Adam's question via email about Laplace Transforms

In summary: You are, of course, right. I don't know what I did. I made the same mistake you did! I got [tex]C= 5- 3/121= 605/121- 3/121= 602/121[/tex] and then, for some reason, wrote [tex]602/121- 3/121= 602/121- 3/121= 599/121[/tex]. I must have been thinking "subtract 3/121" and then wrote "minus 3/121" twice.I don't know why I didn't check that. I always check to make sure the answer satisfies the equation but I didn't do
  • #1
Prove It
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Solve the following IVP using Laplace Transforms:
$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} + 11\,y = 3\,t, \quad y\left( 0 \right) = 5$

Take the Laplace Transform of the equation:

$\displaystyle \begin{align*} s\,Y\left( s \right) - y\left( 0 \right) + 11\,Y\left( s \right) &= \frac{3}{s^2} \\
s\,Y\left( s \right) - 5 + 11\,Y\left( s \right) &= \frac{3}{s^2} \\
\left( s + 11 \right) Y\left( s \right) &= \frac{3}{s^2} + 5 \\
Y\left( s \right) &= \frac{3}{s^2 \,\left( s + 11 \right) } + \frac{5}{s + 11} \end{align*}$

Apply Partial Fractions:

$\displaystyle \begin{align*}
\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + 11} &\equiv \frac{3}{s^2 \,\left( s + 11 \right) } \\
A\,s\left( s + 11 \right) + B \left( s + 11 \right) + C\,s^2 &\equiv 3
\end{align*}$

Let $\displaystyle s = 0 \implies 11\,B = 3 \implies B = \frac{3}{11}$

Let $\displaystyle s = -11 \implies 121\,C = 3 \implies C = \frac{3}{121}$

Then $\displaystyle A\,s\left( s + 11 \right) + \frac{3}{11} \left( s + 11 \right) + \frac{3}{121}\,s^2 \equiv 3$

Let $\displaystyle s = 1$

$\displaystyle \begin{align*}
12\,A + \frac{36}{11} + \frac{3}{121} &= 3 \\
12\,A + \frac{396}{121} + \frac{3}{121} &= \frac{33}{121} \\
12\,A &= -\frac{366}{121} \\
A &= -\frac{61}{242}
\end{align*}$

So

$\displaystyle \begin{align*}
Y\left( s \right) &= -\frac{61}{242}\left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{3}{121} \left( \frac{1}{s + 11} \right) + 5 \left( \frac{1}{s + 11 } \right) \\
Y\left( s \right) &= -\frac{61}{242} \left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{608}{121} \left( \frac{1}{s + 11} \right) \\
\\
y\left( t \right) &= -\frac{61}{242} + \frac{3}{11}\,t + \frac{608}{121}\,\mathrm{e}^{-11\,t}
\end{align*}$
 
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  • #2
Let me, yet again, state my dislike for the "Laplace Transform Method"! And, in fact, Prove It made a slight arithmetic error
(easy to do with something as complicated as "Laplace Transform")
that resulted in an incorrect answer:
if [tex]y(t)= 61/242+ (3/11)t+ (608/121)e^{-11t}[/tex] then
[tex]y'(t)= 3/11- (608/11)e^{-11t}[/tex]
[tex]11y= 61/22+ 3t+ (608/11)e^{-11t}[/tex]

so [tex]y'(t)+ 11t= 6/22+ 61/22+ 3t= 67/22+ 3t[/tex], not "3t".

It is far easier just to recognize that this is a linear differential equation with constant coefficients. Its "characteristic equation" is r+ 11= 0 so r= -11. The general solution to the associated homogeneous equation is [tex]y(t)= Ce{-11t}[/tex] where C can be any constant.

Since the "non-homogeous" part is a linear polynomial, 3t, we look for a solution to the entire equation of the form y(t)= At+ B, for constants A and B. Then y'= A so the equation becomes A+ 11(At+ B)= 11At+ A+ 11B= 3t. In order for that to be true for all t, we must have both 11A= 3 and A+ 11B= 0. So A= 3/11 and then 3/11+ 11B= 0. B= -3/121.

The general solution to the entire equation is [tex]y(t)= Ce^{-11t}+ (3/11)t- 3/121[/tex]. Since we want y(0)= 5, we must have [tex]5= C+ 3/121[/tex] so [tex]C= 5- 3/121= 605/121- 3/121= 602/121.

The solution is
[tex]y(t)= (602/121)e^{-11t}+ (3/11)t- 3/121[/tex].

Check: [tex]y'(t)= -(602/11)e^{-11t}+ 3/11[/tex] while [tex]11y(t)= (602/11)e^{-11t}+ 3t- 3/11[/tex]

[tex]y'(t)+ 11t= 3t[/tex].
 
