What is the general solution of the differential equation? : # (d^2y)/(dx^2) + dy/dx - 2y = -6sin2x-18cos2x#

We have:

# (d^2y)/(dx^2) + dy/dx - 2y = -6sin2x-18cos2x# ..... [A]

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'' + y' - 2y = 0 #

And it's associated Auxiliary equation is:

# m^2 + m - 2 = 0 #
# (m-1)(m+2) = 0 #

Which has two real and distinct solutions #m=-2,1#

Thus the solution of the homogeneous equation is:

# y_c = Ae^(-2x)+Be^(1x) #
# \ \ \ = Ae^(-2x)+Be^(x) #

Particular Solution

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

# y = acos(2x)+bsin(2x) #

Where #a# and #b# are constants to be determined by substitution

Let us assume the above solution works, in which case be differentiating wrt #x# we have:

# y' \ \= -2asin(2x)+2bcos(2x) #
# y'' = -4acos(2x)-4bsin(2x) #

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

# -4acos(2x)-4bsin(2x) -2asin(2x)+2bcos(2x) - 2acos(2x) - 2bsin(2x) = -6sin2x-18cos2x#

Equating coefficients of #cos(2x)# and #sin(2x)# we get:

#cos(2x): -4a+2b-2a=-18#
#sin(2x): -4b-2a-2b=-6 #

Solving simultaneously we get:

# a=3# and #b=0#

And so we form the Particular solution:

# y_p = 3cos(2x)#

General Solution

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Ae^(-2x)+Be^(x) + 3cos(2x)# ..... [B]

Initial Conditions

We are given the initial conditions:

# y(0)=2, y'(0)=2 #

Putting #x=0# in [B] we get:

# 2 = Ae^0+Be^0 + 3cos0 => A+B = -1 #

Differentiating [B] wrt #x# we get:

# y'(x) = -2Ae^(-2x)+Be^(x) -6 sin(2x) #

Putting #x=0# we get:

# 2 = -2Ae^0+Be^0 -6 sin0 => -2A + B = 2#

Solving these two new equations simultaneously we get:

# A=-1# and #B=0#

leading to the specific solution:

# y(x) = -e^(-2x) + 3cos(2x)#