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

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 #