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  • #3
HallsofIvy said:
Let me, yet again, state my dislike for the "Laplace Transform Method"! And, in fact, Prove It made a slight arithmetic error
(easy to do with something as complicated as "Laplace Transform")
that resulted in an incorrect answer:
if [tex]y(t)= 61/242+ (3/11)t+ (608/121)e^{-11t}[/tex] then
[tex]y'(t)= 3/11- (608/11)e^{-11t}[/tex]
[tex]11y= 61/22+ 3t+ (608/11)e^{-11t}[/tex]

so [tex]y'(t)+ 11t= 6/22+ 61/22+ 3t= 67/22+ 3t[/tex], not "3t".

It is far easier just to recognize that this is a linear differential equation with constant coefficients. Its "characteristic equation" is r+ 11= 0 so r= -11. The general solution to the associated homogeneous equation is [tex]y(t)= Ce{-11t}[/tex] where C can be any constant.

Since the "non-homogeous" part is a linear polynomial, 3t, we look for a solution to the entire equation of the form y(t)= At+ B, for constants A and B. Then y'= A so the equation becomes A+ 11(At+ B)= 11At+ A+ 11B= 3t. In order for that to be true for all t, we must have both 11A= 3 and A+ 11B= 0. So A= 3/11 and then 3/11+ 11B= 0. B= -3/121.

The general solution to the entire equation is [tex]y(t)= Ce^{-11t}+ (3/11)t- 3/121[/tex]. Since we want y(0)= 5, we must have [tex]5= C+ 3/121[/tex] so [tex]C= 5- 3/121= 605/121- 3/121= 602/121.

The solution is
[tex]y(t)= (602/121)e^{-11t}+ (3/11)t- 3/121[/tex].

Check: [tex]y'(t)= -(602/11)e^{-11t}+ 3/11[/tex] while [tex]11y(t)= (602/11)e^{-11t}+ 3t- 3/11[/tex]

[tex]y'(t)+ 11t= 3t[/tex].

Thanks for pointing out my error Hallsofivy. Adam is one of my students, and the topic they are learning is Laplace Transforms, so he will have to use that method.
 
  • #4
HallsofIvy said:
The general solution to the entire equation is [tex]y(t)= Ce^{-11t}+ (3/11)t- 3/121[/tex]. Since we want y(0)= 5, we must have [tex]5= C+ 3/121[/tex] so [tex]C= 5- 3/121= 605/121- 3/121= 602/121.

You also have an arithmetic error. When $\displaystyle t = 5$ you end up with $\displaystyle 5 = C - \frac{3}{121} \implies C = 5 + \frac{3}{121} = \frac{608}{121}$.

I also see where my mistake was in the initial Laplace Transform. The easiest way to evaluate A is to look at the coefficient of $\displaystyle s^2 $, which gives

$\displaystyle A + \frac{3}{121} = 0 \implies A = -\frac{3}{121} $.

Thus

$\displaystyle \begin{align*}
Y\left( s \right) &= -\frac{3}{121}\left( \frac{1}{s} \right) + \frac{3}{11}\left( \frac{1}{s^2} \right) + \frac{3}{121} \left( \frac{1}{s + 11} \right) + 5 \left( \frac{1}{s + 11 } \right) \\
&= -\frac{3}{121} \left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{608}{121} \left( \frac{1}{s + 11 } \right) \\
\\
y\left( t \right) &= -\frac{3}{121} + \frac{3}{11}\,t + \frac{608}{121}\,\mathrm{e}^{-11\,t}
\end{align*}$

and this is definitely correct.
 
  • #5
Argh! Arithmetic! I never was any good at that!
 

1. What are Laplace Transforms?

Laplace Transforms are a mathematical tool used to solve differential equations. They transform a function of time into a function of complex frequency, making it easier to solve problems involving time-dependent systems.

2. How are Laplace Transforms used in science?

Laplace Transforms are used in a variety of scientific fields, including physics, engineering, and mathematics. They are particularly useful in solving problems involving electrical circuits, control systems, and heat transfer.

3. What is the difference between a Laplace Transform and a Fourier Transform?

While both Laplace Transforms and Fourier Transforms are used to transform functions from one domain to another, the main difference is that Laplace Transforms are used for functions that are defined for all positive time values, while Fourier Transforms are used for functions that are defined for all time values.

4. How do you perform a Laplace Transform?

To perform a Laplace Transform, you need to apply the Laplace operator to the function you want to transform. This involves integrating the function with respect to time and multiplying it by a complex exponential term. There are also tables and formulas that can be used to simplify the process.

5. What are the applications of Laplace Transforms?

Laplace Transforms have many applications in science and engineering, including solving differential equations, analyzing electrical circuits, and studying the behavior of control systems. They are also used in signal processing, image processing, and data compression.

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