Hello. Today we are solving an initial value problem with the Laplace transform method. So we have the differential equation which is given here. So the third order differential equation in our dependent variable X or independent variable t. And on the right-hand side, we have forcing function, which is also dependent on t, the independent variable. Since this is a third order equation, we need three initial conditions. So we have an initial condition for the value at t = 0. And for the first derivative, x = 0, they're both 0. And then we also have a third equation, third condition, which is the value of the second derivative at t = 0, that is given as 1. So in the first part of this, we are asked to calculate the Laplace transform solution and show that this is the given formula in capital X (s). So in this case, we have two ways to do this. We can either use directly the formula given in the lecture or we can go through it step-by-step. So today I will go through this step-by-step. So first, we apply the Laplace transform to our differential equation. So in this case we can just right: L4 Laplace transform. Then we have the third derivative plus two. So we can, since a Laplace transform is linear, we can take the two out of this, and so on for the remaining terms. So now we can use the property for the time derivative for the Laplace transform. So when we do this for this first part here, so from this it follows that we get x^3 of capital X(s). So third derivative gives us s^3 And then we have to add all the different initial values. So we have minus s^2. So you can see we go from s^3 to s^2 here. And this is the first initial value. Minus no is s. So again, we decrease the order here. And this is the first derivative now I indicated the first derivative with this prime here and minus the second derivative. And here we have the s^3, s^2, s^1. And this will be s to the power of 0 (s^0) which is equal to 1. And then we add all the other terms. So we have plus,  we have the 2 and then for the second derivative, so we get s^2, X(s) minus sx(0) minus the first derivative, Plus 4 times the Laplace transform of the first derivative, so it's just S of x(s) minus 0 plus 8 times the Laplace transform x. So x is our solution in the time domain which we are looking for. So in the Laplace domain, this is just capital X of S and this needs to be equal to, and let's just leave it as previously L{te^(2t)} So in this equation and we will handle this term here a bit later. So first, let's have a look at this. So we have all these the capital X(s), 1,2,3, and 4 terms. So we can collect all these terms. And in the same step, we will insert all our initial conditions. The initial conditions for x(0), x'(0) are 0. So we can basically cancel all these out, and we're just left with this one here. So we insert initial conditions and collect terms of X(s). So we do this, let's collect the terms. So we have capital X(s). So we have s^3 for this term here. Plus 2s^2 for this term here, plus 4s, this term here, plus eight. So then as I said, instead the initial conditions, but x(0) is 0, so all these terms are 0. The first derivative was also 0. So these two terms also 0. So we're just left with minus this term here. And the second derivative at t equals 0 is 1. So we just get one. And this is equal to the Laplace transform of the right-hand side. So now let's handle this part here. We can solve this by using the Laplace transform tables and the frequency shift. In the equation we have t, which is the ramp, and we are multiplied with an exponential. In the equation. We have t times exponential. Every time you see an exponential in the time domain, you should think about the frequency shift property. So the Laplace transform of just t, So the ramp is just 1/s^2. And we shift this in the frequency domain by 2. So our Laplace transform of this is 1 over (s-2)^2. This is the, this is the shift. And the other one is the Laplace transform of t. So now we can insert this into our equation. And in the same step we just move this 1 to the right hand side. This is now plus 1. Since we moved that over. So now to get the solution in the Laplace domain. So basically we just want to have this X(s) on the left hand side. We divide everything by this term in the square brackets. And this is just the characteristic polynomial of our original equation. And then we get X(s) is equal to 1 over (x-2)^2 plus 1, over s^3 + 2s^2 + 4s + 8 So this step is basically the solution, as it is always in this case. We have the characteristic polynomial, at denominator, at the numerator, we have the Laplace transform of the right-hand side, which is often also called the forcing, plus the terms for the initial conditions. So now to bring it into the required form, we multiply 1 with (s-2)^2 So we can bring this into a common fraction. And then we just move this to the bottom. This is the same as having 1 + (s-2)^2 over (s-2)^2 times our characteristic polynomial. So the only thing left here is to expand this term and to factorize this. So you can do some forms of long division. You can look at this, you can try some values. So in this case, we have quite a lot of twos, fours, eights. So it seems like a good guess to think that we have zeros at. Two or minus two. So, and as it works, I'll we have one at minus two. So once we do this, we're left with 1, so solve this bracket, so it's s^2 - 4s + 4. And at the bottom of the equation we have one term in (s+2). And if we do long division of this third-order polynomial by (s+2), we will get the term (s^2 + 4). And obviously, we can combine the 1 and 4 to get +5. And then we have it in the form as asked in this question. So this one is basically done. So in the second part of this, we're asked to solve this. So part B would be (this is our X(s)). So we want to find the time domain solution for this X(s) and the time domain solution, this will be the solution to our original initial value problem. So we can do this by splitting our X(s) into partial fractions. Just right the X(s) again. s^2 minus 4s plus 5, (s-2)^2 , (s + 2), (s^2 + 4) And this, as we can see, we have five zeros. So we get three from our differential equation because what the third-order equation. And we actually get two from the right hand side or the forcing of the function. So we can split it into A/(s-2) plus B/(s-2)^2, plus C/(s+2) two plus Ds + E over (s^2 + 4) So now, after we have split this, so the first part, because this is a root of multiplicity two, we need to have two terms, A and B, just as the individual root here. And then we have these non reducible factors, which would give us complex roots, which we just split into, or we don't split. But we have also two parameters  in the numerator. Having done this, we now need to find the parameters A, B, C, D, and E. So there's different ways to do this. We could just add values of s. So we can add s equals 2, s equals minus two to remove couple of these terms. And then we would get equations for A, B, C, D, and E. Or we can put in some values and set up a linear equation system and compare powers of s. So, if we do this, so I'm not doing this here. We would get values of A is equal to minus 3/128. B is 1 over 32. C is 17/128. D is minus 7/64, and E is equal to 18/64. Doing this, we can now write our solution in the Laplace domain with these partial fractions up here with the parameters given them. So we have our X(s) is    So having split our solution, the Laplace domain into the five fractions, we can now calculate the inverse of this. So the first term, we know, this is the Laplace transform of the exponential. So we can directly invert this. So we would get our small x of t is equal to Now we keep the constant here. And we can change this term as directly the e^(2t). Next term here. We know this from our right hand side. So this is the frequency shifted ramp function. By using this, we would get again t and e^(2t), So we have this handled, This term we can handle directly from the tables as well, because this is the negative exponential. So we get plus 17/128 and e^(-2t). So the last two terms,  let's handle these, So this term here is just the Laplace transform of the Cosine. So we can handle this directly. And this is minus 7/64 x cos(2t), This term here on the other hand is not directly the Laplace transform of the Sine. For this. We need to modify this to have a 2 here. So to balance this, divide that by two. So we multiply this by 2, divided this by 2. And now this term here is in the form of the Laplace transform of the Sine. So in total we will get 9/64. We basically cancelled the, (we divide 18 by two) times sin(2t) And all this is only valid for t larger or equal than 0. So this is the solution in the time domain calculated through the Laplace transform method.