Variation of Parameters #y'' + 9y = 9sec^2(3x)# ; y = ?

1 Answer
Steve M
Jun 16, 2017

# y(x) = Acos3x + Bsin3x + sin3x \ ln|tan3x+sec3x| -1 #

Explanation:

We have:

# y'' + 9y = 9sec^2(3x) # ..... [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.

Complimentary Function

The homogeneous equation associated with [A] is

# y'' + 9y = 0 #

And it's associated Auxiliary equation is:

# m^2 + 9 = 0 #

Which has pure imaginary solutions #m=+-3i#

Thus the solution of the homogeneous equation is:

# y_c = e^0(Acos3x + Bsin3x) #
# \ \ \ = Acos3x + Bsin3x #

Particular Solution

With this particular equation [A], the interesting part is find the solution of the particular function. We would typically use practice & experience to "guess" the form of the solution but that approach is likely to fail here. Instead we mist use the Wronskian. It does, however, involve a lot more work:

Once we have two linearly independent solutions say #y_1(x)# and #y_2(x)# then the particular solution of the general DE;

# ay'' +by' + cy = p(x) #

is given by:

# y_p = v_1y_1 + v_2y_2 \ \ #, which are all functions of #x#

Where:

# v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx #
# v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx #

And, #W[y_1,y_2]# is the wronskian; defined by the following determinant:

# W[y_1,y_2] = | ( y_1,y_2), (y'_1,y'_2) | #

So for our equation [A]:

# p(x) = 9sec^2(3x) #
# y_1 \ \ \ = cos3x => y'_1 = -3sinx #
# y_2 \ \ \ = sin3x => y'_2 = 3cos3x #

So the wronskian for this equation is:

# W[y_1,y_2] = | ( cos3x,,sin3x), (-3sin3x,,3cos3x) | #

# " " = (cos3x)(3cos3x) - (sin3x)(-3sin3x) #
# " " = 3cos^2 3x +3sin^2 3x#
# " " = 3 #

So we form the two particular solution function:

# v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx #

# \ \ \ = -int \ (9sec^2(3x) sin3x)/3 \ dx #
# \ \ \ = -3 \ int \ 1/(cos^2 3x) sin3x \ dx #
# \ \ \ = -3 \ int \ sec3xtan3x \ dx #
# \ \ \ = -sec3x #

And;

# v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx #

# \ \ \ = int \ (9sec^2 3x cos3x )/3\ dx #
# \ \ \ = 3 \ int \ 1/(cos^2 3x) cos3x \ dx #
# \ \ \ = 3 \ int \ sec3x \ dx #
# \ \ \ = ln|tan3x+sec3x| #

And so we form the Particular solution:

# y_p = v_1y_1 + v_2y_2 #
# \ \ \ = (-sec3x)(cos3x) + (ln|tan3x+sec3x|)(sin3x) #
# \ \ \ = sin3x \ ln|tan3x+sec3x| -1 #

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Acos3x + Bsin3x + sin3x \ ln|tan3x+sec3x| -1 #

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