In this problem, \(\displaystyle{y}={c}_{{1}}{e}^{{x}}+{c}_{{2}}{e}^{{-{x}}}\) is a two-parameter

use the Laplace transform to solve the initial value problem.
$y"-3y^{\prime }+2y=\{\begin{array}{ll}0& 0\le t<1\\ 1& 1\le t<2\\ -1& t\ge 2\end{array}$
$y(0)=-3$
$y^{\prime }(0)=1$

Solve differential equation $y^3{y}^{\prime }+\frac{1}{x}{y}^4=\frac{\sin x}{x^4}$

Find the Laplace transform of the following:
${\displaystyle h\left(t\right)=t^2{e}^{-t}h(t-2)}$

Write first-order differential equations that express the situation: The increase in the number of people who have heard a rumor in a town of population 20,00020,000 is proportional to the product of the number PP who have heard the rumor and the number who have not heard the rumor at time tt .

I have to find the form of the function $h(t)$ in this equation:
$h^{\prime }(t)-\frac{\theta h(t)}{2}+1=0$
with $h(T)=C$ where $T$ and $C$ are constants and $T\ge 0$
At first I thought it was a Bernoulli differential equation and I tried:
$V(t)=\frac{1}{h(t)}\Rightarrow V^{\prime }(t)=\frac{h^{\prime }(t)}{h(t{)}^2}$
My equation becomes worse:
$V^{\prime }(t)-\frac{\theta }{2}+\frac{1}{V(t{)}^2}=0$
Then I tried with the integrating factor:
$(e^{t}h(t){)}^{\prime }=2e^{t}h(t)-e^t$
$e^{t}h(t)=\int 2e^{t}h(t)-e^t$
I'm stuck here because $h(t)$ is not a variable, is a function and I have to find how it looks like.
Edit: I forgot: $\theta$ is a constant.
How can I solve this problem?

Explain First Shift Theorem & its properties?

I have a Laplace Transform problem which Im