# Laplace Transform Of Derivatives And Integra Ls Of Expressions With E Homework

## Homework 6 collaboration area

Question from [[User:AP

Section 6.1, problem 1 and 2, I solved with Integration by Parts.

Is this the least time consuming way to solve them? What is the best way?

Remark from Steve Bell: Feel free to use a table of integrals for any tricky integrals. (For example, the integral of exp(at)sin(bt) requires two integration by parts and then some algebra to compute. After you do that once as a freshman, you don't have to do it again. You can use a table.)

Also, 6.1, #5, when I solve with integration by parts, I get a solution that always requires another integration by Parts so I am not getting anywhere. Can anyone help on this?

I used integration by parts for 1,2, and 5.  They weren't too terrible.  Obviously, these transforms are covered in the table but going through the motions to transform them was the point.  You can check your answers using the table.  For #5, I broke sinh(t) into its exponential components and then you get two seperate integrals. you should have one transform of e3t and then a transform of et.  You should get F(s) = .5/(s-3) - .5/(s-1). You get common denominators and this becomes:

1/(s-3)(s-1)  = 1/(S^2-4S+3)  you then add 1 and subtract 1 on the denominator to complete the square.

Response from AP

Thanks Mickey!

I am sort of stuck on Lesson 19 #26:

First of all, are the instructions correct when they say "...if L(F) equals:" --- Is it F or f(t)?

Remark from Steve Bell: Yes, that's a typo. It should read, "Find f(t) if L[f(t)] equals:" Back to student:

Secondly, I am not sure how to proceed with this. Thm 3 says the inverse Laplace is 1/s times the F(s) function. Therefore I can factor out a 1/s^3 to get : (1/s^3)*(1/(s^2-1)) ...Not sure where to go from here, or if my approach is wrong...

Thanks - Mac

I think you need to factor out s^2 to get (1/(s^2)*(s^2-1)), also question says inverse transform by integral which is done in example 3. If you will integrate inverse transform of (1/(s^2-1)) i.e., sinh t twice you will get inverse transform of (1/(s^4-s^2)) i.e., sinh t - t. (I am not sure, if I am right)

-- Kunal

I did the same thing and got the same answer. After the first integration of sinh(t) you get cosh(t) -1 and integrate that to find the solution of sinh(t )- t.

Question from T. Roe 21:28, 6 October 2013 (UTC):

I have a question on Lesson 19 #19. I have set up the equations as follows:

$\mathcal{L}(f')=s\mathcal{L}(f)-f(0)$

f(0) = 0

so

$\mathcal{L}(f)=\mathcal{L}(f')/s.$

Using the information from class I get:

$\mathcal{L}(f')=\frac{2\omega}{(s^2+4\omega^2)}$

and

$\mathcal{L}(f)=\frac{2\omega}{s(s^2+4\omega^2)},$

but the book as well as mathematica output a solution of

$\mathcal{L}(f)=\frac{2\omega^2}{s(s^2+4\omega^2)}.$

Can someone explain to me where the extra ω is coming from?

I started by changing sin2(wt)  into 1/2-cos(2wt)/2  I took f'(t) to be wsin(2wt) with f'=0.  f''(t) = 2w2 * cos(2wt)

Then, I took L[f(t)] to get s2F=2w2(-2F + L[1]).  The trick was to recognize that if f(t) = 1/2-cos(2wt)/2 you can rearrange and find cos(2wt) = -2*f(t) + 1.  Then when you set the right side of the equation of the Laplace transform of the second derivative to be 2w2 * cos(2wt) you can substitute and make the right side 2w2(-2*f(t) + 1) which transforms to 2w2(-2F + 1/s). Set is equal to your earlier obtained s2F

(s2 + 4w2)F = 2w2/s

Response from Seth Beckley beckleys

I believe you can also solve this problem without the conversion to 1/2-cos(2wt)/2.  Using the Chain Rule, find f'(t) to be 2w*sinwt*coswt.  (I think the w on the left side of the derivative was left out when we derived it in class, which would explain the missing w, T. Roe.) We then can find the Laplace transform of f'(t) to be:

