Solution of $ty'' +(2t+3)y' +(t+3)y = 3e^{-t}$ via Laplace transform A recent question which was put on hold due to lack of context by the OP was the following:

Solve the following ODE using Laplace transforms.
$$ty'' +(2t+3)y' +(t+3)y = 3e^{-t}, \qquad y(0)=0$$

Putting the equation into the form
$$t(y^{\prime\prime}+2y^\prime+y)+3(y^\prime+y)=3e^{-t} $$
which has $y_c=ce^{-t}$ as a solution of its complementary equation immediately suggests $y=Ate^{-t}$ as a particular solution.
And this is borne out by substitution, with $A=1$. Applying the initial condition yields the solution
$$ y=te^{-t} $$
So why would the original OP want the equation solved using Laplace transforms?
Is there a shorter path using Laplace transforms than the following?
Use the fact that $(3e^{-t})^\prime+3e^{-t}=0$ to get the homogeneous equation
$$ [t(y^{\prime\prime}+2y^\prime+y)+3(y^\prime+y)]^\prime+[t(y^{\prime\prime}+2y^\prime+y)+3(y^\prime+y)]=0 $$
This simplifies to the homogeneous equation
$$ 
t(y^{\prime\prime\prime}+3y^{\prime\prime}+3y^\prime+y)+4(y^{\prime\prime}+2y^\prime+y)=0 $$
Taking the Laplace transform of this involves quite a bit of tedium which I will spare the reader, but yields the following:
\begin{eqnarray}  
   (s+1)^3Y^\prime+2(s+1)^2Y&=&0\\\  
   (s+1)^2Y^\prime+2(s+1)Y&=&0\\\  
   \left[(s+1)^2Y\right]^\prime&=&0\\\  
   Y&=&\frac{c}{(s+1)^2}\\\  
   y&=&cte^{-t}  
   \end{eqnarray}
So, with $c=1$, yielding the same solution found much more easily by inspection.

Keeping in mind that
$$
\mathcal{L}(t\mathcal{D}(y)) = -\frac{d}{ds}\mathcal{L}(\mathcal{D}(y)) 
$$
assuming null initial conditions and applying this in
$$
t \mathcal{D}^2y + \mathcal{D}y = 3e^{-t}
$$
we have
$$
-\frac{d}{ds}\left((s+1)^2 Y\right)+(s+1)Y = \frac{3}{s+1}
$$
or
$$
-Y'+\frac{1}{s+1}Y = \frac{3}{(s+1)^3}
$$
the homogeneous gives
$$
Y_h = \frac{c_1}{s+1}
$$
and the particular
$$
Y_p = \frac{1}{(s+1)^2}
$$
hence
$$
y(t) = (c_1+t)e^{-t}
$$
not difficult at all. Now if $y(0) = 0$ then
$$
y(t) = te^{-t}
$$
