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now I think is a good time to add some notation and techniques to our Laplace transform toolkit so the first thing I want to introduce is just it's just kind of a quick way of doing something and that is if I had the Laplace transform let's say I want to take the Laplace transform of the second derivative of Y well we've proved several videos ago that if I wanted to take the Laplace transform of the first derivative of Y that is equal to s times the Laplace transform of Y minus y of 0 right and we use this property in the last couple of videos to actually figure out the Laplace transform of the second derivative because if you just you know if you say this is y prime this is the antiderivative of it then you can just pattern match and you can say well the last Laplace transform of Y prime prime that's just equal to s times the Laplace transform of Y prime minus y prime of 0 right the this is the derivative of this just like this is the derivative of this right I'll draw a line here just so you don't get confused so the plas transform of Y prime prime is this thing and now we can use this which we proved several videos ago to resub stitute it and get in terms of the Laplace transform of Y so we can expand this part write the Laplace transform of the derivative of Y that's just equal to s times the Laplace transform of Y minus y of 0 and then we have the outside right we have s minus y prime of 0 and then when you expand it all out and we've done this before you get s squared times the Laplace transform of Y minus s times y of 0 minus y prime of 0 now there's something interesting to to note here and if you learn this it'll make it a lot faster you won't have to go through all this and risk making careless mistakes when you have when you have scarce time I am and paper on your tests just notice that when you take when you take the Laplace transform of the second derivative what do we end up we end up with s squared right this was the second derivative with s squared times the Laplace transform of Y minus s times y 0 minus 1 times y prime is 0 so every term we started with s squared and then every term we lower the degree of s 1 and then everything except the first term is a negative sign and then we started with the Laplace transform of Y and then you can almost view the Laplace transform it's a kind of integral so we kind of take the derivative so then you get Y and then you take the drivet in you get Y Prime and of course every other term is negative and these aren't the actual functions these are those functions evaluated 0 but that's a good way to help you hopefully remember how to do these and once you get this the hang of it you can take the Laplace transform of any arbitrary function very very quickly or any arbitrary derivative so let's say we wanted to take the Laplace transform of I don't know this should hit the point home the fourth derivative of Y that 4 in parentheses means the fourth derivatives I could have drawn four prime marks but either way so what is this equal to if we use this technique and substituted we were bound to make some form of careless mistake or other and it would take us forever and it would waste a lot of paper but now we see the pattern and so we can just say well the Laplace transform of this in terms of the Laplace transform of Y right that's what we want to get to is going to be s to the fourth s to the fourth times the Laplace transform of Y now every other term is going to have a minus in front of it - lower the degree on the s - s to the third and then you can kind of say let's take the you know some form of derivative so that you get Y of 0 it's not a real derivative the Laplace transform really isn't the antiderivative of Y of 0 but anyway I think you get the idea and then we lower the degree on s again - s squared take the derivative and of course these aren't functions but we're taking we're evaluating the derivative of that function now at zero so Y prime of zero - now we lower the degree one more minus s times this is an S times y prime prime of zero we have one more term lower the degree on the S one more time then you get s to the zero which is just one so - and one is the coefficient and then you have Y the third derivative of Y let me scroll over a little bit the third derivative of Y evaluated at zero so I think you see the pattern now and this is a much faster way of evaluating of evaluating the Laplace transform of an arbitrary derivative of Y as opposed to keep going through that pattern over and over again another thing I want to introduce you to is just a notational savings it's just something that you'll see so you might as well get used to it it actually saves time or keep you know keep writing this curly L in this bracket if I the Laplace transform of Y of T I can write as and people tend to write it as what's going to be a function of s and they you what they use is they use a capital Y to denote the function of s and that well it makes sense because normally when we're when we're doing anti derivatives you just take you know when we take we often when you learn the fundamental theorem of calculus learn that the integral of F with respect to DX you know from zero to X is equal to capital f of X so it's kind of borrowing that notation because the this function of s is kind of an integral of Y of T the Laplace transform to some degree is a kind of it's like a special type of integral where you have a little exponential function in there to mess around with things a little bit anyway I just want you to get used to this notation when you see capital y of s that's the same thing as the Laplace transform of Y of T and you might also see see it this way the Laplace transform of F of T is equal to capital f of s and and the clue that tells you that this isn't just a normal antiderivative is the fact that they're using that s the as the independent variable because in general s represents the frequency domain and if people were to use s with just a general antiderivative people would get confused etc etc anyway I'm trying to think whether I have time to teach you more fascinating concepts of Laplace transform well sure I think we do so my next question for you and now we'll teach you a couple more properties and this will be helpful in taking Laplace transforms what is the Laplace transform of e to the a T times F of T fascinating well let's just go back to our definition of the Laplace transform it is the integral from 0 to infinity of e to the minus st times whatever we have between the curly brackets so with the curly brackets we have e to the 8e f of T DT and now we can add these exponents right we have a similar base so this is equal to what this is equal to the integral from 0 to infinity and let's see I want to write it as I could write it minus the S Plus a but I'm going to write it as minus s minus a T and you can expand this out right it becomes minus s plus a which is exactly what we have here times F of T F of T DT now let me show you something if I were to just take the Laplace transform of F of T the Laplace transform of just F of T that is equal to some function of s right it is equal to some function of s whatever we essentially have right here for s it becomes some function of that so this is interesting this is some function of s here all we did went to get to go from actually let me rewrite this little plus which is equal to 0 to infinity e to the minus s T f of T DT right the Laplace transform of just f of T is equal to this which is some function of s right well the Laplace transform of e to the a T times F of T it equals this and what's the difference between this and this what's the difference between the two well it's not much here wherever I have an S I have an S minus a here right so if this is a function of s what's this going to be it's going to be that same function whatever the Laplace transform of F was it's going to be that same function but instead of s it's going to be a function of s minus a and and once again how did I get that what is the Laplace transform of f is a function of s and it's equal to this well if I just replaced an S with an S minus a I get this which is a function of s minus a which was the Laplace transform of e to the a T times F of T maybe that's a little confusing let me show you an example so if I let's just take the Laplace transform the Laplace transform of cosine of 2t we've shown is equal to well I'll write the notation it's equal to some function of s and that function of s is s over s squared plus 4 we've shown that already and so the Laplace transform of e to the I don't know 3t times cosine of 2t is going to be equal to the same function but instead of s is going to be a function of s - a so s minus three which is equal to s minus three over s minus three squared plus four notice when you get when you just multiply something by this e to the 3t and then or e to the 80 you take the Laplace transform of it you just it's the same thing as a Laplace transform of this function but everywhere where you had an S you replace it with an S minus this a anyway I hope I didn't confuse you too much with that last part I think my my power adapter act you just went on I hope the video keeps recording I'll see you in the next one