I've got the equation which I am supposed to use laplace transforms to solve:

$$ y'' + y = e^t + tsin(t), ~~~~~y(0) = 3/2,~~~~~y'(0) = 5/2 $$

Using laplace transforms, I come up with the equation:

$$ Y(s) = {1 \over (s-1)(s^2+1)} + {2s \over (s^2+1)^3} + {3s+5 \over 2(s^2+1)}$$

I can use partial fraction and inverse laplace to solve most if it:

$$ y(t) = \frac{1}{2}e^t +cos(t) + 2sin(t) + L^{-1} \left[\frac{2s}{(s^2+1)^3} \right]$$

I've been trying to use convolution:

$$ L^{-1} \left[ \frac{s}{(s^2+1)^2} \right] = \frac{t}{2}sin(t) ~~~~~L^{-1} \left[ \frac{2}{(s^2+1)} \right] = 2sin(t) $$

$$ \left(2sin(t) * \frac{t}{2} sin(t)\right) = \int_0^t \frac{\tau}{2}sin(\tau) \cdot 2sin(t-\tau) d\tau$$

I just can't seem to get the two terms I'm missing. Is there a better way to approach the laplace or convolution? Or do I just need to make sure I'm doing the integral correctly?

Symbolab answer to check for correctness:

$$y=-\frac{1}{4}t^2\cos \left(t\right)+2\sin \left(t\right)+\frac{e^t}{2}+\frac{1}{4}t\sin \left(t\right)+\cos \left(t\right)$$

  • $\begingroup$ I didn't show the complete partial fractions, but that is what I got as well. If you take the inverse laplace of the first 3 terms here you will get the same as my third equation. The problem area is the term $ \frac{2s}{(s^2 +1)^3}$ $\endgroup$ – Arod529 Nov 19 '16 at 23:19

What if we write:

$$q(s) = \dfrac{2s}{(s^2+1)^3} = -\dfrac{1}{2} \dfrac{d}{ds} \dfrac{1}{(s^2+1)^2}$$

Now, we ca write:

$$\mathcal{L}^{-1} (q(s)) = \mathcal{L}^{-1} \left(-\dfrac{1}{2} \dfrac{d}{ds} \dfrac{1}{(s^2+1)^2} \right)$$

Let's do it with a Laplace Table first, items $11$ and $30$:

$$ \mathcal{L}^{-1} \left(-\dfrac{1}{2} \dfrac{d}{ds} \dfrac{1}{(s^2+1)^2} \right) = \dfrac{1}{4} t\left(\sin t - t \cos t\right)$$

We can also use the convolution method that you tried.

$$ \mathcal{L}^{-1} \{ F(s) G(s) \} = (f * g)(t)$$

We set (try other variants):

$F(s) = \dfrac{2}{(s^2+1)^2} \implies \mathcal{L}^{-1}(F(s)) = f(t) = \sin t -t \cos t \\ G(s) = \dfrac{s}{s^2+1} \implies \mathcal{L}^{-1}(G(s)) = g(t) = \cos t$

We now have:

$$\displaystyle \int_0^t f(t-v) g(v)~dv = \int_0^t f(v) g(t-v)~dv = \dfrac{1}{4} t( \sin t- t \cos t)$$

Looks like you know how to do the rest of the problem.


For the integral, we have $f(v) g(t-v) = \cos (t-v) (\sin v -v \cos v)$. When we integrate, we get:

$$\displaystyle \int \cos (t-v) (\sin v -v \cos v)~dv = \dfrac{1}{8} \left(-2 v^2 \cos t+2 v (\sin (t-2 v)+2 \sin t)-3 \cos (t-2 v)\right)$$

Evaluating this at $v = 0$ and accounting for subtraction, we get:

$$\dfrac{3}{8} \cos t$$

Evaluating this at $v = t$ and accounting for subtraction, we get:

$$\dfrac{1}{8} \left(-2 t^2 \cos t+2 t \sin t-3 \cos t\right) $$

Adding those two, gives the result above.

  • $\begingroup$ Apparently the the equivalent to item 30 in my book is not written in general form... Using the tables is much more convenient than convolution. I'm making a mistake somewhere amongst all the trig identities and integration by parts; I'll just have to keep working through it. $\endgroup$ – Arod529 Nov 20 '16 at 0:53
  • $\begingroup$ After doing the integration $\int_0^t f(v)g(t-v) dv = cost~cosv~sinv+sint~sin^2v-v~cost~cos^2v-v~sint~sinv~cosv~dv$ I am left with two terms I can't seem to cancel: $-\frac{t~sin2t}{4}$ and $\frac{t~sint~cos2t}{4}$. They are part of the result of the 3rd and 4th inintegration terms respectively. $\endgroup$ – Arod529 Nov 20 '16 at 2:54

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