$$xy'' + (2x + 3)y' + (x + 3)y = 3e^{–x} \\ y(0) = 0 $$
I know I can plug in $L[y''],L[y']$ and get a first order linear differential equation, but I just cannot get the right answer.
I use the universal equation to solve the first order equation, is there anything wrong with my steps? Grateful for any answer!!!
Not using Laplace, one can factorize the equation as the sum of the outer coefficients is the inner coefficient $$ x(D^2+D)+(x+3)(D+1)=(xD+x+3)(D+1)\\ =(x(D+1)+3)(D+1)=(D+1)(x(D+1)+2) $$ This allows to solve the second order equation as a sequence of two first-order equations $$ z'+z=3e^{-x},\\ xy'+(x+2)y=z. $$ Other such decompositions are possible.
The DE is regular singular at $x=0$, so that the implied boundedness there is a second boundary condition. There might be strange effects on the Laplace transform.