# Laplace transform with 2s(s^2+9) in denominator

• bmed90
In summary: K(t).In summary, the conversation discusses how to expand a function in order to find the inverse Laplace Transform. The person attempted to use partial fraction decomposition, but had trouble with a term in the denominator. Others suggest using integration or complex numbers to break down the function further. Ultimately, the conversation ends with the person still unsure of how to find the constants and the inverse Laplace Transform for the given function.
bmed90

## Homework Statement

Hi I would like to know how to expand 20/2s(s^2+9) in order to find the inverse Laplace Transform of the function K(s) to gt k(t). The (s^2+9) term in the denominator is throwing my calculations off for the constants because of the s^2 term.

## The Attempt at a Solution

I attempted to use partial fraction decomposition and expanded like so

c1/2 + c2/s + c3/(s^2+9)

from there I multiplied both sides by the common denominator, attempted to distribute and collect like terms to equate coefficients in order to find the proper system of equations to find the constants in order to take the inverse laplace of each term to find k(t) however I don't think I am expanding this problem right in the partial decomp and s^2+9 is not factorable any further. Please help!

Replace c3 with c3s, and lose the first term.

Chet

Chestermiller said:
Replace c3 with c3s, and lose the first term.

Chet

Ok, so I expanded it like this

c1/2 + c2/s + (c3(s)+c4)/(s^2+9)

got my system of eqs got my constants

c1=0
c2=-20/7
c3=20/7
c4=0

so

-20/7*(1/s) --> L^-1[ -20/7*(1/s) ] = -20/7*Us(t) from table

but I am a little unsure how to find inverse laplace for the next term

L-1[ 20/7*(s) / (s^2+9) ] where 20/7 is c3 the const found c4 was 0 so its gone.

Nevermind we got it. It's simply 20/7(s/s^2+3^2) which is 20/7cos(3t)...do you agree?

bmed90 said:

## Homework Statement

Hi I would like to know how to expand 20/2s(s^2+9) in order to find the inverse Laplace Transform of the function K(s) to gt k(t). The (s^2+9) term in the denominator is throwing my calculations off for the constants because of the s^2 term.

## The Attempt at a Solution

I attempted to use partial fraction decomposition and expanded like so

c1/2 + c2/s + c3/(s^2+9)

from there I multiplied both sides by the common denominator, attempted to distribute and collect like terms to equate coefficients in order to find the proper system of equations to find the constants in order to take the inverse laplace of each term to find k(t) however I don't think I am expanding this problem right in the partial decomp and s^2+9 is not factorable any further. Please help!

You could also do it by integrating, using the well-known fact that if ##g(s)## is the LT of ##f(t)##, then ##g(s)/s## is the LT of ##\int_0^t f(u) \, du##. So, you can get the inverse transform of ##1/(s^2+9)##, then integrate it.

Thanks for all the help, let me ask you guys this,

If you had F(s)=1/((s)^2+16)^2 could you expand it like so to find your constants

[ (c1(s)+c2) / ((s)^2+16)] + [(c3(s)+c4) / ((s)^2+16)^2]

bmed90 said:
Thanks for all the help, let me ask you guys this,

If you had F(s)=1/((s)^2+16)^2 could you expand it like so to find your constants

[ (c1(s)+c2) / ((s)^2+16)] + [(c3(s)+c4) / ((s)^2+16)^2]

If you restrict yourself to real constants, the function 1/(s^2 + 16)^2 is already in 'partial fraction' form. The only way you can break it down further is to use complex numbers, writing s^2 + 16 = (s+4*i)*(s-4*i), where i = √(-1). You would then get:
$$\frac{1}{(s^2+16)^2} = \frac{i}{256}\frac{1}{s+4i} -\frac{i}{256}\frac{1}{s-4i} -\frac{1}{64(s-4i)^2}-\frac{1}{64(s+4i)^2}$$

Ray Vickson said:
If you restrict yourself to real constants, the function 1/(s^2 + 16)^2 is already in 'partial fraction' form. The only way you can break it down further is to use complex numbers, writing s^2 + 16 = (s+4*i)*(s-4*i), where i = √(-1). You would then get:
$$\frac{1}{(s^2+16)^2} = \frac{i}{256}\frac{1}{s+4i} -\frac{i}{256}\frac{1}{s-4i} -\frac{1}{64(s-4i)^2}-\frac{1}{64(s+4i)^2}$$

Thanks for your reply Vickson, but to be honest I have no idea how to understand that as of right now...how would you recommend I tackle this problem to find the inverse laplace to get the function back to the time domain?

Does anyone know how to find the constants If you had

F(s)= 1 / ((s)^2+16)^2 ? I need to find the inverse laplace transform of F(s)

## 1. What is the Laplace transform of 2s(s^2+9) in the denominator?

The Laplace transform of 2s(s^2+9) in the denominator is equal to 2/s^3 + 18/s. This can be calculated using the standard Laplace transform formula and properties.

## 2. How do you solve for the inverse Laplace transform of 2s(s^2+9) in the denominator?

To solve for the inverse Laplace transform of 2s(s^2+9) in the denominator, we can use partial fraction decomposition to break down the expression into simpler forms. Then, we can use the inverse Laplace transform table to find the corresponding function.

## 3. What is the purpose of using the Laplace transform with 2s(s^2+9) in the denominator?

The Laplace transform is a mathematical tool used to simplify and solve differential equations. By using 2s(s^2+9) in the denominator, we can convert the original differential equation into an algebraic equation that is easier to solve.

## 4. Can the Laplace transform be used for any type of function, including 2s(s^2+9) in the denominator?

Yes, the Laplace transform can be used for any type of function, including 2s(s^2+9) in the denominator. However, it is most commonly used for functions that are piecewise continuous and have exponential order.

## 5. Are there any limitations to using the Laplace transform with 2s(s^2+9) in the denominator?

There are some limitations to using the Laplace transform, including the fact that it can only be applied to functions that are piecewise continuous and have exponential order. Additionally, some functions may have more complex Laplace transforms that are difficult to solve for the inverse transform.

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