I would like to solve the IVP
$$y'' + 2y' + 3y = 3t$$
with $y(0)=0, y'(0)=1$ using transform methods. Could someone check my work is correct, as my solution differs from that in the text?
Solution.
Taking Laplace transforms on both sides,
\begin{align*} L[y''] + 2L[y'] + 3L[y] &= \frac{1}{s^2} \\ s^2L[y]-sy(0)-y'(0)+2(sL[y]-y(0))+3L[y]&=\frac{1}{s^2}\\ (s^2 + 2s + 3)L[y] + 1 &= \frac{1}{s^2} \\ L[y] &= \frac{1}{s^2(s^2 + 2s + 3)} - \frac{1}{s^2 + 2s + 3}\\ L[y] &= \frac{1}{s^2[(s+1)^2 + (\sqrt{2})^2]} - \frac{1}{(s+1)^2 + (\sqrt{2})^2} \tag{1} \end{align*}
For the second term on the RHS, we have:
$$L^{-1}\left[\frac{1}{(s+1)^2 + (\sqrt{2})^2}\right]=\frac{1}{\sqrt{2}}e^{-t}\sin(\sqrt{2}t)\tag{2}$$
Let's write the first term on the RHS as:
\begin{align*} \frac{1}{s^2[(s+1)^2 + (\sqrt{2})^2]} &= \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2 + 2s + 3}\\ 1 &= As(s^2 + 2s + 3) + B(s^2 + 2s + 3) + (Cs + D)s^2 \end{align*}
Putting $s = 0$, yields $3B = 1$, so $B = \frac{1}{3}$.
Expanding out the terms, we have:
\begin{align*} 1 &= A(s^3 + 2s^2 + 3s) + B(s^2 + 2s + 3) + Cs^3 + Ds^2\\ 1 &= s^3(A + C) + s^2(2A + B + D) + s(3A + 2B) + 3B \end{align*}
Comparing the coefficients,
\begin{align*} A + C &= 0\\ 2A + B + D &= 0 \\ 3A + 2B &= 0 \end{align*}
$$A = -\frac{2}{3}\times B = -\frac{2}{9} \tag{3}$$
$$C = -A = \frac{2}{9}\tag{4}$$
\begin{align*} 2A + B + D &= 0\\ 2 \times \frac{-2}{9} + \frac{1}{3} + D &= 0\\ -\frac{4}{9}+\frac{3}{9} + D &=0\\ D&= \frac{1}{9} \tag{5} \end{align*}
So:
\begin{align*}\small L^{-1}\left[\frac{1}{s^2[(s+1)^2 + (\sqrt{2})^2]}\right] &= -\frac{2}{9} L^{-1}\left[\frac{1}{s}\right] + \frac{1}{3}L^{-1}\left[\frac{1}{s^2}\right] + \frac{2}{9}L^{-1}\left[\frac{s}{(s+1)^2 + (\sqrt{2})^2}\right] + \frac{1}{9\sqrt{2}}L^{-1}\left[\frac{\sqrt{2}}{(s+1)^2 + (\sqrt{2})^2}\right]\\ &=-\frac{2}{9} + \frac{t}{3} + \frac{2}{9}e^{-t}\cos(\sqrt{2}t) + \frac{1}{9\sqrt{2}}e^{-t}\sin(\sqrt{2}t) \end{align*}
Putting everything together, we find:
$$ y(t) = -\frac{2}{9} + \frac{t}{3} + \frac{2}{9}e^{-t}\cos(\sqrt{2}t) - \frac{8}{9\sqrt{2}}e^{-t}\sin(\sqrt{2}t)$$
whilst the answer provided in the text is:
$$ y(t) = \frac{2}{3}e^{-t}\cos(\sqrt{2}t) + \frac{\sqrt{2}}{3}e^{-t}\sin(\sqrt{2}t) + t - \frac{2}{3}$$