$$(t+2)x''+(t+1)x'-x=0, \qquad x(0)=0, \qquad x'(0)=2$$ and this $$ tx''-tx+4x=2e^t :: x(0)=sinh 1 ; $$
with using Derived from Laplace we get : $$ \begin{split} L(ty'') &= −s2Y'(s)−2sY(s)+y(0) \\ L(ty') &= -sY'(s)−Y(s)\\ L(ty) &= −Y'(s)\\ L(y') &= sY(s)−y(0) \end{split} $$
and when we put them in equation we get :
$$L[x](2s^2-s-2)-L'[x](s+s^2)=4s+4$$
how to solve the rest ?
how to solve this equation with laplace transform;
$$ tx''-tx+4x=2e^t :: x(0)=sinh 1 ; $$
Here is another way to approach for the sake of curiosity.
According to the substitution $u = x' + x$, we reduce the proposed ODE as follows:
\begin{align*} (t+2)x'' + (t+1)x' - x = 0 & \Longleftrightarrow (t+2)(x' + x)' - (x' + x) = 0\\\\ & \Longleftrightarrow (t+2)u' - u = 0 \end{align*}
Can you take it from here?