y"+4y=3sin(x).
The characteristic solution is obtained by solving y"+4y=0, that is, y"=-4y.
Let yc=Asin(2x)+Bcos(2x); yc'=2Acos(2x)-2Bsin(2x); yc"=-4Asin(2x)-4Bcos(2x)=-4yc.
The particular solution is probably of the form y=axsin(2x)+bxcos(2x):
Let yp=axsin(2x)+bxcos(2x); yp'=2axcos(2x)+asin(2x)-2bxsin(2x)+bcos(2x);
yp"=-4axsin(2x)+2acos(2x)+2acos(2x)-4bxcos(2x)-2bsin(2x)-2bsin(2x).
yp"+4yp=
-4axsin(2x)+2acos(2x)+2acos(2x)-4bxcos(2x)-2bsin(2x)-2bsin(2x)
+4axsin(2x)+4bxcos(2x)=4acos(2x)-4bsin(2x)=3sin(2x).
Therefore a=0, 4b=-3, b=-¾.
Complete solution is: yc+yp=y:
y=Asin(2x)+Bcos(2x)-¾xcos(2x) (A and B are constants).