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 #

Presuming we've got:

# 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 Role

The equation homogeneous linked to [A] is

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

Additionally, the related 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#

Consequently, the homogeneous equation's solution is:

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

Specific Resolution

For this specific equation [A], the following is a likely solution:

# 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#

Thus, we arrive at the specific solution:

# z_p = -8 #

Overall Resolution

which ultimately results in the GS of [A}

# z(x) = z_c + z_p # # \ \ \ \ \ \ \ = Ae^(5x)+Bcos(5x)+Csin(5x) - 8 #
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Answer 2

The general solution of the given differential equation is:

z(t) = c₁e^(5t) + c₂te^(5t) + c₃t^2e^(5t) + 20t^3 + 300t + 400

Where c₁, c₂, and c₃ are arbitrary constants.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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