So, here's a problem using the Dirac Delta function. It's an impulsive force. So, the inhomogeneous term is modeled by Dirac Delta function that hits at t equals zero. So, physically if this was a mass on a spring or a pendulum, you would have a hammer hitting the mass at t equals zero. The mass is initially at rest with zero velocity, and the impulsive force will then impart some momentum to the object. So, we assume that this will act at a time slightly larger than zero. How do we solve this we have to take the Laplace transform of this equation. So, the easiest thing is to have a look at the table. So, we have homogeneous initial conditions. So, x of zero and x dot of zero equals zero. We apply a line 16 to take the Laplace transform of the second derivative so we get S squared times capital X. We apply line 15 to take the Laplace transform of the first derivative. So, we have a 3X dot, so we end up with a 3S times x. Then we take the Laplace transform of 2X which is just two capital X that's equal to the Laplace transform of the Dirac Delta function. The Laplace transform of Delta of t with c equals zero then would just be one from line 14. So, we have the Laplace transform of the differential equation. This is our algebraic equation we can solve for X. So, X equals one over, we have S squared plus 3S plus two that factors into S plus one times S plus two. That's our solution in S space for X of S. Then we have to take the inverse Laplace transform of this to get back to t space. Again this type of expression needs a partial fraction expansion. So, we can write one over S plus one times S plus two equals a over S plus one plus b over S plus two. We can use the cover up method again. So to determine a, we multiply both sides by S plus one. We said that S equal to minus one. So, we get a equal to one to determine b. We multiply both sides by S plus two and said S equal to minus two, so that b is equal to minus one. Then we need to take the inverse Laplace transform of this to get X of t. So, let's go back to the table. The inverse Laplace transform of one over S plus one. We can use line three where a is equal to negative one. So, that would be e to the minus t. The inverse Laplace transform of b over S plus two is also from line three with a equals minus two. So, that will be e to the minus two t. That term gets multiplied by minus one because b is equal to minus one. So, we've taken the Laplace transform of capital x using the partial fraction expansion and we get e to the minus t minus e to the minus 2t. If you write that in a nicer way, you would factor out the e to the minus t term which is the dominant term, and you would have a one minus e to the minus t. There's one point that is worth making here. X of zero is equal to zero right. That's our initial condition. But here X dot of zero before the hitting with the impulsive force is equal to zero. But after you hit with the impulsive force so that's at zero plus, it's the derivative of this expression evaluated at t equal to zero. That would be minus one minus minus two or minus one plus two would be one. So, the impulse force causes a discontinuity in the velocity of X. Before the impulse hits X dot is zero after the impulse hits X dot equals one. X is still at zero before and after the impulse hits. So let me summarize. We're trying to solve an equation with an impulsive inhomogeneous term that's modeled by a Dirac Delta function. We use the table to take the Laplace transform of the equation and solve for X of s, then we can use a partial fraction expansion to write down x of t. Pretty straightforward. The Dirac Delta function makes it very easy to do integrals. The only thing that's tricky here is that you notice that the impulse force causes a discontinuity in the velocity. I'm Jeff Chasnov. Thanks for watching, and I'll see you in the next video.