Please check the attached image and answer the math questions. I really need help. 8. Determine the Laplace transform off(t) = tª – t²e3t + 29. Verify that if y(t) is a solution of the differential equation y" + 8y' + 12y = 42te*, y(0) = 0, y'(0) = 0then the Laplace transform Y(s) of y(t) is42Y (s) =(s – 12) (s + 2) (s + 6)It can be shown, (you are not required to do so), that the right hand side of the above equation splitsinto partial fractions as4227+(s – 12) (s + 2) (s + 6)(s -1)(8 – 1)*2 (s + 2)2 (s + 6)Use this to solve the given equation by the method of Laplace transforms.unacceptable.Any other method is

Answered: Determine the Laplace transform of f(t)…

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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Please check the attached image and answer the math questions. I really need help.

8. Determine the Laplace transform of
f(t) = tª – t²e3t + 2
9. Verify that if y(t) is a solution of the differential equation y" + 8y' + 12y = 42te*, y(0) = 0, y'(0) = 0
then the Laplace transform Y(s) of y(t) is
42
Y (s) =
(s – 12) (s + 2) (s + 6)
It can be shown, (you are not required to do so), that the right hand side of the above equation splits
into partial fractions as
42
2
7
+
(s – 12) (s + 2) (s + 6)
(s -
1)
(8 – 1)*
2 (s + 2)
2 (s + 6)
Use this to solve the given equation by the method of Laplace transforms.
unacceptable.
Any other method is

Expert Solution

Step 1

Since there are multiple questions posted, we will answer the first question (8). If you want the second question (9) to be answered, then please resubmit that question only or mention the question number in your message.

The given function is $f (t) = t^{4} - t^{2} e^{3 t} + 2$ .

Find the Laplace transform of $f (t) = t^{4} - t^{2} e^{3 t} + 2$ as shown below.

Step 2

It is known that,

$L \{1\} = \frac{1}{s} L \{t^{n}\} = \frac{n!}{s^{n + 1}} L \{t^{n} e^{a t}\} = \frac{n!}{{(s - a)}^{n + 1}}$

Then,

$\begin{array}{rcl} L \{f (t)\} & = & L \{t^{4} - t^{2} e^{3 t} + 2\} \\ = & L \{t^{4}\} - L \{t^{2} e^{3 t}\} + L \{2\} \\ = & L \{t^{4}\} - L \{t^{2} e^{3 t}\} + 2 L \{1\} \\ = & \frac{4!}{s^{4 + 1}} - \frac{2!}{{(s - 3)}^{2 + 1}} + 2 (\frac{1}{s}) \\ = & \frac{24}{s^{5}} - \frac{2}{{(s - 3)}^{3}} + \frac{2}{s} \end{array}$

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