Question

Check my answer!
Laplace transform of

Integral 0 to t of (1-e^au)/u du

i got
(1/s) ln [(s-a)/s]

Answer 1

\(\Large L[\int \limits_0^t \dfrac{1-e^{au}}{u}du ] = \dfrac{1}{s}\ln \dfrac{s-a}{s}\)

is this correct ?
can be simplified more?

Answer 2

@amistre64
@ganeshie8

Answer 3

there is this web site mathway it can help u @hartnn

Answer 4

thank you :)
i know about mathway
but i like OpenStudy better :)

Answer 5

amistre, if you want, i can upload my work...

Answer 6

ok sorry ya i like openstudy better i don't blame u LOL

Answer 7

yeah, uploading your work may help out alot :)

\[\Large L[\int \limits_0^t \frac 1u-\frac{e^{au}}{u}du ] \]
\[\Large L[\int \limits_0^t \frac {du}u]-L[\int \limits_0^t\frac{e^{au}}{u}du ] \]
\[\Large L[ln(t)-ln(0)]-L[e^a\int \limits_0^t\frac{e^{u}}{u}du ] \]

Answer 8

i sorta did it some other way...

Answer 9

some other way is fine ... my way isnt getting me anywhere special lol

Answer 10

LOL

Answer 11

you seem to be applying thrms that im just to rusty to verify :/ but the good news is that there are plenty of people smarter than me about :) ... basically anyone with a smartscore above 10 as i see it ;)

Answer 12

@hartnn, I'm getting the same

Answer 13

Thanks :)

quick question :
did i start the 2nd one correctly ?
any better method?

Answer 14

Just checked your work - I had a slightly different method using a trig sub somewhere along the way, but that's only because I'm not as familiar with the theorems as you are :)

Answer 15

As for the second one, give me a sec... It doesn't look finished to me.

Answer 16

yup, i am solving it right now, as we speak
will give you my answer in a minute...

Answer 17

my answer for 2nd looks ugly :(

Answer 18

I'm getting something much messier...
\[\begin{align*}\mathcal{L}\left\{t^2\sinh^22t\right\}&=\frac{1}{4}\mathcal{L}\left\{t^2e^{4t}\right\}-\frac{1}{2}\mathcal{L}\left\{t^2\right\}+\frac{1}{4}\mathcal{L}\left\{t^2e^{-4t}\right\}\\
&=\frac{1}{4}\cdot\frac{2!}{(s-4)^3}-\frac{1}{2}\cdot\frac{2!}{s^3}+\frac{1}{4}\cdot\frac{2!}{(s+4)^3}\\
&=\frac{1}{2(s-4)^3}-\frac{1}{s^3}+\frac{1}{2(s+4)^3}
\end{align*}\]
It's messy if you try to write it as one fraction.

Answer 19

oh shoot! i just differentiated once :P

Answer 20

yessss
now i am getting the same result! :)
thank you vey much, i appreciate it :)

Answer 21

you're welcome!