Laplace Transform F(s) = s/((s^2)+2s+10)

• krnhseya
In summary, the homework statement is to reduce the inverse laplace transform of 1/s to s/(((s+1)^2)+9). However, the student can't find a way to do this and is asking for help. Another question asks how to solve for the derivative of a function multiplied by an exponential function. The first additional question is to rewrite the function as a function of t. The second additional question asks how to find the derivative of a function multiplied by an exponential function. The main problem is that the student is new to the inverse laplace transform and is lost. The student asks for help with the main problem and is told to take the derivative of the function.
krnhseya

Homework Statement

Inverse Laplace Transform of given, F(s) = s/((s^2)+2s+10)

n/a

The Attempt at a Solution

I reduced it down to s/(((s+1)^2)+9) but I can't find the proper form of it, meaning I have to reduce is a little bit more but I can't think of a way to do this.

Another questions
1)For inverse laplace transform of 1/s...it's 1(t), which 1 is in terms of t.
If there's a constant, 3, can I write it as 3(t)?

2)Derivation of the Laplace transform of, f(t) = t(e^-2t)
How should I approach this problem?

Thanks!

For the main problem, I think you should use impartial fractions to solve it.

zeros: 0
poles: use quadratic formula for s^2+2s+10. roots might be complex numbers.

Then once you get the residue, apply inverse laplace.

1)inverse laplace transform of 1/s is F(t)=1 by F(t)=k ---> F(s)=k/s and F(t)=kt -----> F(s) = k/s^2

This stuff is new to me right now but I will try to put out some thoughts.

2) derivative of f(t)=t(e^-2t), use product rule.

Last edited:
pooface said:
For the main problem, I think you should use impartial fractions to solve it.

zeros: 0
poles: use quadratic formula for s^2+2s+10. roots might be complex numbers.

Then once you get the residue, apply inverse laplace.

1)inverse laplace transform of 1/s is F(t)=1 by F(t)=k ---> F(s)=k/s and F(t)=kt, F(s) = k/s^2

This stuff is new to me right now but I will try to put out some thoughts.

2) derivative of f(t)=t(e^-2t), use product rule.

i am lost 1) and 2)...
how you get F(s) = k/s^2?
isnt inverse of laplace transform of (1/s) 1?
so since i have k=-3, which is a constant, i should just write -3?
Why does book have inverse laplace transform of 1/s as 1(t) instead of just 1?

Last edited:
You shouldn't use partial fractions, the method you proposed is indeed the right approach. However in studying Laplace transforms you certainly have encountered the following formula's:

$$L\left(sin(at)\right)=\frac{a}{s^2+a^2}$$
$$L\left(cos(at)\right)=\frac{s}{s^2+a^2}$$
$$L\left(e^{at}F(t)\right)=f(s-a)$$

Can you proceed from here by rewriting your expression:

$$f(s)=\frac{s}{(s+1)^2+9}$$

to something that fits the above formulas?

Last edited:
good one coomast. I didnt see that. I am new to this because of my control systems theory course. thanks.

EDIT: OP, K is a constant by the way.

Last edited:

The first is correct, but I wouldn't write it explicitely as a function of t, it is independent of it, The value is the constant function ranging from 0 to infinity (if your laplace integral also has this range, there are other definitions). So in short I would write it as:

$$F(t)=1 \qquad \rightarrow \qquad f(s)=\frac{1}{s}$$
$$F(t)=3 \qquad \rightarrow \qquad f(s)=\frac{3}{s}$$

For the second additional question try to see if it fits the third formula I gave and the fact that:

$$F(t)=t \qquad \rightarrow \qquad f(s)=\frac{1}{s^2}$$

That should do the job.

thank you very much. so 1) questions is answered
i need to work on main problem and 2)...
can anyone give my rough guidance to how to derive that?
i am guessing that i can't use any of those "derived formulas"
once again thank you!

for 2) you want to conduct laplace transform then take the derivative?

F(t)=k[e^(-at)] -----> f(s)= k/(s+a)

k,a are constants.

For the second additional question, there is no derivative involved here. Look at the third formula I gave, It says that the Laplace transformation of a function multiplied by an exponential function is the Laplace transform of the function and adapt the value of s by s-a. This means that you need to take the transformation of t, and change the s to s-a, which is in this case s-(-2)=s+2.

For the main one, how did you rewrite what you had in order to fit the formulas I gave?

OK, I see the problem now. Assume the Laplace transform of F(t) is f(s) then you have:
$$L\left(e^{at}F(t)\right)=f(s-a)$$
Maybe this clears it up.

Last edited:
main one...not really sure how i can break that into s^2 + a^2...

You don't need to break it up like that. I meant to say is write it as:

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

Looking now at the formula's I gave, can you recognize what this is?

For the second additional question, it is exactly the same method:

$$L\left(t\right)=\frac{1}{s^2}$$

using the third formula:

$$L\left(te^{-2t}\right)=\frac{1}{(s-(-2))^2}=\frac{1}{(s+2)^2}$$

Hope that helps in order to find the main one. Normally I'm not in the habit of posting a solution, but it seems that you have to go back to page one of your course text. I'm under the impression that it's all very strange to you. The way to solve this is by studying the basic formulas of basic functions and look on how they are derived and also to proof the (few) calculation rules involved in the Laplace transform, without going into the complex domain, that's for later and more advanced study.

I don't want to sound too harsh, please come back if you have any more questions, I'm happy to help you.

yeah lol

(e^-t)(cos3t)-(1/3)(e^-t)(sin3t)
now time to derive...

 just saw your last post...
i have a test tomorrow and we covered this last class which was on wednesday...
it's very new to me but things are making sense now...i've been working on it since yesterday ;)
thank you very much for your help!

Very good, that's the correct one. This gives me a warm feeling, you saying: oooooh...I see. That's why I'm coming to the forum :-)

Do the test well tomorrow.

best regards,
Coomast

coomast said:
Very good, that's the correct one. This gives me a warm feeling, you saying: oooooh...I see. That's why I'm coming to the forum :-)

Do the test well tomorrow.

best regards,
Coomast

yeah i am set for that homework.
time to study for my test! :)
have a great one!

1. What is the Laplace Transform of the given function?

The Laplace Transform of F(s) = s/((s^2)+2s+10) is given by F(s) = 1/(s+1) - 1/(s+3).

2. How is the Laplace Transform useful in solving differential equations?

The Laplace Transform is useful in solving differential equations because it transforms a differential equation into an algebraic equation that is easier to solve.

3. What is the significance of the 's' in the Laplace Transform?

The 's' in the Laplace Transform represents the variable in the transformed domain, which is typically used to represent time in the original function. It allows for the transformation of a function from the time domain to the frequency domain.

4. Can the Laplace Transform be used for functions with discontinuities?

Yes, the Laplace Transform can be used for functions with discontinuities, as long as the function is piecewise continuous. The transformation may involve using partial fractions to handle the discontinuities.

5. Are there any limitations to using the Laplace Transform?

Yes, the Laplace Transform may not exist for all functions, and it may not be applicable to functions with exponential growth or decay. Additionally, it may be difficult to apply the Laplace Transform to functions with complex or imaginary roots.

• Calculus and Beyond Homework Help
Replies
1
Views
49
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
871
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
778
• Calculus and Beyond Homework Help
Replies
2
Views
147
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
796