Pls help on these two questions. PLS I BEG DO BOTH OF THEM. Use Laplace transform to solve the following initial value problemy" + 4y = 4, y(0) = –5, y'(0) = –4Express your answer as the Laplace transform of y and use partial fraction decomposition where possible. 2s + 2Find the inverse Laplace transform of(s – 3)2 + 4est cos(t) + et sin(t)e* cos(2t) + e* sin(2t)2et cos(2t) + 4et sin(2t)2et cos(3t) + 4et sin(3t)

Answered: Use Laplace transform to solve the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Pls help on these two questions. PLS I BEG DO BOTH OF THEM.

Use Laplace transform to solve the following initial value problem
y" + 4y = 4, y(0) = –5, y'(0) = –4
Express your answer as the Laplace transform of y and use partial fraction decomposition where possible.

2s + 2
Find the inverse Laplace transform of
(s – 3)2 + 4
est cos(t) + et sin(t)
e* cos(2t) + e* sin(2t)
2et cos(2t) + 4et sin(2t)
2et cos(3t) + 4et sin(3t)

Expert Solution

Step 1

Given differential equation is $y^{''} + 4 y = 4$ and $y (0) = - 5, y^{'} (0) = - 4$

We have to solve the given initial value problem by using Laplace transformation.

Use $L \{y^{''}\} = s^{2} Y (s) - s y (0) - y^{'} (0)$ and $L \{y\} = Y (s)$

Now, use Laplace transform of both side in $y^{''} + 4 y = 4$

$\begin{matrix} L \{y^{''} + 4 y\} = L \{4\} \\ L \{y^{''}\} + 4 L \{y\} = \frac{4}{s} (Since, L \{4\} = \frac{4}{s}) \\ s^{2} Y (s) - s y (0) - y^{'} (0) + 4 Y (s) = \frac{4}{s} \\ s^{2} Y (s) - s (- 5) - (- 4) + 4 Y (s) = \frac{4}{s} (Use y (0) = - 5, y^{'} (0) = - 4) \\ s^{2} Y (s) + 5 s + 4 + 4 Y (s) = \frac{4}{s} \\ (s^{2} + 4) Y (s) + 5 s + 4 = \frac{4}{s} \\ (s^{2} + 4) Y (s) = \frac{4}{s} - (5 s + 4) \\ Y (s) = \frac{4}{s (s^{2} + 4)} - \frac{(5 s + 4)}{(s^{2} + 4)} \end{matrix}$

Hence, $Y (s) = \frac{4}{s (s^{2} + 4)} - \frac{(5 s + 4)}{(s^{2} + 4)}$

Now, taking a pretrial fraction of $\frac{4}{s (s^{2} + 4)}$ , which is given below

$\frac{4}{s (s^{2} + 4)} = \frac{1}{s} - \frac{s}{s^{2} + 4}$

And taking a pretrial fraction of $\frac{(5 s + 4)}{(s^{2} + 4)}$ , which is given below

$\frac{(5 s + 4)}{(s^{2} + 4)} = \frac{5 s}{s^{2} + 4} + \frac{4}{s^{2} + 4}$

Therefore,

$\begin{matrix} Y (s) = \frac{1}{s} - \frac{s}{s^{2} + 4} - (\frac{5 s}{s^{2} + 4} + \frac{4}{s^{2} + 4}) \\ Y (s) = \frac{1}{s} - \frac{s}{s^{2} + 4} - \frac{5 s}{s^{2} + 4} - \frac{4}{s^{2} + 4} \end{matrix}$

Now, taking inverse Laplace transform, which is given below

$\begin{matrix} L^{- 1} \{Y (s)\} = L^{- 1} \{\frac{1}{s} - \frac{s}{s^{2} + 4} - \frac{5 s}{s^{2} + 4} - \frac{4}{s^{2} + 4}\} \\ y (t) = L^{- 1} \{\frac{1}{s}\} - L^{- 1} \{\frac{s}{s^{2} + 4}\} - 5 L^{- 1} \{\frac{s}{s^{2} + 4}\} - L^{- 1} \{\frac{4}{s^{2} + 4}\} (Since, L^{- 1} \{Y (s)\} = y (t)) \\ y (t) = H (t) - \cos (2 t) - 5 \cos (2 t) - 2 \sin (2 t) \end{matrix}$