What is the solution to the Differential Equation #(d^2y)/(dx^2) - 6dy/dx = 54x + 18#?

1 Answer
Jul 21, 2017

# y(x) = A + Be^(6x)-9/2x^2-9/2x#

Explanation:

We have:

# (d^2y)/(dx^2) - 6dy/dx = 54x + 18 #

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''-6y'+0y = 0 #

And it's associated Auxiliary equation is:

# m^2-6m+0 = 0 #
# m^2-6m = 0 #
# m(m-6) = 0 #

Which has two real and distinct solutions #m0,6#

Thus the solution of the homogeneous equation is:

# y_c = Ae^(0x) + Be^(6x) #
# \ \ \ = A + Be^(6x)#

Particular Solution

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

# y = ax^2+bx +c#

Where #a,b,c# are constants to be determined by substitution

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

# y' \ \= 2ax+b #
# y'' = 2a #

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

# 2a - 6(2ax+b) = 54x + 18 #
# :. 2a - 12ax-6b = 54x + 18 #

Equating coefficients of #x^0# and #x# we get:

#x^0: 2a-6b=18 #
#x^1: -12a=54 #

Solving simultaneous we have:

# a = -9/2, b = -9/2 #

And so we form the Particular solution:

# y_p = -9/2x^2-9/2x#

General Solution

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = A + Be^(6x)-9/2x^2-9/2x#

As we have a linear combination of three linearly independent solutions, this is the GS.