Answer
y = -$\frac{1}{13}$sin2x + $\frac{3}{26}$cos2x + $C_{1}$ + $C_{2}$$e^{3x}$
Work Step by Step
y"-3y'=sin2x
r²-3r=0
r(r-3)=0
r=0,3
$y_{c}$= $C_{1}$$e^{0x}$ + $C_{2}$$e^{3x}$
$y_{c}$= $C_{1}$ + $C_{2}$$e^{3x}$
$y_{p}$= Acos2x + Bsin2x
$y_{p}$'= -2Asin2x + 2Bcos2x
$y_{p}$"= -4Acos2x -4Bsin2x
Substitute in y"-3y'= sin2x
6A-4B=1 and -4A-6B=0
-4B=6A
B=-$\frac{3}{2}$A
substitute in 6A-4B=1
A=-$\frac{1}{13}$
B=$\frac{3}{26}$
$y_{p}$= -$\frac{1}{13}$sin2x + $\frac{3}{26}$cos2x
y = $y_{c}$+$y_{p}$
y = -$\frac{1}{13}$sin2x + $\frac{3}{26}$cos2x + $C_{1}$ + $C_{2}$$e^{3x}$