$\mathcal{L}(f')=\frac{2\omega^2}{(s^2+4\omega^2)}$

Recognizing that the Laplace transform of f'(t) is also equal to s*F(s) - f(0), and that f(0) = sin^2(0) = 0, we can set both forms of the Laplace transform of f'(t) equal to each other to get:

$s\mathcal{}F(s)=\frac{2\omega^2}{(s^2+4\omega^2)}$

We can then solve for F(s) to find:

$F(s)=\frac{2\omega^2}{s(s^2+4\omega^2)}$

Question 13 on pg. 223: I think the answer of the book is not correct >> 2[1+u(t-�pi)]sin3t. Instead of "+" between 1 and u(t-pi) it should be "-". Am I right? Thanks!

I made the same mistake the fist time through, but I checked the answer in Maple so I reworked it and found my mistake.  When you use the triginometric difference formula, you will have one term vanish from sin (3pi) and the other has a cos(3*pi).  This is a (-1) and it multiplies times the -2 * u(t-pi) to make it positive.  --Mick

Question from Tlouvar

On Lesson 19 #26, building on above:  I understand factoring out the s^2, but I'm confused why you need to integrate twice.  Maybe someone can explain that part a little better.  Thanks, Tim.

Tlouvar,

You need to "differentiate" twice.  The method to solving these problems is to get a funtion on the right that looks like the original function so you can substitute in an f(t), but it is derivative of some order.  With exponentials, this can usually be accomplished with one differentiation, however, with triginometric functions, it takes two differentiations to produce a function that looks like the original.  You need to get a function on the right you can sub in f(t) for so that when you perform the Laplace transform you get F.   The trick, or the big fact (as Dr. Bell likes to say) is that you can do away with all the t's and come up with right and left sde of all s's.  Once you are dealing with all s's it becomes an algebra problem.  I hope this explanation is sufficient...  -Mick

Response from T. Roe 18:00, 9 October 2013 (UTC) Thanks Seth! That is exactly what I was doing wrong. I did not even think to double check the derivative from class. Such a quick fix.

Question 25 on pg 223: The system is equal to t for 0<t<1, this is  t*[1-u(t-1)] so it is a ramp, and it is zero everywhere else. So, if it is 0 for t>1 do we need to compute something else here? The back of the book gives you an answer for t > 1, but I don't understand why.  - G.Noriega

It is not a ramp because there is a second derivative in there.  the answer in the back is what I get, and from 0 to 1 the function is t-sin(t).  Then for t > 1 you add in [cos(t-1} + sin(t-1) -t)

From Mnestero:

On Prob 10 of 6.3, on the switch u(t-2)sinh((t-2)+2), I took the +2, knowing that sinh = (e^t - e^-t)/2, and just factored out an e^2 so that e^2*u(t-2)sinh(t-2). Is this correct, or am I way off?

You should start with u(t)*sinh(t) - u(t-2)*sinh(t).  The first part is easy to transform, however to use s-shifting you are correct to add and subtract two.  Your way won't work though because you will have e(t-2+2) - e-(t-2+2) / 2 and then what you have is an e2 in the left term and an e-2 in the right term because you need the e-t+2 term to keep your sinh(t-2).

Also, it will be much more difficult to solve this problem with exponentials because you get an e-t and you will have a hard time getting the t-2 you need in the exponential.

When you transform the sinh(t) you get 1/s2-1. Use the difference identity to turn sinh((t-2)+2) into sinh and cosh expressions.  You will then transform u(t-2)*cosh(t-2)*sinh(2) and u(t-2)sinh(t-2)cosh(2).   sinh(2) and cosh(2) are constants and you should end up with something like F(s) = (1 - e-2s( s*sinh(2) + cosh(2)))/(s2-1)

From Mnestero:

Ok, that makes more sense. I did not realize that the hyperbolic sine and cosine functions had similar difference identities to the regular sine and cosine functions. For anyone who is also hairy on the sinh and cosh difference identities see: http://math.uchicago.edu/~vipul/teaching-0910/153/hyperbolicfunctions.pdf

From Steve Bell:

Find some hints about p. 223: 39. at the very bottom of

[Aftermath of Lesson21]

From Eun Young:

Example 4 (P.222) is very similar to this problem. It might be helpful.

from Jayling: any clues for Page 210 Question 23 (I hate the theoretical questions!)

