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The Laplace transform of the non-homogeneous second order differential equation is [tex]\mathcal {L} \{f(t)\} = \frac{4\cdot (s+6)}{s\cdot (s-4)\cdot (s+1)\cdot (s-5)} +\frac{1}{s} -\frac{1}{s\cdot (s-5)}[/tex].
How to determine the Laplace transform of a non-homogeneous second order differential equation
A Laplace transform is a frequency-based algebraic substitution method used to determine the solutions of differential equations in a quick and efficient manner.
In this question we shall use the following Laplace transforms:
[tex]\mathcal {L} \{f(t) + g(t)\} = \mathcal {L} \{f(t)\} + \mathcal {L}\{g(t)\}[/tex] (1)
[tex]\mathcal {L} \{\alpha\cdot f(t)\} = \alpha\cdot \mathcal {L} \{f(t)\}[/tex] (2)
[tex]\mathcal{L} \left\{y^{(n)} \right\} = s^{n}\cdot \matcal {L}\{f(t)\}-s^{n-1}\cdot y(0) -...-y^{(n)}(0)[/tex] (3)
[tex]\mathcal {L} \{e^{-a\cdot t}\} = \frac{1}{s+a}[/tex] (4)
Now we proceed to derive an expression fo the Laplace transform of the solution of the differential equation:
[tex]y'' -5\cdot y' = 8\cdot e^{4\cdot t}-4\cdot e^{-t}[/tex]
[tex]s^{2}\cdot \mathcal {L}\{f(t)\}-5\cdot y(0) - y'(0) - 5\cdot s \cdot \mathcal {L} \{f(t)\} +5\cdot y(0) = \frac{8}{s-4}-\frac{4}{s+1}[/tex]
[tex]\mathcal {L} \{f(t)\} = \frac{4\cdot (s+6)}{s\cdot (s-4)\cdot (s+1)\cdot (s-5)} +\frac{1}{s} -\frac{1}{s\cdot (s-5)}[/tex]
The Laplace transform of the non-homogeneous second order differential equation is [tex]\mathcal {L} \{f(t)\} = \frac{4\cdot (s+6)}{s\cdot (s-4)\cdot (s+1)\cdot (s-5)} +\frac{1}{s} -\frac{1}{s\cdot (s-5)}[/tex]. [tex]\blacksquare[/tex]
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The Laplace transform Y'' − 5y' = 8e4t − 4e−t when y(0) = 1, y'(0) = −1 for f(t) = 8e4t − 4e−t would be [tex]\frac{4(s+6)}{s.(s-4)(s+1).(s-5)} + \frac{1}{s} - \frac{1}{s.(s-5)}[/tex].
What is Laplace transform?
A Laplace transform is a frequency-based algebraic substitution method used to determine the solutions of differential equations quickly and efficiently.
The Laplace transform
L[ f(t) + g(t) ] = Lf(t) + Lg(t)
Also,
L [[tex]y^{n}[/tex]] = [tex]s^{n} . L[ f(t) ] - s^{n-1} .y(0) .....y^{n}(0)[/tex]
We have
Y'' − 5y' = 8 . e^4t − 4 . e−t
[tex]s^{2} . L[ f(t) ] - 5. y(0)- y'(0) .....y^{n}(0)[/tex]
y(0) = 1, y'(0) = −1
for f(t) = 8e4t − 4e−t.
= [tex]\frac{8}{s-4} - \frac{4}{s+1}[/tex]
L [f(t)] = [tex]\frac{4(s+6)}{s.(s-4)(s+1).(s-5)} + \frac{1}{s} - \frac{1}{s.(s-5)}[/tex]
Learn more about Laplace transform ;
https://brainly.com/question/14487937
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