Question

What is the general solution of the differential equation? `z'''-5z''+25z'-125z=1000`

Answer 1

`z(x) = e^(5x)+Acos(5x)+Bsin(5x) - 8`

Explanation:

Assuming that we have:

`z'''-5z''+25z'-125z=1000` .... [A]

This is a third order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, `z_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, `z_p` of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

`z'''-5z''+25z'-125z=0`

And it's associated Auxiliary equation is:

`m^3-5m^2+25m-125 = 0`

The hardest part with higher order DE is solving this equation. If we consider the graph `y = x^3-5x^2+25x-125`:
graph{y = x^3-5x^2+25x-125 [-10, 10, -30 30]}

We note there is one real solution at `x=5`, with this in mind we can factorise the cubic auxiliary equation:

`(m-5)(m^2+25)`

And so we have one real real `m=5` and two pure imaginary roots `m=+-5i`

Thus the solution of the homogeneous equation is:

`z_c = Ae^(5x)+Bcos(5x)+Csin(5x)`

Particular Solution

With this particular equation [A], a probable solution is of the form:

`z = a`

Where `a` is a constant coefficient to be determined. Let us assume the above solution works, in which case be differentiating wrt `x` we have:

`z' = z'' = z''' = 0`

Substituting into the initial Differential Equation `[A]` we get:

`=0-0+0-125a=1000 => a = -8`

And so we form the Particular solution:

`z_p = -8`

General Solution

Which then leads to the GS of [A}

`z(x) = z_c + z_p`
`\ \ \ \ \ \ \ = Ae^(5x)+Bcos(5x)+Csin(5x) - 8`