Hint from Steve Bell:

Write out the definition of L[f(ct)] and then perform a change of variables, letting tau=ct. The result will just pop out after that.

From Jayling: Thanks Steve

From Jake Eppehimer:

Steve, I am getting an error message when I try to open lesson 21 aftermath, whether from that link or from the link on the homepage. Not sure if this was just me, so I thought I'd give a heads up.

Question From User:AP:

In lecture 20 @ 36:02

Should the first term in the equation be at the bottom have an S^3 not S^2?

If we start with U(t-3)(t-3), then say the laplace transform of U(t-3)=e^-3s/s, then we say (t-3)=f(t-3)=t, the the laplace transform of t is 1/S^2. Finally multiplying the two Laplace transforms gives ((e^-3s)/s)(1/S^2)=((e^-3s)/s^3).

Also, why does (t-3)=f(t-3)=t?

From Eun Young:

In general, L(fg) is not equal to L(f)L(g). So, L( u(t-3)f(t-3)) is not equal to L(u(t-3))L(f(t-3)).

In the example, we want to find L(u(t-3)t) . We know that $L( u(t-3) f(t-3)) = e^{-3s} F(s)$ where L(f(t)) = F(s). So, we need to make u(t-3)t in the form of u(t-3)f(t-3).

u(t-3) t = u(t-3) ( t-3 +3 ) = u(t-3)(t-3) + u(t-3) 3. Look at the first term u(t-3) (t-3) = u(t-3) f(t-3). So, f(t-3) = t-3. Hence, f(t) = t . $L(u(t-3)f(t-3)) = e^{-3s} F(s)$ where F(s)= L(f(t)) = L(t) = 1/s^2. Therefore, $L(u(t-3)f(t-3)) = e^{-3s}/s^2$.

Back to MA527, Fall 2013

This section is a continuation of our development of the Laplace Transform in .

Let be continuous for , and of exponential order. Then

,

where .

If    are of exponential order, then

.

Show that  .

Solution.

If we let  , then   and  ,  and is to be determined.  Because   ,  we have  ,  and Theorem 12.13 implies that

Thus, we have  ,  and    from which it follows that

.

Let be continuous for , and of exponential order and , then

.

Show that  ,    Show that  .

Solution.

Part (a). Use the fact that  ,  and apply Theorem 12.14, with     and  ,
then obtain

.

Part (b).  Now we can use this first result as a fact,  .  This time we apply Theorem 12.14, with     and  ,
and obtain

.

One of the main uses of the Laplace transform is its role in the solution of differential equations.  The utility of the Laplace transform lies in the fact that the transform of the derivative corresponds to multiplication of the transform by and then the subtraction of .  This permits us to replace the calculus operation of differentiation with simple algebraic operations on transforms.

This idea is used to develop a method for solving linear differential equations with constant coefficients.  Let's consider the initial value problem

,

with initial conditions   and .  We can use the linearity property of the Laplace transform to obtain

.

If we let and and apply Theorem 12.13 and Corollary 12.1 then we have

.

this in turn can be rearranged to obtain the form

.

The Laplace transform of the solution is easily found to be

.

For many physical problems involving mechanical systems and electrical circuits, the transform is known, and the inverse of   can easily be computed.  This process is referred to as operational calculus and has the advantage of changing problems in differential equations into problems in algebra.  Then the solution obtained will satisfy the specific initial conditions.

Solve the initial value problem

A graph of the solution.

Solution.

The right side of the differential equation is , so we have .  The initial conditions yield    and Equation (12.30)

becomes    which simplifies, and we get

.

Solving we get  .

We then solve with the help of Table 12.2 to compute

Solve the initial value problem

A graph of the solution.

Solution.

As in Example 12.15, the right side of the differential equation is , so we have .  The initial conditions yield    and ,  and Equation (12.30) becomes  ,  which can be rewritten as    which simplifies, and we get  .

This time we use Equation (12.31) and obtain  , which simplifies, and we get  .  Now use the partial fraction expansion    to get the solution