Question

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

Answer 1

`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, `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'' -y = 0`

And it's associated Auxiliary equation is:

`m^2 -1 = 0`

Which has distinct real solutions `m=+-1`

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 `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) = 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 `a,b,c` are constants to be determine. Using substitution we have:

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)`