Question

case closed

Answer 1

im lost

Answer 2

you're mostly done!

Answer 3

just notice that the integral in the last line is same as the laplace transform of cos(at)

Answer 4

it keeps on coming back.. hehe

Answer 5

replace that integral by \(\mathcal{L}(\cos (at))\)

Answer 6

you mean?

Answer 7

The entire last line equals \(\mathcal{L}(\cos(at))\), yes ?

Answer 8

ah yes

Answer 9

then what would be the next step hehe?

Answer 10

I'm just giving the idea here. You can interpret last line in your attached pic as :
\[\mathcal{L}(\cos(at)) ~~=~~ (\cdots ) - \dfrac{a}{s^2} \mathcal{L}(\cos(at))\]

Answer 11

Notice that integrals are gone

Answer 12

just bring that \(\mathcal{L}(\cos(at))\) term on right hand side to left side, isolate it, and you're done

Answer 13

also, you have forgot to evaluate the bounds... you will need to fix that first

Answer 14

totally lost here

Answer 15

\[\large\rm \color{orangered}{\mathscr L\{\cos(at)\}=\int\limits \cos(at) e^{-st}dt}\]So far you've determined that,\[\large\rm \mathscr L\{\cos(at)\}=-\frac{1}{s}\cos(at)e^{-st}+\frac{a}{s}\sin(at)e^{-st}-\frac{a^2}{s^2}\int\limits \cos(at)e^{-st}dt\]Something like that...

Answer 16

yes :)

Answer 17

\[\large\rm \mathscr L\{\cos(at)\}=-\frac{1}{s}\cos(at)e^{-st}+\frac{a}{s}\sin(at)e^{-st}-\frac{a^2}{s^2}\color{orangered}{\int\limits\limits \cos(at)e^{-st}dt}\]
So rsadhvika was trying to help you realize that this integral that you `end up with` is the same integral that you started with.\[\large\rm \mathscr L\{\cos(at)\}=-\frac{1}{s}\cos(at)e^{-st}+\frac{a}{s}\sin(at)e^{-st}-\frac{a^2}{s^2}\color{orangered}{\mathscr L\{\cos(at)\}}\]

Answer 18

So from that point,
you can do some basic algebra, "combining like-terms"
We want to get both of these laplace transforms on the same side.
So let's add a^2/s^2 L to each side,\[\large\rm \frac{a^2}{s^2}\mathscr L\{\cos(at)\}+L\{\cos(at)\}=-\frac{1}{s}\cos(at)e^{-st}+\frac{a}{s}\sin(at)e^{-st}\]

Answer 19

Multiply through by s^2 to make things easier.\[\large\rm a^2\mathscr L\{\cos(at)\}+s^2L\{\cos(at)\}=-(s)\cos(at)e^{-st}+(a)\sin(at)e^{-st}\](Woops the fraction on the sine should have been a/s^2 so the s^2 cancels out completely.)

Answer 20

Factor our your laplace thingy,\[\large\rm \mathscr L\{\cos(at)\}(a^2+s^2)=-(s)\cos(at)e^{-st}+(a)\sin(at)e^{-st}\]

Answer 21

And then a few more goofy steps from there.
That's maybe the confusing part of it though.

What do you think? :o

Answer 22

the second term.. it should be a/s^2

Answer 23

yes, my bad :) was typo.
But when i multiplied through by s^2, i fixed it from that point.

Answer 24

What do you mean with the goofy steps here? Haha

Answer 25

\[\large\rm \mathscr L\{\cos(at)\}(a^2+s^2)=-(s)\cos(at)e^{-st}+(a)\sin(at)e^{-st}\]Well um, I guess you'd next have to divide by a^2+s^2, right?\[\large\rm \mathscr L\{\cos(at)\}=\frac{-s}{a^2+s^2}\cos(at)e^{-st}+\frac{a}{a^2+s^2}\sin(at)e^{-st}\]

Answer 26

And then plug in your bounds and see what you end up with.

Answer 27

Oh so my instinct was right about dividing the a^2+s^2. Thank you so much @zepdrix :))

Answer 28

Ahm @zepdrix not sure ha but i know that the exponential term will vanish for the infinite sign.. and for 0, its automatically, 1.

Answer 29

I got -s/a^2+s^2.. ahm it shud be s/a^2+s^2..

Answer 30

Ahm can you please please show me the correct steps in substituting the bounds?

Answer 31

\[\large\rm \mathscr L\{\cos(at)\}=\frac{-s}{a^2+s^2}\color{orangered}{\cos(at)e^{-st}|_0^{\infty}}+0\]

Answer 32

\[\large\rm \mathscr L\{\cos(at)\}=\frac{-s}{a^2+s^2}\color{orangered}{\left[\lim_{b\to\infty}\cos(ab)e^{-sb}-\cos(a0)e^{-s0}\right]}+0\]Which, as you noticed already, gives us 0 as this thing blows up to infinity,
and 1 for the other term,\[\large\rm \mathscr L\{\cos(at)\}=\frac{-s}{a^2+s^2}\color{orangered}{\left[0-1\right]}+0\]

Answer 33

Which gives us a positive s/(a^2+s^2) in the end, ya?

Answer 34

Yes Thank you so much @zepdrix yaaaay :)