# Help with solving Laplace Transform problem

• Ric-Veda
In summary: And why the cos(-3) and cos(7)? And how do you suppose those two cosines are related to the original problem? Why did you introduce a ##t## in the first place?In summary, the Laplace transform of t*sin(2t)*sin(5t) cannot be solved using the given equations and trig identity. The attempt at a solution provided is incorrect and does not address the original problem.
Ric-Veda

## Homework Statement

Solve Laplace Transform L{tsin(2t)sin(5t)}

## Homework Equations

cos(bt)=s/s^2+b^2
trig identity (product identity): sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)
t^nf(t)=(-1)^nd^n/ds^nF(S)
(the template is complicated for me to use. Srry for the inconvinience)

## The Attempt at a Solution

So far I have 1/2L{cos(-3)-1/2L{cos(7)}
1/2(s/s^2+4)-1/2(s/s^2+49)

Am I right?

Ric-Veda said:

## Homework Statement

Solve Laplace Transform L{tsin(2t)sin(5t)}

## Homework Equations

cos(bt)=s/s^2+b^2
trig identity (product identity): sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)
t^nf(t)=(-1)^nd^n/ds^nF(S)
(the template is complicated for me to use. Srry for the inconvinience)

## The Attempt at a Solution

So far I have 1/2L{cos(-3)-1/2L{cos(7)}
1/2(s/s^2+4)-1/2(s/s^2+49)

Am I right?

No, not even close.

BTW: you wrote
$$(1)\hspace{1.5cm}1/2 \left(\frac{s}{s^2}+4 \right) - 1/2 \left( \frac{s}{s^2} + 49 \right),$$
which equals ##(1/2)(-45)##. If you mean
$$(2) \hspace{1.5cm}1/2 \left(\frac{s}{s^2+4} \right) - 1/2 \left( \frac{s}{s^2 + 49} \right),$$
then you need to use parentheses, like this: 1/2 s/(s^2+4), etc. (When I said your answer was wrong I assumed you meant (2), not (1).)

Ric-Veda said:

## Homework Statement

Solve Laplace Transform L{tsin(2t)sin(5t)}

## Homework Equations

cos(bt)=s/s^2+b^2
This equation makes no sense, with the varable ##t## on one side and ##s## on the other. They aren't equal.
trig identity (product identity): sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)
t^nf(t)=(-1)^nd^n/ds^nF(S)

Same comment

(the template is complicated for me to use. Srry for the inconvinience)

## The Attempt at a Solution

So far I have 1/2L{cos(-3)-1/2L{cos(7)}
What happened to the ##t## variable in that last line?

## 1. What is a Laplace Transform?

A Laplace Transform is a mathematical tool used to solve differential equations. It transforms a function of time into a function of a complex variable, making it easier to solve differential equations by converting them into algebraic equations.

## 2. How is a Laplace Transform calculated?

The Laplace Transform is calculated using an integral formula, which involves integration of the function of time multiplied by an exponential term. This integral is evaluated from 0 to infinity.

## 3. What are the properties of Laplace Transform?

The properties of Laplace Transform include linearity, time-shifting, scaling, and differentiation. These properties make it a powerful tool for solving differential equations.

## 4. When is a Laplace Transform used?

A Laplace Transform is used in various fields of science and engineering, such as control systems, signal processing, and circuit analysis. It is particularly useful for solving linear differential equations with constant coefficients.

## 5. What are the common challenges in solving Laplace Transform problems?

Some common challenges in solving Laplace Transform problems include the complexity of the integral formula, determining the appropriate limits of integration, and finding the inverse Laplace Transform. Additionally, understanding the properties of Laplace Transform and their applications can be challenging for some students.

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