to find the inverse Laplace transform of the given function. F(s)=\frac{2^{n+1}n!}{s^{n+1}}

Solve differential equation $(x^2\ln x)y^{\prime }+xy=0$

Solving separable differential equations by formal computations
When solving a separable differential equation we move from the original equation
$N(y){\displaystyle \frac{dy}{dx}}=M(x)$
to
$N(y)dy=M(x)dx$
Then we integrate both sides. My question is how precise is the expression $N(y)dy=M(x)dx$ ? is it a formal writing to simplify computation and get quickly into integrating both sides or is it a precise mathematical expression that has a precise mathematical meaning but goes beyond an introductory course on differential equations? Thanks for your help !

How to find the particular the solution for this differential equation $x{y}^{\prime }(x)-y(x)(1+x^2)=\frac{x^2}{2}$ first i solved it as homogenous differential equation as follows $x{y}^{\prime }(x)-y(x)(1+x^2)=0$ thrn i got $y(x)=c{e}^{\frac{x^2}{2}}x$

Solve the following initial value problem by using Laplace transform
$y^{″}+4y=4t+12,y(0)=4,y^{\prime }(0)=0$

Obtain Laplace transforms for the function ${\displaystyle e^t}$ step(t−3)

Find the Laplace transform $L\{u_3(t)(t^2-5t+6)\}$
$a)F(s)=e^{-3s}(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2})$
$b)F(s)=e^{-3s}(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s})$
$c)F(s)=e^{-3s}\frac{2+s}{s^4}$
$d)F(s)=e^{-3s}\frac{2+s}{s^3}$
$e)F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$

Find the inverse Laplace Tranformation by using convolution theorem for the function $\frac{1}{s^3(s-5)}$