Show that # xsin2x # is a solution to the DE # y'' + 4y = 4cos2x #?

To show this you need to derive twice your solution #y=xsin(2x)# and substitute for #Y' '# and #y# into your equation and see if it works.
Let us derive:
#Y'=sin(2x)+2xcos(2x)#
again:
#Y' '=2cos(2x)+2cos(2x)-4xsin(2x)#

let us substitute #Y' '# and the given result #y=xsin(2x)# into our equation#:

#2cos(2x)+2cos(2x)-4xsin(2x)+4(xsin(2x))=4cos(2x)#

that gives #0=0# implying that the function #y=xsin(2x)# is solution of our equation.

We seek to show that:

# y=xsin2x #

is a solution to the DE:

# y'' + 4y = 4cos2x # ..... [A]

Given that we have the solution we can just differentiate (twice) and substitute:

# y' = (x)(d/dxsin2x)+(d/dxx)(sin2x) #
# \ \ \ \ = 2xcos2x+sin2x #

# y'' = (2x)(d/dxcos2x) + (d/dx2x)(cos2x) + 2cos2x #
# \ \ \ \ \ = -4xsin2x + 2cos2x + 2cos2x #
# \ \ \ \ \ = -4xsin2x + 4cos2x #

Substituting into the DE [A], we get:

# y'' + 4y = (-4xsin2x + 4cos2x) + 4(xsin2x) #
# " " = -4xsin2x + 4cos2x + 4xsin2x #
# " " = 4cos2x # QED.

Note: this solution, is known as the Particular Solution. We could go on and form the General Solution as follows:

Complementary Function

The homogeneous equation associated with [A] is

# y'' + 4y = 0 # ..... [B]

And it's associated Auxiliary equation is:

# m^2 + 4 = 0 #

Which has pure imaginary solutions #m=+-2i#

Thus the solution of the homogeneous equation [B] is:

# y_c = e^0(Acos(2x) + Bsin(2x)) #
# \ \ \ = Acos2x + Bsin2x #

Which then leads to the GS of [A}

# y(x) = y_c + y_p =Acos2x + Bsin2x # #+ xsin2x #