Solve $$ t \, x''-t \, x' + 4 \, x = 2 \, e^t, \hspace{5mm} x(0)=\sinh(1) ; $$
by using the Laplace transform. Using the transform of a derivative we get: \begin{split} L(tx'') &= −s^2X'(s)−2sX(s)+x(0) \\ L(tx') &= -sX'(s)−X(s)\\ L(tx) &= −X'(s)\\ L(x') &= sX(s)−x(0) \end{split}
and by applying that we get :
$$X(4-2s) - X'(s^2-1) = \frac{2}{s-1} - \sinh(1).$$
What to do next, or is there anyway else to solve this, with the Laplace transformed equation ?
The LT of $x''(t)$ is $s^2X(s)-sx(0)-x'(0).$ Multiplying the $x''(t)$ by $t$ is tantamount to taking the negative derivative of the result w.r.t. $s,$ thus: \begin{align*} L[t x''(t)]&=-2sX(s)-s^2X'(s)+x(0)\\ L[x'(t)]&=s X(s)-x(0)\\ L[t x'(t)]&=-X(s)-s X'(s)\\ L[e^t]&=\frac{1}{s-1}, \end{align*} so that the LT of the entire equation is \begin{align*} -2sX(s)-s^2X'(s)+\sinh(1)+X(s)+s X'(s)+4X(s)&=\frac{2}{s-1}\\ (s-s^2)\,X'(s)+(5-2s)\,X(s)&=\frac{2}{s-1}-\sinh(1). \end{align*} Solving this is rather annoying, but it's first-order linear, so you can apply the integrating factor method. The result is this: $$X(s)=\dfrac{(1-s)^3\left(-\dfrac{\left(e^2-1\right)(s-1)-\dfrac{2(e-3)(1+3e)}{s-1}+\dfrac{2-2(e-6)e}{(s-1)^2}+\dfrac{1-(e-16)e}{3(s-1)^3}+\dfrac{e}{(s-1)^4}+4((e-1)e-1)\ln(s-1)}{e}+2 c_1\right)}{2s^5}.$$ Good luck getting the inverse LT of that. Incidentally, Mathematica can solve the original equation exactly. The solution is: $$x(t)=\frac{1}{12}\left(e^t(t((t-11)t+26)(1-2\sinh(1))+12\sinh(1))+\frac{t\left((t-6)^2 t-24\right)(((e-1)e-1)\operatorname{Ei}(t)+12\,e\,c_1)}{e}\right),$$ where $$\operatorname{Ei}(t):=-\int_{-t}^\infty \frac{e^{-\tau}}{\tau}\,d\tau,$$ using the principal value, and $c_1$ is a constant.