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

##### 1 Answer

# 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,

**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

graph{y = x^3-5x^2+25x-125 [-10, 10, -30 30]}

We note there is one real solution at

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

And so we have one real real

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

# z' = z'' = z''' = 0#

Substituting into the initial Differential Equation

# =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 #