Find the inverse Laplace transform of frac{6s+15}{(s^2+25)} s>0

Solve the equation:

$L^{-1}=\frac{s}{(s^2-a^2{)}^2}$

How to deal with two interdependent integrators?
I have two functions, f(t,x) and g(t,u), where ${\displaystyle \frac{d}{dt}u=f(t,x)\text{and}\frac{d}{dt}x=g(t,u)}$.
I am trying to discretize the integral of this system in order to track x and u. I have succeeded using Euler integration, which is quite simple, since x(t) and u(t) are both known at t:
${\displaystyle u(t+h)=u\left(t\right)+hf(t,x\left(t\right))}$
${\displaystyle x(t+h)=x\left(t\right)+hg(t,u\left(t\right))}$
However, I am now trying to implement mid-point integration to get more accurate results. (Eventually Runge-Kutta but I am stuck here for now.)

Solve the following homogeneous differential equations
${\displaystyle \frac{dy}{dx}=\frac{x-y}{x+y}}$

Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.
${\displaystyle 6t^4{e}^{-6t}-t^4+\cos 6t}$

Solve $y"+4y=f(t)\text{ subject to }y(0)=0,y^{\prime }(0)=2\text{ and }$
$f(t)=\{\begin{array}{ll}0& t<\pi \\ 1& t\ge \pi \end{array}$

Use the Laplace transform to solve the given initial value problem.y"+2y'+5y=0;y(0)=2,y'(0)=−1

Find the Laplace transform Y(s), of the solution of the IVP
$y"+3y^{\prime }+2y=\cos (2t)$
$y(0)=0$
$y^{\prime }(0)=1$
Do not solve the IVP