Question

Consider the following initial-value problem. Y'' − 5y' = 8e4t − 4e−t, y(0) = 1, y'(0) = −1 find ℒ{f(t)}, for f(t) = 8e4t − 4e−t. (write your answer as a function of s. )

Answer 1

The Laplace transform of the non-homogeneous second order differential equation is $\mathcal {L} \{f(t)\} = \frac{4\cdot (s+6)}{s\cdot (s-4)\cdot (s+1)\cdot (s-5)} +\frac{1}{s} -\frac{1}{s\cdot (s-5)}$.

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:

$\mathcal {L} \{f(t) + g(t)\} = \mathcal {L} \{f(t)\} + \mathcal {L}\{g(t)\}$   (1)

$\mathcal {L} \{\alpha\cdot f(t)\} = \alpha\cdot \mathcal {L} \{f(t)\}$   (2)

$\mathcal{L} \left\{y^{(n)} \right\} = s^{n}\cdot \matcal {L}\{f(t)\}-s^{n-1}\cdot y(0) -...-y^{(n)}(0)$   (3)

$\mathcal {L} \{e^{-a\cdot t}\} = \frac{1}{s+a}$   (4)

Now we proceed to derive an expression fo the Laplace transform of the solution of the differential equation:

$y'' -5\cdot y' = 8\cdot e^{4\cdot t}-4\cdot e^{-t}$

$s^{2}\cdot \mathcal {L}\{f(t)\}-5\cdot y(0) - y'(0) - 5\cdot s \cdot \mathcal {L} \{f(t)\} +5\cdot y(0) = \frac{8}{s-4}-\frac{4}{s+1}$

$\mathcal {L} \{f(t)\} = \frac{4\cdot (s+6)}{s\cdot (s-4)\cdot (s+1)\cdot (s-5)} +\frac{1}{s} -\frac{1}{s\cdot (s-5)}$

The Laplace transform of the non-homogeneous second order differential equation is $\mathcal {L} \{f(t)\} = \frac{4\cdot (s+6)}{s\cdot (s-4)\cdot (s+1)\cdot (s-5)} +\frac{1}{s} -\frac{1}{s\cdot (s-5)}$. $\blacksquare$

To learn more on Laplace transforms, we kindl invite to check this verified question: brainly.com/question/2272409