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I'm trying to solve the following IVP (differential equations) with the Laplace Transform method:

\begin{cases} y''+9y=36t\sin(3t)\\ y(0) = 0\\ y'(0) = 3 \end{cases}

After taking the Laplace Transform of both sides, I obtain

$$s^2Y(s) - 3 + 9sY(s) = \frac{216}{(s^2+9)^2}$$

where $Y(s) = \mathcal{L}[y(t)]$.

Solving for $Y(s)$, I get the following equation: $$Y(s) = \frac{216}{(s^2+9)^2(s^2+9s)} + \frac{3}{s^2+9s}$$ When I try to take the inverse Laplace transform of both sides to solve for $y(t)$, I can't figure out what to do with the first term on the RHS (second term is a simple partial fraction decomposition). The algorithm for partial fractions with linear terms is problematic because the numerator of the first term has a $\frac{1}{0}$ issue. Without partial fractions, the only way I can think to do this one is through Reduction of Order, but my book only gives examples of the case where we only have $(s^2+b^2)^{k+1}$ in the denominator. The thing that's messing me up so much is the $s^2+9s$ term. Any help or insight in how to solve this problem from here would be much appreciated.

EDIT: After probably 2 hours of mulling over this problem, I realized that I had forgotten the $s$ after 216 because of the chain rule when taking the derivative of transforms. That makes this so much easier. I think I can do it.

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  • $\begingroup$ Factor $s^2+9s=s(s+9)$ and then write the (rather complicated) partial fraction decomposition $(As+B)/(s^2+9)^2 + (Cs+D)/(s^2+9) + E/s + F/(s+9)$. Besides, have you remarked that the LHS accounts for an harmonic oscillator with $\omega=3$ and that the RHS accounts for an excitation with the same $\omega$ ? What happens in this case ? What solution can you foresee ? $\endgroup$
    – Jean Marie
    Commented Mar 2, 2016 at 11:09
  • $\begingroup$ Apologies. I had forgotten to include the $s$ when taking the Laplace transform of $36t\sin(3t)$, so the $s$ terms will cancel out, making the partial fraction much easier. On another note, however, I haven't yet learned any physics applications (I assume that's what you were referencing) in DEs as of yet, so I'm not even sure what the variable $\omega$ means. My DEs course is very proofy. Thank you very much for your time, by the way! $\endgroup$ Commented Mar 2, 2016 at 11:13
  • $\begingroup$ I understand. Have a look on a book about elementary wave physics. $\endgroup$
    – Jean Marie
    Commented Mar 2, 2016 at 11:24

1 Answer 1

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Notice:

  • $$\mathcal{L}_{t}\left[y'(t)\right]_{(s)}=sy(s)-y(0)$$
  • $$\mathcal{L}_{t}\left[y''(t)\right]_{(s)}=s^2y(s)-sy(0)-y'(0)$$
  • $$\mathcal{L}_{t}\left[t\right]_{(s)}=\frac{1}{s^2}$$
  • $$\mathcal{L}_{t}\left[\sin(t)\right]_{(s)}=\frac{1}{1+s^2}$$
  • $$\mathcal{L}_{t}\left[\sin(at)\right]_{(s)}=\frac{a}{a^2+s^2}$$
  • $$\mathcal{L}_{t}\left[t\sin(at)\right]_{(s)}=\frac{2as}{\left(a^2+s^2\right)^2}$$

Solving your question:

$$y''(t)+9y(t)=36t\sin(3t)\Longleftrightarrow$$ $$\mathcal{L}_{t}\left[y''(t)+9y(t)\right]_{(s)}=\mathcal{L}_{t}\left[36t\sin(3t)\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_{t}\left[y''(t)\right]_{(s)}+9\mathcal{L}_{t}\left[y(t)\right]_{(s)}=36\mathcal{L}_{t}\left[t\sin(3t)\right]_{(s)}\Longleftrightarrow$$ $$s^2y(s)-sy(0)-y'(0)+9y(s)=36\cdot\frac{6s}{\left(9+s^2\right)^2}\Longleftrightarrow$$


Now, set $y(0)=0$ and $y'(0)=3$:


$$s^2y(s)-s\cdot0-3+9y(s)=\frac{216s}{\left(9+s^2\right)^2}\Longleftrightarrow$$ $$s^2y(s)-3+9y(s)=\frac{216s}{\left(9+s^2\right)^2}\Longleftrightarrow$$ $$s^2y(s)+9y(s)=\frac{216s}{\left(9+s^2\right)^2}+3\Longleftrightarrow$$ $$y(s)\left[s^2+9\right]=\frac{216s}{\left(9+s^2\right)^2}+3\Longleftrightarrow$$ $$y(s)=\frac{\frac{216s}{\left(9+s^2\right)^2}+3}{s^2+9}\Longleftrightarrow$$ $$y(s)=\frac{3\left(72s+\left(9+s^2\right)^2\right)}{\left(9+s^2\right)^3}\Longleftrightarrow$$ $$\mathcal{L}_{s}^{-1}\left[y(s)\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{3\left(72s+\left(9+s^2\right)^2\right)}{\left(9+s^2\right)^3}\right]_{(t)}\Longleftrightarrow$$ $$y(t)=3\mathcal{L}_{s}^{-1}\left[\frac{72s+\left(9+s^2\right)^2}{\left(9+s^2\right)^3}\right]_{(t)}\Longleftrightarrow$$ $$y(t)=3\mathcal{L}_{s}^{-1}\left[\frac{72s}{\left(9+s^2\right)^3}+\frac{1}{s^2+9}\right]_{(t)}\Longleftrightarrow$$ $$y(t)=3\left[72\mathcal{L}_{s}^{-1}\left[\frac{s}{\left(9+s^2\right)^3}\right]_{(t)}+\mathcal{L}_{s}^{-1}\left[\frac{1}{s^2+9}\right]_{(t)}\right]\Longleftrightarrow$$ $$y(t)=3\left[\frac{t\left(\sin(3t)-3t\cos(3t)\right)}{3}+\frac{\sin(3t)}{3}\right]\Longleftrightarrow$$ $$y(t)=t\left(\sin(3t)-3t\cos(3t)+\sin(3t)\right)\Longleftrightarrow$$ $$y(t)=(1+t)\sin(3t)-3t^2\cos(3t)$$

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