Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Transformation of initial value problems AX double prime T plus BX prime T plus CX of T equal F of T. If we use our Laplace transformation, we're going to see that we could rewrite this as a times the Laplace transformation of the second derivative of T plus B times the Laplace transformation of the first derivative of t + C times the Laplace transformation of the original function equaling the Laplace of the new function F of T. So we're going to transform the derivatives. Suppose that a function F of T is continuous and piecewise smooth for T greater than or equal to 0 and is of exponential order as T goes to Infinity. So that there exists non negative constants MC and T such that the absolute value of F of T is less than or equal to ME to the CT for all little T greater than or equal to some constant T. Then the Laplace transformation of the first derivative of the function F of T exists for S greater than some constant C. And the Laplace of the first derivative of F of t = s times the Laplace of the original function F of t -, F of naughty equal s * F of S minus little F of SF of 0. Sorry. So what this is really saying is the Laplace transformation of a first derivative really equals S times the capital F of S minus the initial condition little F of 0. Now we can figure out the the Laplace transformation of that first derivative by our definition goes zero to Infinity of E to the negative S TF prime TDT. Now, using our integration by parts, if I did a tabular integration, I can see that that's really E to the negative S TFT from zero to Infinity plus S the integral zero to Infinity of E to the negative STFTDT. Because remember, when we do tabular, we have two columns. In the first column we're going to take the derivative of and in the second column we're going to take the anti derivative of. We multiply them diagonally, changing the signs from plus to minus as we go. So let G of T equals some F prime of T. So let a new function equal the first derivative in terms of T. So then the Laplace of F double prime of T is going to just really equal the Laplace of the G function of the first derivative by using the definition we just established. That's going to be South of the Laplace of G of t -, g of 0. So that's really just equaling S the Laplace of F prime of t -, F prime of 0. So that's really just s * s the Laplace F of t -, F of 0 -, F prime of 0. So from there we can see that the Laplace of the second derivative is really just s ^2 times the capital F of S function minus S times the initial condition of F of 0 minus the first derivative of F of 0. These are that one's a capital F and these are little FS. These are the initial condition values. So if we look at an example, X double prime minus X prime -6 X equals 0 when X of 0 equal to and X prime of 0 equal -1. So the Laplace for the first derivative says it's just S the Laplace of X of t -, X not. So that in this case is just SX of s -, t two. Then the Laplace and the second derivative by the formula we just about evaluated developed s ^2 Laplace of X of T minus SX of not minus X prime of 0. So we're going to get s ^2 of XS -2 S minus a -1. Now when we stick all of this back into the original condition or the original equation, we get s ^2 XS -2 S plus one minus that X prime. So in this case, it's the SXS -2 -, 6 X. Remember the X is in terms of the new function S. So it's going to be -6 X of S equaling 0. So we're going to solve for X of S So we're going to get all the terms with X of S on one side. We're going to factor out the coefficients, and we're going to take anything without an S of S to the other X of S to the other side. So we can see that X of S equal 2 S -3 / s -, 3 * s + 2. Now we don't know how to deal with that the way it is, so we're going to use our concept of partial fractions. So we're going to let this 2 S -3 over s - 3 S +2 turn into a / s - 3 + b / s + 2. If we multiply everything through by the common denominator, we can see 2 S -3 equal a * s + 2 + b * s - 3. Solving for A&B, we get a is 3/5 and B is 7 fifths. So now I'm just going to literally substitute in. So 3/5 / s - 3 + 7 fifths over s + 2. So our X of S is really the Laplace transformation of X of T So in this case we get 3 fifths, s -, 3 + 7 fifths over s + 2. Recall that the Laplace to the -1 of 1 / s -, a is really just E to the AT, so the Laplace inverse transformation. So here we can see that it's going to be 3/5 is our coefficient E to the three t + 7 fifths is the coefficient E to the -2 T. Now we can do the same concept with solving systems, IE we now have two functions instead of just one. So we're going to have two s ^2 X of S minus SX naughty minus X prime of 0 equaling -6 X of s + 2 Y of S. Then that's going to simplify to two s ^2 X of S equaling -6 X of s + 2 Y of S If I solve that for Y of SI can see that that's really just s ^2 XS plus 3XS. So then if we keep going, the second equation is going to turn into s ^2 y of S minus SY not minus Y prime, Y not equal or Y prime is 0 equal to X of s - 2 Y of s + 40, sine of 3T. But remember the Laplace transformation of sine of 3T is the coefficient that's on the T, in this case 3 / s ^2 plus that a ^2 or s ^2 + 9. So that last term is going to turn into 440 * 3 / s ^2 + 9. We're going to keep manipulating this next one, and we're going to get s ^2 y of S equaling 2X of s - 2 Y of s + 120 / s ^2 + 9. Now I'm going to go back to the first equation and substitute it in for the second so that I can solve for my X equation. So s ^2 instead of our Y of S, we're going to go up and we're going to put in what that equaled. So s ^2, s ^2 XS plus 3XS equal to XS -2. Once again, instead of the Y of S, we're going to put in what it equaled. So s ^2 X of s + 3, X of S plus the 120 / s ^2 + 9. We want to get the X of S all by itself on one side and solve. So we're going to do some manipulation, and we're going to see that X of s = 120 divided by. If we factored the S to the fourth plus five s ^2 + 4, we'd get s ^2 plus one s ^2 + 4, and we already had an s ^2 + 9, so we're going to do partial fractions again. So 120 over all of that is going to equal A s + b / s ^2 + 9 plus CS plus D / s ^2 + 1 plus ES plus F / s ^2 + 4. Now when we do our multiplication by our common denominator, we can actually see pretty easily that AC and E are all going to have to be 0, because when we multiply those out, we would get those all to be odd powers, A s ^3 plus our AS to the 5th, etcetera. And there is no odd power on the left hand side, it's only a constant. So my, AC and E are all going to go to 0, going to be able to solve for my BD and F When I substitute my BD and F back in, we're going to see that these are really the inverse Laplace transformation of a / s ^2 + a ^2. So these are going to all equal sine of AT. So I needed a three on top because my a ^2 was 9. So this first one's just going to be sine 3T. This next one I need a one on top. So the five is going to be a coefficient 5 sine T This last one I need a 2 on top. So we're going to have a coefficient of -4 because -4 * 2 gave us the -8. Now if we go back, we have to do our Y of S equation. And we knew that Y of S equaled s ^2 X of s + 3, X of SI could think of factoring out the X of S and getting that as X of s * s ^2 + 3. So we just take this fraction we had a minute ago and multiply by s ^2 + 3. And then we're going to do the same thing that we did before with our partial fractions. And we're going to realize that AC and E are all going to equal 0 again. Because when I multiply these out, the A and the C and the E would all have odd S powers to them. So our B is -18 our D is 10 and our F is 8. When we substitute that in, we can see that once again, we're going to use that in inverse Laplace transformation. For sine. I need to have a three on top in this first term. So that's going to be a -6 sine 3T plus I need a one on top. So the 10s just going to be a coefficient here. 10 sine T plus I need a 2 on top. So 4 * 2 four is going to be a coefficient sine 2T. So our X of T from a minute ago and our Y of T here. So we've just solved the system and this is frequently used in physics with mass and spring systems. Thank you and have a wonderful day.