[Math] Solve 2nd order ordinary differential equation by Laplace transforms and convolution of their inverse functions. (5.6-39)

calculusintegrationlaplace transformordinary differential equationssolution-verification

Synopsis:

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.

Given:

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

$$y''+y=1.5sin(2t)$$

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 $Y$…

$$Y(s^2+1)=\frac{3}{2}\cdot\frac{2}{2(s^2+4)}=\frac{3}{4}\cdot\frac{2}{s^2+4}$$

$$Y=\frac{3}{4}\cdot\frac{2}{s^2+4}\cdot\frac{1}{s^2+1}$$

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…

$$=-\frac{3}{4}\cdot\frac{1}{2}\int_0^t[cos(\tau+t)-cos(3\tau-t)]d\tau$$

$$=-\frac{3}{8}\left[sin(\tau+t)-\frac{sin(3\tau-t)}{3}\right]_0^t$$

$$=-\frac{3}{8}\left[sin(2t)-\frac{sin(2t)}{3}-sin(t)+\frac{sin(-t)}{3}\right]$$

$$=-\frac{3}{8}\left[sin(2t)-\frac{sin(2t)}{3}-sin(t)-\frac{sin(t)}{3}\right]$$

$$=-\frac{3}{8}\left[\frac{2}{3}sin(2t)-\frac{4}{3}sin(t)\right]$$

$$=-\frac{1}{4}sin(2t)+\frac{1}{2}sin(t)$$

Answer in Text:

$$=-\frac{1}{2}sin(2t)+sin(t)$$

Question:

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?

Best Answer

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.