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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.
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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.
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.
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