^2x
y=Ae^2x
y'=2Ae^2x
y"=4Ae^2x
now plug into y"+529y=e^2x
4Ae^2x+529Ae^2x=e^2x
e^2x[4A+529A]=e^2x
533A=1
A=1/533
Particular solution=(1/533)e^2x
now to find the general solution
y(x)=C1cos(23x)+C2sin(23x)+(1/533)e^2x
y'(x)=-23C1sin(23x)+23C2cos(23x)+(2/53…
y(0)=0
0=C1+1/533
C1=-1/533
y'(0)=0
0=23C2+2/533
-2/533=23C2
-2/12259=C2.
Answer = (1/533)cos(23x)-(2/12259)sin(23x)+(1/533…

do your own homework slacker.

Try A =1/533 and B =2/(533*529)