# Solving y''-6y+9y=2 with Laplace Transforms

• f00lishroy
In summary, the conversation discusses finding a solution for y''-6y+9y=2 with initial conditions y(0)=y'(0)=0 without using partial fractions. The suggested approach involves using a convolution or the Bromwich integral to invert Y(s) into y(t).
f00lishroy

## Homework Statement

y''-6y+9y=2 y(0)=y'(0)=0

*Note* Professor will NOT allow use of partial fractions, so please don't use it.

## Homework Equations

Laplace transform table
Y=[y'(0)+sy(0)+ay(0)+R]/[(s^2)+as+b]

## The Attempt at a Solution

Y=L(2)/(s-3)^2
L(2)=2/s
Y=(2/s)[1/(s-3)^2]
Y=2*1/[s^3-6s^2+9s]
From here I cannot figure out how to continue without using partial fractions since I can't get the roots, from which I would be able to invert and use the Laplace Transform table.

so you have
$$Y(s) = F(s)G(s)$$
with
$$F(s)=\frac{2}{s}$$
$$G(s)=\frac{1}{(s-3)^2}$$

how about consider f(t) and g(t) and using a convolution, if you don't know how to do this have a look at 1.7 in
http://www.vibrationdata.com/math/Laplace_Transforms.pdf

You could also invert Y(s) using the Bromwich integral
$$y(t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} Y(s)e^{st}\,ds$$.

## 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 the equation.

## 2. How do you solve a differential equation using Laplace Transforms?

To solve a differential equation using Laplace Transforms, you first need to take the Laplace Transform of both sides of the equation. This will transform the equation into an algebraic equation, which can then be solved for the unknown function. Once the solution is found, the inverse Laplace Transform is taken to find the solution to the original differential equation.

## 3. How do you use Laplace Transforms to solve y''-6y+9y=2?

To solve the differential equation y''-6y+9y=2 using Laplace Transforms, we first take the Laplace Transform of both sides: s^2Y(s)-6Y(s)+9Y(s)=2. Then, we solve for Y(s) by factoring out the common term Y(s): Y(s)(s^2-6+9)=2. This gives us Y(s)=2/(s^2-6+9)=2/(s^2+3). Finally, we take the inverse Laplace Transform to find the solution to the original differential equation: y(t)=2sin(√3t).

## 4. What is the advantage of using Laplace Transforms to solve differential equations?

The advantage of using Laplace Transforms to solve differential equations is that it can simplify complex equations into algebraic equations, making them easier to solve. It also allows for the use of algebraic techniques, which are often simpler and more familiar than the techniques used to solve differential equations.

## 5. Are there any limitations to using Laplace Transforms to solve differential equations?

Yes, there are some limitations to using Laplace Transforms. They can only be used to solve linear differential equations with constant coefficients. They also cannot be used to solve equations with discontinuous or discontinuous derivatives. In some cases, taking the inverse Laplace Transform may be difficult or impossible, making it challenging to find the solution to the original differential equation.

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