English: If we evaluate the integral, we will obtain $L^{- 1}\left\{ \frac{1}{\left( s^{2} + 4 \right)^{2}} \right\} = \frac{1}{16}sin\ 2t\ - \frac{t}{8}cos\ 2t\ .$ $L^{- 1}\left\{ \frac{1}{\left( s^{2} + 4 \right)^{2}} \right\} = \frac{1}{16}sin\ 2t\ - \frac{t}{8}cos\ 2t\ $అని So that, now we can tell the value of y(t) as $= - \frac{5}{9}sin\ t\ + \frac{5}{18}sin\ 2t\ + \frac{8}{3}\left( \frac{1}{16}sin\ 2t\ - \frac{t}{8}cos\ 2t\ \right)$ This is the required solution of the given ODE.
Telugu: మనం సమగ్రతను మదింపు చేస్తే, మనం $L^{- 1}\లెఫ్ట్\{ \frac{1}{\లెఫ్ట్{1}{{2} + 4 \కుడి)^{2}} \కుడి\} = \frac{1}{16}sin\ 2t\ - \frac{t}{8}cos\ 2t\ .$ $L^{- 1}\లెఫ్ట్\{ \frac{1}{\{2} లెఫ్ట్\{ {frac{1}{16}sin+ 4 \కుడి)^{2}} \frac{1}{16}sin\ 2t\ - \frac{t}{8}cos\ 2t{5}{18}{5}{9}\ frac{8}{3}\ఎడమ( \frac{1}{16}sin\ 2t\ - \frac{t}{8}cos\ 2t\ \కుడి)$ ఇది ఇవ్వబడ్డ ODE యొక్క అవసరమైన పరిష్కారం.