Inverse Laplace Transforms: ay"+by′+cy=0, y(0)=0, y′(0)=0; Y(s) = Theta(s)*F(s)

[drivera001]

Inverse Laplace Transforms: ay"+by′+cy=0, y(0)=0, y′(0)=0; Y(s) = Theta(s)*F(s)

1. Consider the initial value problem for \(\displaystyle 0\, <\, t\, <\, \infty.\)

. . . . .\(\displaystyle a\, y"\, +\, b\, y'\, +\, cy\, =\, 0\). . .\(\displaystyle y(0)\, =\, 0\). . .\(\displaystyle y'(0)\, =\, 0\)

...where a, b, c are constants and f (t ) is a known function. We can view this problem as defining a linear system, where f (t ) is the known input and the corresponding solution y (t ) is the output. Laplace transforms of the input and output functions satisfy the multiplicative relation:

. . . . .\(\displaystyle Y(s)\, =\, \Theta(s)\, F(s)\)

...where \(\displaystyle \Theta(s)\, =\, \dfrac{1}{as^2\, +\, bs\, +\, c}\) is the system transfer function.

Suppose an input f (t ) = 8t, when applied to the linear system above, produces the output:

. . . . .\(\displaystyle y(t)\, =\, 5\, (e^{-3t}\, -\, 1)\, +\, t\, (e^{-3t}\, +\, 14),\, t\, \geq\, 0\)

a. Find \(\displaystyle Y(s)\, =\, \mathcal{L}\{y(t)\}\) and \(\displaystyle F(s)\,=\, \mathcal{L}\{f(t)\}\).

b. Use your answer to part (a) to find the system transfer function, \(\displaystyle \Theta(s).\)

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2. Solve the IVP

. . . . .\(\displaystyle \dfrac{d^2 y}{dt^2}\, +\, 16\, \dfrac{dy}{dt}\, +\, 80y\, =\, 0,\, y(0)\, =\, 0,\, y'(0)\, =\, 6\)

c. The Laplace transform of the solutions is:

d. By completing the square in the denominator, we see that this is the Laplace transform of:

e. ...shifted by the rule:

f. Therefore, the solution is y=:

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For the second question, i got Y(s) equals (39s+126)/(s^2(s+3)^2) but it says its wrong. The first one i got the answers except i dont know what rule they are referring too... Any help would be greatly appreciated.

 

[HallsofIvy]

The Laplace transform of function f(t) is defined as \(\displaystyle F(s)= \int_0^\infty e^{-st}f(t)dt\). You are given that the "input function", f(t), is 8t. So its Laplace transform is \(\displaystyle F(s)= 8\int_0^\infty e^{-st}t dt\). Using "integration by parts" with \(\displaystyle u= t\), \(\displaystyle dv= 8e^{-st}\), \(\displaystyle du= dt\), \(\displaystyle v= -\frac{8}{s}e^{-st}dt\) so the integral becomes \(\displaystyle \left[-\frac{8}{s}te^{-st}\right]_0^\infty+ \frac{8}{s}\int e^{-st}dt\). The first term is 0 at both 0 and infinity and the second term is \(\displaystyle -\frac{8}{s^2}\left[e^{-st}\right]_0^\infty\). The exponential is 0 at infinity and 1 at t= 0 so that is \(\displaystyle \frac{8}{s^2}\). Is that what you got?

You are told that the "output function, y(t), \(\displaystyle 5(e^{-3t}- 1)+ t(e^{-3t}+ 14)\) so that the Laplace transform, Y(s), is \(\displaystyle \int_0^\infty 5e^{-3t- st}- 5e^{-st}+ te^{-3t- st}+ 14te^{-st}dt\). We can do those "one at a time": \(\displaystyle 5\int_0^\infty e^{t(-3-s)}dt- 5\int_0^\infty e^{-st}dt+ \int_0^\infty te^{t(-3- s)}dt+ 14\int_0^\infty te^{-st}dt\).

\(\displaystyle 5\int_0^\infty e^{t(-3- s)}dt= 5\left[\frac{-1}{s+ 3}e^{t(-3-s)}\right]_0^\infty= \frac{-5}{s+ 3}\).
\(\displaystyle -5\int_0^\infty e^{-st}dt= -5\left[\frac{-1}{s}e^{-st}\right]_0^\infty= -\frac{5}{s}\)

To do the last two, which have a "t" multiplied in, use integration by parts, the first with \(\displaystyle u= t\), \(\displaystyle dv= e^{t(-3- x)}dt\), the second with \(\displaystyle u= t\), \(\displaystyle dv= e^{-st}dt\).

