There does exist a definition of the inverse Laplace Transform as a definite integral. Unfortunately, it is a complex contour integral and treats frequency as a complex number (the real part is conventional frequency we all know, the imaginary part has no physical meaning, but is necessary to do the math). That said, electrical engineers always ignore this and just use tables to look up inverses, because they don't get paid to do integrals, they get paid to design circuits that work (/sarcasm). The function that zeroes the original function for negative values of time has a name, it's called the Heaviside Stepfunction, named after English telegraph engineer and self-taught math whiz Oliver Heaviside. It may seem a bit odd to name such a simple function after a person, but interestingly, the Heaviside Stepfunction is closely related to one of the most famous functions of all time, the Error Function.
I've tried evaluating the inverse Laplace transform integral, and it is a hell of a lot easier to just use algebra to line up the expression with standard Laplace transform pairs, than it is to evaluate that integral. It seems like an interesting thing for curiosity sake, but not really practical to use the inverse Laplace transform integral. Are you familiar with any practical cases where the inverse Laplace transform integral has an advantage?
it sort of is! assuming v = e^rt and solving for r reduces the problem to just solving a quadratic equation, called the differential equation’s “characteristic polynomial”, and its exactly the equation you think it is from looking at the DE. you have to use euler’s identity to get real valued functions when the quadratic equation gives complex roots, but otherwise yeah its basically just solving the polynomial. so long as the coefficients are constant numbers, this method can always give you the general solution (even if theres a double root we have a way of getting the other solution). tldr: your intuition that the motivating example here is solved by a quadratic equation is right, the problem does reduce to one, it just also works as a motivating example of the content of this video
I beg anyone please tell me why at 4:15 the "solution" is the sum? I know if you have a quadratic formula (x-1)(x-2) = 0, then x MUST be EITHER 1, OR 2. X=1,2. However, here you don't explain why it's the SUM between the 2 different functions? Basically you're saying that x = 1+2, which for me makes no sense. I am aware that the real factor is (D-1)(D-2)v = 0 and the D's aren't really representing "x" and v is being multiplied at the outside but am soo eager to understand why it's the sum. I encountered this in a lecture long ago as well and never understood why. It was never explained to me. If the explanation is too long, it's okay for anyone reading this not to explain it, but to provide me with where to find a proper explanation. If the explanation is easy I beg anyone to explain it to me, thank you so much guys!!
Good question! The reason why it works is simply because plugging in their sum (or any linear combination of the solutions) works when we plug in these functions. So (D-1)(D-2)(c1 e^t + c2e^(2t)) = (D-1)(D-2)c1 e^t + (D-1)(D-2)e^(2t) = 0 + 0 = 0 A more interesting question might be whether it even solves our initial conditions. So that is: c1 e^t0 + c2 e^(2t0) = f(t0) c2 e^t0 + 2 * c2 e^(2t0) = f'(t0) This is a matrix equation. In the case of differential equations we call the determinant the Wronskian, W. As you might know from linear algebra if the Wronskian is nonzero then this equation has a solution. By calculating the Wronskian you can show that we would need the sum of the solutions to solve our initial conditions. Hope this helps and thank you for your feedback.
