# What is the general solution of the differential equation? : # (d^2y)/(dx^2)- y = 1/(1+e^x) #

##### 1 Answer

# y(x) = Ae^x+Be^(-x) -1/2xe^x - 1/2 - 1/2e^xln(1+e^x) -1/2 e^-x ln(1+e^x)#

#### Explanation:

We have:

# (d^2y)/(dx^2)- y = 1/(1+e^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'' -y = 0 #

And it's associated Auxiliary equation is:

# m^2 -1 = 0 #

Which has distinct real solutions

Thus the solution of the homogeneous equation is:

# y_c = Ae^x+Be^(-x) #

**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 (or variation of parameters). 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) = 1/(1+e^x) #

# y_1 \ \ \ = e^x \ \ \ => y_1' = e^x #

# y_2 \ \ \ = e^(-x) => y_2' = -e^-x #

So the wronskian for this equation is:

# W[y_1,y_2] = | ( e^x,,e^-x), (e^x,,-e^-x) | #

# " " = (e^x)(-e^-x) - (e^x)(e^-x) #

# " " = -2 #

So we form the two particular solution function:

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

# \ \ \ = -int \ (1/(1+e^x) (e^-x))/(-2) \ dx #

# \ \ \ = 1/2 \ int \ e^-x/(1+e^x) \ dx #

We can evaluate this integral using a substitution:

Let

#u=e^x => (du)/dx = e^x # and#e^(-x)=1/u# then

# v_1 = 1/2 \ int \ (1/u)/(1+u) \ (1/u) \ du#

# \ \ \ = 1/2 \ int \ 1/(u^2(1+u)) \ du#

No we decompose the integrand into partial fractions:

# 1/(u^2(1+u)) -= a/u + b/u^2+c/(1+u) #

# " " = (au(1+u) + b(1+u) + c u^2) / (u^2(1+u)) #

# => 1 -= au(1+u) + b(1+u) + c u^2 #

Where

Put

# u=0 => 1 = b #

Put# u=-11 => 1 = c #

Coeff#(u^2): 0 = a+c => a=-1 #

Thus:

# v_1 = 1/2 \ int \ -1/u + 1/u^2+1/(1+u) \ du#

# \ \ \ = 1/2 \ { -ln|u| - 1/u+ln|1+u| }#

# \ \ \ = 1/2 \ { -ln|e^x| - 1/e^x+ln|1+e^x| }#

# \ \ \ = 1/2 \ { -x - e^-x+ln(1+e^x) }#

# \ \ \ = -1/2x - 1/2e^-x + 1/2ln(1+e^x) #

And;

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

# \ \ \ = int \ (1/(1+e^x) e^x)/(-2) \ dx #

# \ \ \ = -1/2 \ int \ (e^x)/(1+e^x) \ dx #

We can similarly evaluate this integral using a substitution:

Let

#u=e^x => (du)/dx = e^x # then

# v_2 = -1/2 \ int \ (u)/(1+u) \ (1/u) \ du #

# \ \ \ = -1/2 \ int \ (u)/(1+u) \ (1/u) \ du #

# \ \ \ = -1/2 \ int \ (1)/(1+u) \ du #

# \ \ \ = -1/2 \ ln|1+u| #

# \ \ \ = -1/2 \ ln|1+e^x| #

# \ \ \ = -1/2 \ ln(1+e^x) #

And so we form the Particular solution:

# y_p = v_1y_1 + v_2y_2 #

# \ \ \ = (-1/2x - 1/2e^-x + 1/2ln(1+e^x))(e^x) + (-1/2 \ ln(1+e^x))(e^-x)#

# \ \ \ = -1/2xe^x - 1/2 - 1/2e^xln(1+e^x) -1/2 e^-x ln(1+e^x)#

Which then leads to the GS of [A}

# y(x) = y_c + y_p #

# \ \ \ \ \ \ \ = Ae^x+Be^(-x) -1/2xe^x - 1/2 - 1/2e^xln(1+e^x) -1/2 e^-x ln(1+e^x)#