What is the general solution of the differential equation # y'' + y = cot(x) #?

1 Answer
Oct 24, 2017

# Acosx + Bsinx -sinxln|cscx+cotx| #

Explanation:

We have:

# y'' + y = cot(x) # ..... [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'' + 9y = 0 #

And it's associated Auxiliary equation is:

# m^2 + 1 = 0 #

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

Thus the solution of the homogeneous equation is:

# y_c = e^0(Acos(1x) + Bsin(1x)) #
# \ \ \ = Acosx + Bsinx #

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 must 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) = cotx #
# y_1 \ \ \ = cosx => y'_1 = -sinx #
# y_2 \ \ \ = sinx => y'_2 = cosx #

So the wronskian for this equation is:

# W[y_1,y_2] = | ( cosx,,sinx), (-sinx,,cosx) | #

# " " = (cosx)(cosx) - (sinx)(-sinx) #
# " " = cos^2 x + sin^2 x#
# " " = 1 #

So we form the two particular solution function:

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

# \ \ \ = -int \ (cotx sinx)/1 \ dx #
# \ \ \ = -int \ cosx/sinx sinx \ dx #
# \ \ \ = -int \ cosx \ dx #
# \ \ \ = -sinx #

And;

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

# \ \ \ = int \ (cotx cosx )/1\ dx #
# \ \ \ = int \ cosx/sinx cosx \ dx #
# \ \ \ = int \ cos^2x/sinx \ dx #
# \ \ \ = int \ (1-sin^2x)/sinx \ dx #
# \ \ \ = int \ cscx-sinx \ dx #
# \ \ \ = -ln|cscx+cotx|+cosx #

And so we form the Particular solution:

# y_p = v_1y_1 + v_2y_2 #
# \ \ \ = (-sinx)(cosx) + (cosx-ln|cscx+cotx|)sinx#

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Acosx + Bsinx -sinxcosx + sinxcosx-sinxln|cscx+cotx| #
# \ \ \ \ \ \ \ = Acosx + Bsinx -sinxln|cscx+cotx| #