# Another laplace differential equation.

• TyErd
In summary, the homework statement is that L{f '(t)} = s F(s) - f(0) and L{f ''(t)} = s^2 F(s) - s f(0) - f '(0)
TyErd

## Homework Statement

question is attached

## Homework Equations

L(1) = 1 /s
L(t^n) = n!/s^(n+1)
L{f '(t)} = s F(s) - f(0)
L{f ''(t)} = s^2 F(s) - s f(0) - f '(0)

## The Attempt at a Solution

okay so i laplace transformed both sides and i got:

(s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s
Y(s) [s^2 - 5s - 6] + 5s - 30 = -84exp(-5s) / s
Y(s) = ((-exp(-5s)(5s(s-6)*exp(5s) +84)) / (s^3 - 5s^2 - 6s)

but that's so long and complicated that I am wondering if I've stuffed up because I can't even seem to find the inverse laplace of the function.

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In your first line, I think I found a problem:

(s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s

The laplace transform of the forcing function -84H(t-5), assuming H is the unit step function, that laplace transform should be different.

The table I have claims that:
$$L{f(t)U(t-a)) = e^{-as}L(F(t+a))$$

Since yout f(t) is basically 1, if you just pull out the constant -84. the laplace of which is 1/s, so the laplace of f(t+a) should be 1/(s+a) I believe.

laplace of f(t+a) isn't 1/(s+a) i think.

I'm looking at the table right now.

The laplace transform of f(t)U(t-1) is e^(-as) times the laplace transform of f(t+a).

Oh wait, I see what you mean.

okay so that would mean laplace of -84 H(t-5) = -84exp(-5s) / (s+5) yea?

That's how I interpret it.

so my working out now is: (s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s+5

therefore Y(s) = -exp(-5s)(5(s-6)(s+5)*exp(5s) +84) / (s-6)(s+1)(s+5)
making y(t) = -1/11exp(-5(t+6))*[3(4exp(11t) - 11exp(4t+35) + 7exp(55) H(t-5)+55 exp(4t+30)] which is way to long to be right.

## 1. What is a Laplace differential equation?

A Laplace differential equation is a type of differential equation that involves the Laplace operator, which is a mathematical operator used to describe the rate of change of a function over time. These equations are often used to model systems in physics, engineering, and other scientific fields.

## 2. How is a Laplace differential equation solved?

Laplace differential equations can be solved using a variety of methods, including separation of variables, variation of parameters, and the Laplace transform. The specific method used will depend on the specific equation and its initial conditions.

## 3. What is the significance of Laplace differential equations?

Laplace differential equations are important because they can be used to model many physical and natural phenomena, such as heat transfer, fluid flow, and electrical circuits. They also have many practical applications in engineering and other fields.

## 4. Are there any limitations to using Laplace differential equations?

While Laplace differential equations are powerful tools, they do have some limitations. For example, they are not always applicable to nonlinear systems and can be difficult to solve analytically for complex systems. In these cases, numerical methods may be used instead.

## 5. How do Laplace differential equations relate to other types of differential equations?

Laplace differential equations are a specific type of differential equation that is closely related to other types, such as ordinary differential equations and partial differential equations. They differ in the mathematical operator used and the types of systems they can model, but they all involve the rate of change of a function over time.

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