Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. If F of S equals the Laplace of F of T exists for some S greater than C, then the Laplace of E to the AT of F of T exists for S greater than a + C, and the Laplace of E to the ATF of t = F of s -, a. Also the Laplace inverse is F of s -, a equaling E to the ATFT. Thus the translation we could think of S going into s -, A and the transform corresponds to a multiplication of the original function of T by E to the AT. If we look at the proof of this, we use the definition of F of S equaling 0 to Infinity, the integral zero to Infinity of E to the negative STFTDT. So for F of S -, A, every time we see that S we're going to put in South minus A. Now we get 0 to Infinity the integral E to the negative S minus AT times TF of TDT. So if we split up that E to the negative quantity S -, A * t into E to the negative St. and E to the ATFTDT, we can then see that we could think of this as our new function in the definition. So now we have the Laplace of E to the ATFT. So a couple formulas. Our F of T if we had E to the ATTN, then our F of S is just N factorial over s -, A to the n + 1. This of course is if S is greater than a, our E to the AT times cosine of KT is just going to be a shift and it's going to be s -, a over the quantity s -, a ^2 + K ^2. And E to the AT times sine of KT is just K over the quantity s -, a ^2 + K ^2 for S greater than a. If we think about our mass spring system where we have our mass equaling 1/2, our K = 17, our C equal 3, and some initial conditions are X of 0 equaling 3 and our X prime of 0 equaling one. We know that our equation would then become 1/2 X double prime plus 3X prime plus 17X equaling zero. We could multiply it all through by two to get rid of the fraction. So X double prime plus 6X prime plus 34 X equals 0. We don't easily know how to solve that. So we're going to use our Laplace transformation. We're going to let our Laplace of X equal X of S and our first derivative then by definition is SL of X -, X naughty. So that's just SX of s -, 3. And our second derivative turns into, for this particular problem, s ^2 X of s -, 3 S -1. So I'm going to substitute my L double prime, my L prime, and my L in for my X double prime, X prime, and X. So we get a new equation that says s ^2 X of s -, 3 S -1 + 6 times the quantity SX of s -, 3 + 34 X of S equaling zero. If I solve for X of S, we can see that that turns into 3S plus 19 divided by s ^2 + 3, six s + 34. And what I'm going to do is I'm going to make that into a perfect square. So I'm going to pull out nine of the 34 to get the s + 3 quantity squared plus 25. And 25 is obviously 5 ^2. So if I have an s + 3 in the denominator, that's quantity squared in the numerator. For it to be a cosine, I need an s + 3. So there's a three in front of the three S So I could think of it as I need 3 * s + 3 or I need 3S plus 9 here. Well, there were 19 total, so if I pull nine of those 19 out, that would leave me 10 for a next term. That 10 because there is no S. We're going to want to think of it as something times the five because this was 5 ^2. So the second term we're going to think of as 2 * 5 in the numerator. So we can pull the three and we could pull the two out in front if we wanted. Three. This term here is going to turn into E to the -3 T cosine 5T, and the second term is going to turn into two E to the -3 T sine of 5T. The E to the -3 TS are coming from this quantity squared in the denominator. If we look at another one, let R of S equaling s ^2 + 1 / s ^3 - 2 S squared -8 S. Well, if we factor that, we can see that that's s ^2 + 1 / s times the quantity s -, 4 times quantity s + 2. Figuring out our partial fractions. So literally going through the steps to find our AB and our C, you can see that a is -1 eighth and B is 1720 fourths and C is five twelfths. So those are just the numbers in the numerator or the coefficients of our terms. So if I have negative 1/8 * 1 / s, we know that the inverse Laplace of 1 / s is just T. So that's terms going to turn into -1 eighth T, the next one, 1720 fourths times 1 / s -, 4. That 1 / s -, 4 is going to turn into E to the 4T. So 1720 fourths E to the 4T, and the last one is going to end up being 5 twelfths E to the -2 T. If we look at a mass spring system now, it's going to be similar to the one we did a minute ago, But now we're going to have our initial condition and our first derivative starting with 0. But we're also going to add an external force, and our external external force is going to be F of T equaling 15 sine 2T. So doing the same steps as a minute ago, we're going to end up with our X double prime plus 6X prime plus 34X equaling 30 sine 2T. So using our definition for Laplace and our first derivative and 2nd derivative and then substituting, we're going to see that we get s ^2 X of s + 6 S of X 6 * X of s + 34 X of S equaling 30. Now remember the Laplace of sine 2T is going to be 2 / s ^2 + 2 ^2 or 4. So solving for X of S, we're going to get 60 divided by that s ^2 + 6, X plus 34, and the s ^2 + 4 is going to be on the denominator. So completing the square like we did a moment ago, we're going to get the quantity of s + 3 ^2 + 5 ^2 in the denominator. Now these are quadratics that can't be factored any further down, so we're actually going to have a linear on top. So we're going to get a s + b over the quantity s + 3 ^2 + 5 ^2 plus CS plus d / s ^2 + 2 ^2. Then we're going to solve for ABC and D So just cross multiply and set the numerators equal or actually multiply through by the common denominator and set the numerators equal if we solve for ABC and D. There's some nice mathematics going on there. I started with realizing a equal negative C. So now I have 3 equations with three unknowns right here. Whenever we have an equation with whenever we have 3 equations with one of them missing a variable, we want to get use the other two with the same variable to get rid of that variable. So like in this case, I saw that the bees we're only in two of the three, so I multiplied the top one through by a -2 and then added the two together. And then what happens is I get a new equation and I / 3 to get it more simplified. But this new equation has the same variables as that equation that we didn't use a minute ago. I divided that equation through by a -6 to simplify it down. Then I combined the two variable 2 equations to solve, and from there I just substituted back to get everything. So then we're going to actually substitute those answers that we found for ABC and D. So we get 129th times the whole thing, and now we're just going to do some algebra. I know that the s + 3 quantity squared. So in the numerator I need to have an s + 3 if it's a cosine. Well, if I have 10 S, then I know I'm going to need plus 30. So if I have a + 30, in order to get the 10 I would have to subtract 20. So +30 and -20 gave me the 10 that I have. Well, this -20 because it was a 5 ^2 here, I could think of as -4 * 5 over here. The -10 S we're going to split up, and then the 50 we're going to put in as a second term because this is a s ^2 + 2 ^2, we just need a plain S in the numerator. So it's going to be a -10 coefficient. Because this one's s ^2 + 2 ^2, I need a two in the numerator, so it's going to turn into 25 * 2. So we get 129th times the quantity 10 times the quantity s + 3 over the quantity s + 3 ^2 + 5 ^2 + -4 * 5 over the quantity s + 3 ^2 + 5 ^2 + -10 S over s ^2 + 2 ^2 + 25 * 2 / s ^2 + 2 ^2. Taking the in first Laplace transformation, we can see that X of t = 120 ninth times the quantity 10 E to the -3 T cosine five t -, 4 E to the -3 T sine 5T. Remember that E to the -3 T is coming from the fact that we have a quantity that's squared that s + 3 quantity squared so -10 cosine 2T plus 25 sine 2T. Thank you and have a wonderful day.