Laplace transform of \(\displaystyle{f{{\left({t}\right)}}}={t}{e}^{{-{t}}}{\sin{{\left({2}{t}\right)}}}\)

Use the transforms in the table below to find the inverse Laplace transform of the following function.
${\displaystyle F\left(s\right)=\frac{5}{s^4}}$

Consider the system of differential equations ${\displaystyle \frac{dx}{dt}=-y\frac{dy}{dt}=-x}$.a)Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation.
b)Solve the equation you obtained for y as a function of t; hence find x as a function of t.

Use the definition of Laplace Transforms to find $L\{f(t)\}$
$f(t)=\{\begin{array}{ll}-1& 0\le t<1\\ 1& t\ge 1\end{array}$
Then rewrite f(t) as a sum of step functions, $u_c(t)$, and show that by taking Laplace transforms, this yields the same answer as your direct computation.

The Laplace transform of $L\left\{\frac{d^{3}x(t)}{d{t}^3}\right\}$ is given by
Select one:
$sX(s)-x(0)$
${\displaystyle s^{3}X\left(s\right)-s^{2}x\left(0\right)-s\dot{x}\left(0\right)-\ddot{x}\left(0\right)}$
${\displaystyle s^{3}X\left(s\right)-s^2\ddot{x}\left(0\right)-s\dot{x}\left(0\right)-x\left(0\right)}$
${\displaystyle s^{2}X\left(s\right)-sx\left(0\right)-\dot{x}\left(0\right)}$

Obtaining Differential Equations from Functions
${\displaystyle \frac{dy}{dx}=x^2-1}$
is a first order ODE,
${\displaystyle \frac{d^{2}y}{{dx}^2}+2{\left(\frac{dy}{dx}\right)}^2+y=0}$
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for
${\displaystyle y=e^x(A\cos x+B\sin x)}$ and the steps that I followed are as follows.
${\displaystyle \frac{dy}{dx}=e^x(A\cos x+B\sin x)+e^x(-A\sin x+B\cos x)}$
${\displaystyle =y+e^x(-A\sin x+B\cos x)}$ (1)
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)}$
${\displaystyle =\frac{dy}{dx}+(\frac{dy}{dx}-y)-y}$
using the orginal function and (1). Finally,
${\displaystyle \frac{d^{2}y}{{dx}^2}-2\frac{dy}{dx}+2y=0}$,
which is the required differential equation.
Similarly, if the function is $y=(A\cos 2t+B\sin 2t)$, the differential equation that I get is
${\displaystyle \frac{d^{2}y}{{dx}^2}+4y=0}$
following similar steps as above.

Solve the following manually using separable differential equations method.
${\displaystyle mydx+nxdy=0}$

I need to calculate a taylor polynomial for a function ${\displaystyle f:R\to R}$ where we know the following
${\displaystyle f{}^{″}\left(x\right)+f\left(x\right)=e^{-x}\mathrm{\forall }x}$
${\displaystyle f\left(0\right)=0}$
${\displaystyle f^{\prime }\left(0\right)=2}$