(D2-4D+3)y=sin2xcosx

Answer 1

Solve the characteristic equation to get the homogeneous solution as
yh(x)=C1*e^x+C2*e^3x, or
y1(x)=e^x, y2(x)=e^(3x)

Use variation of parameters to find the particular solution, namely, assume the particular solution to be
yp(x)=v1(x)y1(x)+v2(x)y2(x)...(0)
which implies:
v1'(x)y1(x)+v2'(x)y2(x)=0 ...(1)
v1'(x)y1'(x)+v2'(x)y2'(x)=sin(2x)cos(x) ...(2)

Substitute y1'(x) and y2'(x) in (2) to get
v2'(x)=(e^(-3x)/2)sin(2x)cos(x)...(3)
and by substituting v2'(x) into (1), we get
v1'(x)=-(e^(-x)/2)sin(2x)cos(x)...(4)

Integrate v1'(x) and v2'(x) to find v1(x) and v2(x).
Substitute v1(x) and v2(x) into (0) to find yp(x).

The general solution is:
y(x)=yh(x)+yp(x)
=C1*e^x+C2*e^(3x) + (3sin(x)+6cos(x)-sin(3x)+2cos(3x))/60

Check my arithmetic or typo.