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Applications of laplace transform in engineering?
Laplace transforms to reduce a differential equation to an algebra problem. Engineers often must solve difficult differential equations and this is one nice way of doing it.
Does every continious function has laplace transform?
There are continuous functions, for example f(t) = e^{t^2}, for which the integral defining the Laplace transform does not converge for any value of the Laplace variable s. So you could say that this continuous function does not have a Laplace transform.
How Laplace Transform is used solve transient functions in circuit analysis?
f(t)dt and when f(t)=1=1/s or f(t)=k=k/s. finaly can be solve:Laplace transform t domain and s domain L.
Using laplace transform find function y given 2nd derivative of y plus y equals 0 with both initial conditions equal to 0?
Solve y''+y=0 using Laplace. Umm y=0, 0''+0=0, 0.o Oh well here it is. First you take the Laplace of each term, so . . . L(y'')+L(y)=L(0) Using your Laplace table you know the Laplace of all these terms s2L(y)-sy(0)-y'(0) + L(y) = 0 Since both initial conditions are 0 this simplifies to. . . s2L(y) + L(y) = 0 You can factor out the L(y) and solve for it. L(y) = 0/(s2+1) L(y) = 0 Now take the inverse Laplace of both sides and solve for y. L-1(L(y)) = L-1(0) y = 0
Y' equals x plus y Explicit solution?
using Laplace transform, we have: sY(s) = Y(s) + 1/(s2) ---> (s-1)Y(s) = 1/(s2), and Y(s) = 1/[(s2)(s-1)]From the Laplace table, this is ex - x -1, which satisfies the original differential eq.derivative of [ex - x -1] = ex -1; so, ex - 1 = ex - x - 1 + xto account for initial conditions, we need to multiply the ex term by a constant CSo y = C*ex - x - 1, and y' = C*ex - 1, with the constant C, to be determined from the initial conditions.
