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## Differential equations

### Course: Differential equations>Unit 3

Lesson 1: Laplace transform

# Part 2 of the transform of the sin(at)

Part 2 of getting the Laplace transform of sin(at). Created by Sal Khan.

## Want to join the conversation?

• pretty sure a small mistake was made at when you factored the -e^-st out and re-wrote the terms inside the brackets, you forgot the 'a' before the cos (the term was (a/s^2)e^(-st)cos(at)

if this is not a mistake please explain where that a went

edit: it was caught at • I don't understand how to get from y+(a^2/s^2)y to (s^2+a^2/s^2)y on the left hand side of the equation. Around time • Actually, this long method is not needed. If you write sinat in euler form, ie. sinat= e^iat-e^-iat/ 2i, things turn out pretty simple and short. • at i still don't understand why there is still exist a boundaries from [0,infinity]. Can anyone explain this? thanks • I also have the same question initially. I think it's because when applying int(u'v)=uv-int(uv'), u(t) and v(t) should have the boundaries from int(u'v), otherwise it does not make sense that the equation generates an expression with t but does not follow the original boundary. The final results should not have t, because t has a boundary of [0, infinity].
• why is -e^-s(0) not equal to -1 • much easier to use complex, doesn't it? • At about in the video Sal made a small mistake and as usual there was a corrective note added to the video. My question is, "Is there a way to make those corrective notes show up when watching the video in full screen mode or on YouTube?" • Does anyone have any explanation of why this is?:
After watching some trig substitution videos, I realized that
L{sin(t)} = d/ds[arctan(s)]
Note: d/ds[arctan(s)] is the derivative with respect to s of arctan(s), in case having an s there instead of an x confused you.
After that, I figured out the more general case:
L{sin(at)} = d/ds[arctan(s/a)]
Why is this? Are there other properties like this for other trig functions?

Also, L{cos(at)} = d/ds[ ln( sqrt[ s^2 + a^2 ] ) ]
Note: sqrt[ s^2 + a^2 ] is the square root of s^2 + a^2.
Does this have any significance? How does it relate to L{sin(at)} = d/ds[arctan(s/a)]?

L{sin(at)} = d/ds[ arctan( s/a ) ]
L{cos(at)} = d/ds[ ln( sqrt[ s^2 + a^2 ] ) ] • Good question. Coincidence maybe? What the Laplace transform actually does is compress an exponential function so that you can analyze it. I would need to learn more to really answer your question. Remind me in an year or 2 ;-)

But based on my current intuition, the transform doesn't seem to do anything useful to periodic functions, except remove the.. 'periodicity'
(1 vote)
• Just a comment on this problem... I feel it is much easier to show and explain if when you use the partial differential formula you take uv|0 to infinity - differential vdu ....in that process it makes things much quicker and simpler without mistakes  