I cannot duplicate the answer in my text although it does appear very similar. This tells me that my method is correct but I am making another kind of error -- perhaps in my integration? The following documents in good detail the steps taken to solve for this so that the root of the error can easily be found. Your input is very graciously welcomed.


Solve the following initial value problem where $y(0)=0$ and $y'(0)=0$ by applying the convolution theorem...


As a reference the convolution is defined as...

$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$

My Solution:

Begin by immediately taking the Laplace transform of entire equation and solving for $Y$...



From this we can easily see that the solution is the product of two isolated transforms and their associated functions...

$$F(s)=\frac{3}{4}\cdot\frac{2}{s^2+4}\implies f(t)=\frac{3}{4}\cdot sin(2t)$$

$$G(s)=\frac{1}{s^2+1}\implies g(t)=sin(t)$$

Now insert functions into the formula $h(t)$ and simplify by switching $t$ and $\tau$ in the sine and changing its sign...

$$h(t)=\frac{3}{4}\cdot sin(2t)*sin(t)$$

$$=\frac{3}{4}\int_0^t sin(2\tau)\cdot sin(t-\tau)d\tau$$

$$=-\frac{3}{4}\int_0^t sin(2\tau)\cdot sin(\tau-t)d\tau$$

We further simplify by applying the following trigonometric product formula identity given here as a reference...

$$sin(\alpha)\cdot sin(\beta)=\frac{1}{2}[cos(\alpha-\beta)-cos(\alpha+\beta)]$$

This yields...







Answer in Text:



It seems that I am off by a factor of 2. Please help me find the root of my discrepancy. Where did I go wrong?


1 Answer 1


You transformed $\sin(2t)$ incorrectly.

The rule says

$y(t) = \sin(at) \implies Y(s) = \frac{a}{s^{2} + a^{2}}$

But you did

$y(t) = \sin(at) \implies Y(s) = \frac{1}{a}\frac{a}{s^{2} + a^{2}}$

The rest of your working is fine though.

  • $\begingroup$ You pulled through for me again. Thank you! I just made a simple error that I should have known not to do. $\endgroup$ Feb 26, 2015 at 22:16
  • $\begingroup$ Glad to help. And it's alright mate, we all make these mistakes. $\endgroup$ Feb 26, 2015 at 23:59

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