# What is the general solution of the differential equation? : # Acosx + Bsinx -sinxln|cscx+cotx| #

##### 1 Answer

# 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,

**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

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

# 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] = | ( 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| #