As for the last part of the problem, you are told that \(\displaystyle Y(s)= \Theta(s)F(s)\) so, of course, the "transfer function", \(\displaystyle \Theta(s)\), is given by \(\displaystyle \Theta(s)= \frac{Y(s)}{F(s)}\).

 

[HallsofIvy]

The usefulness of Laplace Transforms in solving differential equations lies in the fact that the Laplace transform of a derivative of f(t) is an algebraic formula in F(s): \(\displaystyle L(df/dt)= F(s)- f(0)\). Applying that again, \(\displaystyle L(d^2f/dt^2)= s^2F(s)- sf(0)- f'(0)\).

In problem 2, we are given the differential equation \(\displaystyle \frac{d^2y}{dt^2}+ 16\frac{dy}{dt}+ 80y\) with initial condition y(0)= 0, y'(0)= 6. By the formulas above, taking the Laplace transform of that equation gives \(\displaystyle s^2 F(s)- 6+ 16sF(s)+ 80F(s)= (s^2+ 16s+ 80)F(s)- 6= 0\). So \(\displaystyle F(s)= \frac{6}{s^2+ 16s+ 80}\). To find the "y(t)" corresponding to that we need to use a "table of Laplace transforms" and to do that, typically, we need to expand F(s) in "partial fractions". So we need to reduce the denominator. "Completing the square", \(\displaystyle s^2+ 16s+ 80= s^2+ 16s+ 64+ 16= (s+ 8)^2+ 4^2\).

Now consult a table of Laplace transforms- there is one at http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf. I see that the inverse Laplace transform of \(\displaystyle \frac{a}{s^2+ a^2}\) is \(\displaystyle sin(at)\) (number 7 in the table). Further, I see (number 29) that the inverse Laplace transform of \(\displaystyle F(s- c)\) is \(\displaystyle e^{ct}f(t)\). We can write \(\displaystyle Y(s)=\frac{6}{(s- (-8))^2+ 4^2}\) as \(\displaystyle \frac{6}{4}\frac{4}{(s+ 8)+ 4^2}\) to get \(\displaystyle y(t)= \frac{3}{2}e^{-8t} sin(4t)\)

 

[HallsofIvy]

Frankly, I have never liked the "Laplace Transform Method". I have never seen a differential equation that could not be done, more simply, by easier methods. They might be useful for non-homogeneous linear equations but the only functions that can reasonably be done by Laplace Transform can typically be done with easier methods. In my opinion, the only reason "Laplace Transform Methods" are taught at all is to give engineers the feeling that they can solve such problems by "plugging into a formula and grinding"!

This problem, \(\displaystyle y''+ 16y'+ 80y= 0\), with initial conditions y(0)= 0, y'(0)= 6 is a case in point. Its "characteristic equation" is simply \(\displaystyle r^2+ 16r+ 80= 0\) which can be solved by "completing the square" as before: \(\displaystyle r^2+ 16r+ 64+ 16= (r+ 8)^2+ 16= 0\). That is the same as \(\displaystyle (r+ 8)^2= -16\) so that \(\displaystyle r+ 8= \pm 4\), \(\displaystyle r= -8\pm 4\). That tells us that the general solution to this differential equation is \(\displaystyle y(t)= e^{-8t}(A cos(4t)+ B sin(4t)\). From that, \(\displaystyle y'(t)= -8e^{-8t}(A cos(4t)+ B sin(4t))+ e^{-8t}(-4A sin(4t)+ 4B cos(4t))= -8e^{8t}((A+ 4B)cos(t)+ (B- 4A)sin(t)\).

The initial conditions are \(\displaystyle y(0)= A= 0\) and \(\displaystyle y'(0)= -8(A+ 4B)= 6\). Since A= 0, -32B= 6 so B= -6/32= -3/16. The solution to this "initial value problem" is \(\displaystyle y(t)= -\frac{3}{16}e^{-8t}sin(4t)\).

(That is NOT exactly what I got before but I would be willing to bet I made an arithmetic error doing the Laplace Transform Method!)