#y'' + 9y = 2x^2 - 5#, How about solution. (#y_h#=?, #y_p=#?)

1 Answer
May 17, 2018

# y(x) = Acos(3x)+Bsin(3x) + 2/9x^2 -49/91 #

Explanation:

We seek a solution to

# y'' + 9y = 2x^2 - 5 # ..... [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 polynomial 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'' + 9y = 0 # ..... [B]

And it's associated Auxiliary equation is:

# m^2+9 = 0#

And so we have the solutions:

# m = +--1 \ \ \ \ # (pure imaginary)

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

  • Real distinct roots #m=alpha,beta, ...# will yield linearly independent solutions of the form #y_1=Ae^(alphax)#, #y_2=Be^(betax)#, ...
  • Real repeated roots #m=alpha#, will yield a solution of the form #y=(Ax+B)e^(alphax)# where the polynomial has the same degree as the repeat.
  • Complex roots (which must occur as conjugate pairs) #m=p+-qi# will yield a pairs linearly independent solutions of the form #y=e^(px)(Acos(qx)+Bsin(qx))#

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

# y = e^(0x)(Acos(3x)+Bsin(3x) ) #
# \ \ = Acos(3x)+Bsin(3x) #

Particular Solution

In order to find a particular solution of the non-homogeneous equation:

# y'' + 9y = f(x) \ \ # with #f(x) = 2x^2 - 5 #

So, we should probably look for a solution of the form:

# y = ax^2+bx+c # ..... [C]

Where the constants #a,b,c,d# is to be determined by direct substitution and comparison:

Differentiating [C] wrt #x# twice we get:

# y^((1)) = 2ax+b #
# y^((2)) = 2a #

Substituting these results into the DE [A] we get:

# (2a) + 9(ax^2+bx+c) = 2x^2 - 5 #

Equating coefficients we get:

# x^2: 9a = 2 => a = 2/9 #
# x^1: 9b=0=> b =0 #
# x^0: 2a+9c = -5 => c = -49/91 #

And so we form the Particular solution:

# y_p = 2/9x^2 -49/91 #

General Solution

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Acos(3x)+Bsin(3x) + 2/9x^2 -49/91 #

Note this solution has #2# constants of integration and #2# linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution