Why does the Laplace Transform fail to solve algebraic equations?

Question

Given the idea behind the Laplace Transform solving ODEs, I would expect it to be possible to solve algebraic equations via a similar technique. For example, $\sin t = \cos t$ would transform to $\frac{1}{s^2 + 1} = \frac{s}{s^2 + 1}$, which can be manipulated to $1 = \frac{1}{s^2}$. However, the inverse transform of this would be $\delta = t$, which is not a valid answer. What is the reason that this technique fails to solve algebraic equations?

Answer 1

Solving ODEs by Laplace transform solves only linear ODEs. I think you would agree that for algebraic equations, we do not need such technique to solve linear equations.

Your example: What it tells us is: if we have $\sin t = \cos t$ for all $t$, then we get some conclusion. Of course any conclusion we draw from a false hypothesis like that is worthless.

Answer 2

$\sin t =\cos t$ doesn't make any sense, you mean $$\sin t = h\ast \cos( t)$$ Taking the Laplace transform $$\frac{1}{s^2 + 1} = H(s)\frac{s}{s^2 + 1}$$ $$H(s) = \frac1s, \qquad h(t) = 1$$ which is the correct solution $$h\ast \cos(t)=\int_0^t \cos(u)du = \sin t$$ (there are several traps about the support, the convolution and the domain of convergence we are using, replace $h(t)=1$ and $\sin t $ by $h(t)=1_{t >0}$ and $1_{t > 0}\sin t$ to make it clear)