Solve the differential equation # (D^2+2D+5)y=xe^x # ?

We have:

# (D^2+2D+5)y=xe^x #

Where #D# is the linear differential operator #d/dx#. Thus we can write the equation as:

# (d^2y)/(dx^2) + 2dy/dx + 5y = xe^x # ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. 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.

Complimentary Function

The homogeneous equation associated with [A] is

# (d^2y)/(dx^2) + 2dy/dx + 5y = 0#

And it's associated Auxiliary equation is:

# m^2 + 2m+5 = 0 #

Which has two complex solutions #m=-1+-2i#

Thus the solution of the homogeneous equation is:

# y_c = e^(-1x)(Acos2x + Bsin2x) #
# \ \ \ = Ae^(-x)cos2x + Be^(-x)sin2x #

Particular Solution

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

# y = (ax+b)e^x #

Where the constants #a# and #b# are to be determined by direct substitution and comparison:

Differentiating wrt #x# (using the product rule) we get:

# dy/dx = (ax+b)e^x+(a)e^x #
# " " = (ax+a+b)e^x #

Differentiating again wrt #x# (using the product rule) we get:

# (d^2y)/(dx^2) = (ax+a+b)e^x+(a)e^x #
# " " = (ax+2a+b)e^x #

Substituting into the DE [A] we get:

# (ax+2a+b)e^x + 2(ax+a+b)e^x +5(ax+b)e^x = xe^x#

# :. (ax+2a+b) + 2(ax+a+b) +5(ax+b) = x#

Equating coefficients of #x# and constants we get

# x^1: (2a+b) + 2(a+b) +5(b) = 0 => 4a+8b=0#
# x^0: (a) + 2(a) +5(a) = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=> 8a=1#

Solving simultaneously, we have:

#a=1/8#
#1/2+8b=0=>b=-1/16#

And so we form the Particular solution:

# y_p = (1/8x-1/6)e^x #
# \ \ \ = (xe^x)/8 - e^x/16#

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Ae^(-x)cos2x + Be^(-x)sin2x + (xe^x)/8 - e^x/16#