Question

Differential equations.
Question!

Answer 1

Cannot see it if you make it a doc file, it must be an image or just write it down?

Answer 2

@kola908

Answer 3

(a): Laplace transform of\[f(t)=(2014+e^{-t}+e^{2t})\sin 7t\]

Answer 4

I'm gonna use this table: http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx
looks like we want to distribute the sin term to get terms into forms 7 and 19

Answer 5

(b): LT of\[g(t)=e^{14t}(t-5)^2\]expand and distribute and you should be able to apply forms 2 and 23

Answer 6

I'll let you figure out the last one by yourself

Answer 7

@TuringTest

Just so i know im on the right path can you help with part A
i got
\[\frac{ 14098 }{ s ^{2}+49}+\frac{ 7 }{(s+1) ^{2}+49}+\frac{ 7 }{(s-2) ^{2}+49}\]

Answer 8

yup

Answer 9

For part C

i am coming unsure of how to get the right answer/

Answer 10

Somebody help

Answer 11

(c) \[y(t) = e^t(t^5-\sin\tfrac t5)\]

Answer 12

start with the shift theorem

Answer 13

Um...I wanted to know what would be the value of C in A/P=B/Q=C/D

Answer 14

The Shift theorem

\[\large\boxed{\mathcal L\big\{e^{-ax}f(x)\big\}=\mathcal L\big\{f(x)\big\}\Big|_{s\to s+a}}\]

Answer 15

\[y(t) = e^t(t^5-\sin\tfrac t5)\]
So
\[\mathcal L\{y(t)\}=\mathcal L\{e^t(t^5-\sin\tfrac t5)\}\\\qquad\qquad
=\mathcal L\{t^5-\sin\tfrac t5\}\Big|_{s\to s-1}\]

Answer 16

Then use the fact that the laplace transform is linear
\[=\mathcal L\{t^5\}\Big|_{s\to s-1}-\mathcal L\{\sin\tfrac t5\}\Big|_{s\to s-1}
\]

Answer 17

then use your table of transforms

\[\large\boxed{\mathcal L\{t^n\} = \dfrac{n!}{s^{n+1}}}\]
\[\large\boxed{\mathcal L\{\sin nt\} = \dfrac{n}{s^2+n^2}}\]

Answer 18

does that help at all?

Answer 19

@kola908 your answer for part A is correct

@UnkleRhaukus I don't think they want a proof of the sort you recommend, since the doc says 'use the table provided'. Instead I would again just distribute and apply forms 19 and 23 from the table I linked above.

Answer 20

@TuringTest i haven't offered a proof, i was using the table too.

forms 19 and 23 are simply forms 3 and 7 combined with the shift theorem