# Solving Laplace Transform y''-6y'+13y=0 | Help Needed

• andrewdavid
In summary, the Laplace inverse of the given function is -3 e^{3t} \cos(2t) + \frac{-9}{2} e^{3t} \sin (2t). This can be found by completing the square and using the inverse Laplace transforms from the Laplace Table.
andrewdavid
I have this laplace transform that I need to solve: y''-6y'+13y=0 y'(0)=2 y(0)=-3

I figured out my Y(s)=(-3s+20)/(s^2-6s+12). All I need to do is take the inverse laplace of this but I can't figure it. I know I need to split it into two fractions, but after that I'm lost. I'd appreciate any help.

Yes Mr Beagss, it's right.

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Try doing complete the square on the bottom... see what happens

Ok I got -3s/((s-3)^2)+4 + 20/((s-3)^2)+4) after completing the square and splitting up the fraction.

Ok now look at your Laplace Table of inverse and convert them.

For example

$$e^{at} \sin (bt) = \frac{b}{(s-a)^{2} + b^{2}}$$

$$10 \frac{2}{(s-3)^{2} + (2)^{2}} = 10 e^{3t} \sin (2t)$$

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For the other Laplace inverse is:

$$e^{at} \cos (bt) = \frac{s-a}{(s-a)^{2} + b^{2}}$$

$$\frac{-3s}{(s-3)^{2} + 4} = \frac{-3s + 9 - 9}{(s-3)^{2} + 4}$$

thus

$$\frac{-3s + 9 - 9}{(s-3)^{2} + 4} = \frac{-3(s - 3) - 9}{(s-3)^{2} + 4}$$

$$\frac{-3(s - 3) - 9}{(s-3)^{2} + 4} = \frac{-3(s - 3)}{(s-3)^{2} + 4} + \frac{-9}{(s-3)^{2} + 4}$$

and finally

$$\frac{-3(s - 3)}{(s-3)^{2} + 4} = -3 e^{3t} \cos(2t)$$

$$\frac{-9}{2} \frac{2}{(s-3)^{2} + (2)^{2}} = \frac{-9}{2} e^{3t} \sin (2t)$$

so for the end Laplace inverse of

$$\frac{-3s}{(s-3)^{2} + 4} = -3 e^{3t} \cos(2t) + \frac{-9}{2} e^{3t} \sin (2t)$$

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## 1. What is the purpose of solving a Laplace Transform?

The Laplace Transform is a mathematical tool used to solve differential equations, which are equations that involve a function and its derivatives. The purpose of solving a Laplace Transform is to transform the differential equation from the time domain to the frequency domain, making it easier to solve using algebraic methods.

## 2. How do you solve a Laplace Transform equation?

To solve a Laplace Transform equation, you first need to take the Laplace Transform of both sides of the equation. This will transform the differential equation into an algebraic equation in the frequency domain. Then, you can use algebraic methods to solve for the transformed function. Finally, you can take the inverse Laplace Transform to find the solution in the time domain.

## 3. What is the Laplace Transform of a derivative?

The Laplace Transform of a derivative is s times the Laplace Transform of the function, where s is the complex frequency variable. In other words, if the Laplace Transform of a function is F(s), then the Laplace Transform of its derivative is sF(s).

## 4. What are the initial and final value theorems for Laplace Transforms?

The initial value theorem states that the initial value of a function in the time domain can be found by taking the limit of the Laplace Transform as s approaches infinity. The final value theorem states that the final value of a function in the time domain can be found by taking the limit of the Laplace Transform as s approaches zero.

## 5. How is a Laplace Transform used in real-life applications?

Laplace Transforms are used in various fields of science and engineering, such as physics, electrical engineering, and control systems. They are particularly useful in analyzing systems with complex inputs and outputs, as well as in solving differential equations that arise in real-life problems. Some examples of real-life applications include analyzing the behavior of electric circuits, solving differential equations in control systems, and modeling the response of mechanical systems to external forces.

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