This is my first time doing Laplace transform of derivatives and I'm not sure if my transform is correct.

The given expression is $$y^{''}+3y^{'} +2y=t$$ and the given values are $$y(0)=0, \space y^{'}(0)=2$$ What I got was $$S^2Y-0-2+3SY-3(0)+2Y=\frac{1}{s^2}$$ Then solving for $Y$ $$Y=\frac{1}{(S^2)(S^2+3S)}$$ Is my work correct?

  • $\begingroup$ No, you should get $$Y(s) = \dfrac{2s^2+1}{s^2(s^2+3s+2)}$$ $\endgroup$
    – Moo
    May 5, 2020 at 14:48
  • $\begingroup$ @Moo Could you show me what I got wrong? $\endgroup$ May 5, 2020 at 14:49
  • $\begingroup$ Look at how you grouped $Y$ on the LHS, see it? Then, what do you get for $1/s^2 + 2$ when you combine them? $\endgroup$
    – Moo
    May 5, 2020 at 14:50
  • $\begingroup$ @Moo Ah yes I see now, thank you $\endgroup$ May 5, 2020 at 14:51
  • 1
    $\begingroup$ Stupid algebra gets me every time too, so be careful because all the heavy lifting was already correct! $\endgroup$
    – Moo
    May 5, 2020 at 14:52


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