Why do you use laplace transform?

Answer 1

The most generalized reason would be:
"To solve initial-valued differential equations of the 2nd (or higher) order." Laplace is a little powerful for 1st order, but it will solve them as well.
There is a limitation here: Laplace will only generate an exact answer if initial conditions are provided. Laplace cannot be used for boundary-valued problems.

In terms of Electronics Engineering, the Laplace transform is used to get your model into the s-domain, so that s-domain analysis may be performed (finding zeroes and poles of your characteristic equation).
This is particularly useful if one needs to determine the kind of response an RC, RLC, or LC circuit will provide (i.e. underdamped, overdamped, critically damped).
Once in the s-domain, we may begin discussing the components in terms of impedance. Sometimes it is easier to calculate the voltage or current across a capacitor or an inductor in terms of the components' impedances, rather than find it in a t-domain model.

The node-voltage and mesh-current methods used to analyze a circuit in the t-domain work in the s-domain as well.