Help me see if I got it right, let say I got a differential equation $${y}''(t)+4{y}'(t)+3y(t)=tu(t)$$ Transfer function is $$\dfrac{output}{input}$$ that is, $$H(s)=\dfrac{Y(s)}{X(s)}$$ Since I have no initial conditions, I can disregard the ${y}'(0)$ and $y(0)$ part in the Laplace transform: $$s^{2}Y(s)+4sY(s)+3Y(s)=\frac{1}{s^{2}},$$ I am not sure about the $u(t)$ part but I do know $\, \, we\, \, add\, \, it\, \, in\, \, the\, \, end \\\\ Now\, \, in\, \, this\, \, example\, \, the\, \, input\, \, is\, \, also\, \, a\, \, part\, \, of\, \, the\, \, output?\, \, I\, \, do\, \, know\, \, if\, \, \\\\ it \, \, was\, \, x(t)\, \, instead\, \, of\, \, tu(t)\, \, it\, \, would\, \, be\, \, much\, \, easier\, \, since\, \, then\, \, I \, \, would\, \, have\, \, only\, \, X(s) $
$$ Y(s)(s^{2}+4s+3)=\dfrac{1}{s^{2}}$$ $$Y(s)=\dfrac{\dfrac{1}{s^{2}}}{s^{2}+4s+3}=\dfrac{1}{s^{2}(s+1)(s+3)}$$ $ Now\, \, we\, \, use\, \, \frac{1}{s^{2}}\, \, as\, \, input\, \, and\, \, we\, \, get\, \, the\, \, transfer\, \, function \\\\$ $$ H(s)=\dfrac{output}{input}=\dfrac{\frac{1}{s^{2}(s+1)(s+3)}}{\dfrac{1}{s^{2}}}=\dfrac{s^{2}}{s^{2}(s+1)(s+3)}=\dfrac{1}{(s+1)(s+3)}$$ Now for the forced response I take apart $Y(s)$ using partial fractions :
$$Y(s)=\dfrac{1}{s^{2}(s+1)(s+3)}=\dfrac{A}{s}+\dfrac{B}{s^{2}}+\dfrac{C}{s+1}+\dfrac{D}{s+3} $$ Anyway after a bit of solving I get: $$\left(\dfrac{1}{3}t-\dfrac{1}{18}e^{-3t}+\dfrac{1}{2}e^{-t}-\dfrac{4}{9} \right)u(t)$$ Natural response is when $x=0$ Which means $Y(s)=0$ $$m^{2}+4m+3=0 $$ $$m_{1,2}=-1,-3 $$ $$y(t)=C_{1}^{ m_1t}+C_{2}^{ m_2t} $$ $$y(t)=C_{1}^{ -1t}+C_{2}^{ -3t}$$ And for the impulse response I just partial fraction solve the transfer function $H(s)$ and then inverse Laplace.