Initial value problem and laplace transform

The definition of a square wave is not something you can "find out" - it's a mathematical concept that you should be familiar with if you're studying this topic. In summary, the conversation discusses solving an initial value problem with initial values of y(0) = y'(0) = 0. The equation is y''+3y''+2y=f(t) where f(t) is a square wave. The conversation also mentions using Laplace transforms to solve the problem and discusses the steps taken to do so. The conversation concludes with a reminder to consult a textbook for further clarification on the concept of a square wave.
  • #1
azserendipity
10
0

Homework Statement


I understand how to do initial value problems but I'm slightly stuck when the initial values are y(0) = y'(0)=0

The question is Solve:

y''+3y''+2y=f(t), y(0)=y'(0)=0 where f(t) is a square wave.


Homework Equations



[itex]\Im[/itex]{y'} =s[itex]\Im[/itex]{y}-y(0)
[itex]\Im[/itex]{y''}=s[itex]^{2}[/itex][itex]\Im[/itex]-sy'(0)-y'(0)

The Attempt at a Solution



I've gotten so far:

(s[itex]^{2}[/itex][itex]\Im[/itex](y)-sy(0)-y'(0))+3(s[itex]\Im[/itex](y)-y(0))+2[itex]\Im[/itex](y)=F(t)

[itex]\Rightarrow[/itex] (s[itex]^{2}[/itex]-y'(0)-0)+3s-y'(0)+2=F(t)

Its then when I substitute in the initial condition I get

s[itex]^{2}[/itex]+3s+2=F(t)

I'm not sure this is right because I can't then do partial fractions or the inverse of it to get the final answer.

The other thing is I don't understand how F(t) being a square wave affects it.


Any help would be greatly appreciated!
 
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  • #2
azserendipity said:

Homework Statement


I understand how to do initial value problems but I'm slightly stuck when the initial values are y(0) = y'(0)=0

The question is Solve:

y''+3y''+2y=f(t), y(0)=y'(0)=0 where f(t) is a square wave.


Homework Equations



[itex]\Im[/itex]{y'} =s[itex]\Im[/itex]{y}-y(0)
[itex]\Im[/itex]{y''}=s[itex]^{2}[/itex][itex]\Im[/itex]-sy'(0)-y'(0)

The Attempt at a Solution



I've gotten so far:

(s[itex]^{2}[/itex][itex]\Im[/itex](y)-sy(0)-y'(0))+3(s[itex]\Im[/itex](y)-y(0))+2[itex]\Im[/itex](y)=F(t)
To make things easier to type, I'm going to use the usual convention of using a capital letter to denote the Laplace transform and a lower-case letter to denote the corresponding function of time, e.g. F(s) is the Laplace transform of f(t).

You have to take the Laplace transform of both sides of the equation, so you should get
$$[s^2Y(s) - s y(0) - y'(0)] + 3[sY(s)-y(0)] + 2Y(s) = F(s)$$After substituting in the initial values, you're left with
$$(s^2+3s+2)Y(s) = F(s)$$You want to solve for Y(s) and transform back to the time domain to find y(t). In your attempt, you mysteriously dropped the Y(s).
[itex]\Rightarrow[/itex] (s[itex]^{2}[/itex]-y'(0)-0)+3s-y'(0)+2=F(t)

Its then when I substitute in the initial condition I get

s[itex]^{2}[/itex]+3s+2=F(t)

I'm not sure this is right because I can't then do partial fractions or the inverse of it to get the final answer.

The other thing is I don't understand how F(t) being a square wave affects it.


Any help would be greatly appreciated!
You need to find the Laplace transform of a square wave to be able to find Y(s).
 
  • #3
I'm guessing to work out Y(s) you need to move it to the other side of the = sign so its:


(s2+3s+2)=Y(s)?

Then to find the square wave do you then need to work it out? (if so how?) or is it just something you can find out from the internet?
 
  • #4
azserendipity said:
I'm guessing to work out Y(s) you need to move it to the other side of the = sign so its:


(s2+3s+2)=Y(s)?
That doesn't make any sense. You still need to follow the rules of algebra.

Then to find the square wave do you then need to work it out? (if so how?) or is it just something you can find out from the internet?
I suggest you consult your textbook.
 

FAQ: Initial value problem and laplace transform

What is an Initial Value Problem (IVP)?

An Initial Value Problem is a type of differential equation that involves finding a function that satisfies both an equation and a set of initial conditions. These initial conditions typically specify the value of the function at a specific point or a set of points. Solving an IVP involves finding a function that satisfies both the equation and the initial conditions.

What is the Laplace Transform?

The 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. The Laplace Transform is particularly useful for solving Initial Value Problems.

How is the Laplace Transform related to Initial Value Problems?

The Laplace Transform is used to solve Initial Value Problems by transforming the differential equation into an algebraic equation, which can then be solved using standard algebraic techniques. This technique is especially useful for solving linear differential equations with constant coefficients.

What are the advantages of using the Laplace Transform to solve Initial Value Problems?

The Laplace Transform allows for the solution of more complex differential equations that cannot be solved using traditional methods. It also simplifies the solving process by converting the differential equation into an algebraic equation, which can be solved using standard techniques. Additionally, the Laplace Transform can be used to solve Initial Value Problems with non-constant coefficients.

Are there any limitations to using the Laplace Transform for solving Initial Value Problems?

While the Laplace Transform is a powerful tool for solving Initial Value Problems, it does have some limitations. It is not effective for solving non-linear differential equations, and it may not always produce a closed-form solution. In some cases, it may also be difficult to interpret the solution in terms of the original function of time.

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