The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True of False? Explain why.

1. A body falls from rest against resistance proportional to the square root of the speed at any instant. If the body attains speed V1 and V2 feet per second, after 1 and 2 seconds in motion, respectively, find an expression for the limiting velocity.

The Laplace transform of ${\displaystyle u(t-2)}$ is
(a) ${\displaystyle \frac{1}{s}+2}$
(b) ${\displaystyle \frac{1}{s}-2}$
(c) ${\displaystyle e^2\frac{s}{s}\left(d\right)\frac{e^{-2s}}{s}}$??

1 degree on celsius scale is equal to
A) $\frac{9}{5}$ degree on fahrenheit scale
B) $\frac{5}{9}$ degree on fahrenheit scale
C) 1 degree on fahrenheit scale
D) 5 degree on fahrenheit scale

The Laplace transform of $t{e}^t$ is A. $\frac{s}{(s+1{)}^2}$ B. $\frac{1}{(s-1{)}^2}$ C. $\frac{s}{(s+1{)}^2}$ D. $\frac{s}{(s-1)}$

What is the Laplace transform of ${\displaystyle t\cos t}$ into the s domain?

Find the general solution of the given differential equation:
$y^{″}-<!--\; -\; -->2y^{\prime }+y=0$

The rate at which a body cools is proportional to the difference in 
temperature between the body and its surroundings. If a body in air 
at 0℃ will cool from 200℃ 𝑡𝑜 100℃ in 40 minutes, how many more 
minutes will it take the body to cool from 100℃ 𝑡𝑜 50℃ ?

A body falls from rest against a resistance proportional to the velocity at any instant. If the limiting velocity is 60fps and the body attains half that velocity in 1 second, find the initial velocity.

What's the correct way to go about computing the Inverse Laplace transform of this?
$\frac{-<!--\; -\; -->2s+1}{(s^2+2s+5)}$
I Completed the square on the bottom but what do you do now?
$\frac{-<!--\; -\; -->2s+1}{(s+1{)}^2+4}$

How to find inverse Laplace transform of the following function?
$X(s)=\frac{s}{s^4+1}$
I tried to use the definition: $f(t)={\mathcal{L}}^{-<!--\; -\; -->1}\{F(s)\}=\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}i}\underset{T\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->iT}^{\mathrm{\gamma <!--\; \gamma \; -->}+iT}{e}^{st}F(s)ds$or the partial fraction expansion but I have not achieved results.

How do i find the lapalace transorm of this intergral using the convolution theorem? ${\int <!--\; \int \; -->}_0^t{e}^{-<!--\; -\; -->x}\cos <!--\; \; -->xdx$

How can I solve this differential equation? : ${\displaystyle xy dx-(x^2+1) dy=0}$

Find the inverse Laplace transform of $\frac{s^2-<!--\; -\; -->4s-<!--\; -\; -->4}{s^4+8s^2+16}$

inverse laplace transform - with symbolic variables:
$F(s)=\frac{2s^2+(a-<!--\; -\; -->6b)s+a^2-<!--\; -\; -->4ab}{(s^2-<!--\; -\; -->a^2)(s-<!--\; -\; -->2b)}$
My steps:
$F(s)=\frac{2s^2+(a-<!--\; -\; -->6b)s+a^2-<!--\; -\; -->4ab}{(s+a)(s-<!--\; -\; -->a)(s-<!--\; -\; -->2b)}$
$=\frac{A}{s+a}+\frac{B}{s-<!--\; -\; -->a}+\frac{C}{s-<!--\; -\; -->2b}+K$
$K=0$
$A=F(s)\ast <!--\; \ast \; -->(s+a)$

Laplace transform of $(3e{)}^t{\sin }^2<!--\; \; -->t$