what is the laplace transform of y? is it sy(s) or sy(0)? or maybe y(s) or y(0)? ..i think s is just used for derivatives...im so confused
@waterineyes i would appreciate a quick help
so y would be...?
one sec
either it's not there...or there's too many things written that i cant see it
no according the website y is 0 that is the answer I saw as I was scrolling on the website
if you want me to I will copy and past this answer for you.
lol what o.O
here is the break down Translation Property The translation formula states that Y(s) is the Laplace transform of y(t), then where a is a constant. Here is an example. The Laplace transform of the y(t)=t is Y(s)=1/s^2. Hence Laplace Transform of the Derivative Suppose that the Laplace transform of y(t) is Y(s). Then the Laplace Transform of y'(t) is For the second derivative we have For the n'th derivative we have Derivatives of the Laplace Transform Let Y(s) be the Laplace Transform of y(t). Then
Laplace transform of y means: \(\mathcal{L}(y)\)..
^that doesnt answer my question :|
@godorovg i dont get what you said
It's like a parametrization of the function y(t) into y(s).
can everyone stop beating around the bush....it's just a simple question o.O
im asking if the laplace of y is y(s), y(0), sy(s) or sy(0)
im doubting it's the s ones...i think the s ones are used in y'
maybe i should just post the original question...
\[y^\prime + y = e^{2t}; \; y(0) = 0\] i know y' = sy(s) - y(0) my problem is y
so what's the laplace of y?
Yes I got it @lgbasallote
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