Use the method of "undetermined coefficients" to solve the 2nd ODE #y''+4y'+4y=e^(-2x)sin2x# ?

We have:

# y''+4y'+4y = e^(-2x)sin2x #

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

# y''+4y'+4y #

And it's associated Auxiliary equation is:

# m^2+4m+4 = 0 #
# (m+2)^2 = 0 #

Which has a repeated real root #m=-2#

Thus the solution of the homogeneous equation is:

# y_c = (Ax+B)e^(-2x) #

Particular Solution

With this particular equation [A], a probable solution is of the form:

# y = e^(-2x)(acos2x + bsin2x) #

Where #a,b# are constant coefficients to be determined. Let us assume the above solution works, in which case be differentiating wrt #x# we have:

# y' \ \ = e^(-2x)(-2asin2x + 2bcos2x) -2e^(-2x)(acos2x + bsin2x)#
# \ \ \ \ \ = e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)#

And:

# y'' = e^(-2x)(-2(2b-2a)sin2x - 2(2a+2b)cos2x) -2 e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x) #

# \ \ \ \ \ = e^(-2x)(-8b)cos2x + (8a)sin2x) #

Substituting into the initial Differential Equation #[A]# we get:

# {e^(-2x)(-8b)cos2x + (8a)sin2x)} + 4{e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)} + 4{e^(-2x)(acos2x + bsin2x)} = e^(-2x)sin2x #

Equating coefficients of #cos2x# and #sin2x# we get:

#cos2x: -8b + 8b-8a + 4a =0 #
#sin2x: 8a - 8a-8b+4b=1 #

Solving simultaneous we have:

# a = 0, b=-1/4#

And so we form the Particular solution:

# y_p = -1/4e^(-2x)sin2x #

General Solution

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = (Ax+B)e^(-2x) -1/4e^(-2x)sin2x #
# \ \ \ \ \ \ \ = Axe^(-2x)+Be^(-2x) -1/4e^(-2x)sin2x #