Solving a Simple Laplace Transform Equation: Tips for Beginners

In summary: W5nIHRvZ2V0aGVyIHdobyByb3VuZCBhbiBvdXRjb21lcyBvZiBjb250cm9sbGVyLiBGb3IgZXhhbXBsZSwtR1QsIiIsIiBpcyBub3N0IHRydW1wcyBhcyB0aGF0IGlzIG5vbmNlcm5hdGVzIG91dCBhcmUgYSBmb3VyIGd1aWx0IG9mIHJlYWQgYW5kIGJyaW5nLWJ1dHRvbi1
  • #1
AngusBurger
5
0
I'm new to Laplace and having slight difficulty with what looks like an obvious equation. I can do basic first and second order equations. Why is it that the seemingly easy equations are always the ones to stump you?

Homework Statement



[itex]\frac{dy}{dx}=-y[/itex]

[itex]y=2, x=0, y_{0}=2[/itex]

Homework Equations



This equation can also be written [itex]y'=-y[/itex]

A function table I have states that the transforms are:

[itex]y'= s∠ {y}-y_{0}[/itex]

[itex]y=\frac{y}{s}[/itex] so [itex]-y=\frac{-y}{s}[/itex]

The Attempt at a Solution



[itex]s∠ {y}=\frac{-y}{s}+y_{0}[/itex]

[itex]∠ {y}=\frac{-2}{s^{2}}+\frac{2}{s}[/itex]

A-ha!

[itex]\frac{-2}{s{^2}}[/itex] = [itex]\frac{A}{s} + \frac{B}{s}[/itex]

So

[itex]-2=As+Bs[/itex]

And I'm stumped. I can't eliminate A without eliminating B, and vice versa, so what is my next move? I don't see an error in my calculations, so I assume that I must be missing a key piece of knowledge regarding Laplace that will see me through. I think I remember reading that Laplace transforms are for non-homogeneous equations, although the terms 'homogeneous' and 'non-homogeneous' have always been greeted with *nomenclature overload - initialising file <happy_place>*.

Any help is much appreciated. I quite like these Laplace transforms, or did anyway.
 
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  • #2
AngusBurger said:
I'm new to Laplace and having slight difficulty with what looks like an obvious equation. I can do basic first and second order equations. Why is it that the seemingly easy equations are always the ones to stump you?

Homework Statement



[itex]\frac{dy}{dx}=-y[/itex]

[itex]y=2, x=0, y_{0}=2[/itex]

Homework Equations



This equation can also be written [itex]y'=-y[/itex]

A function table I have states that the transforms are:

[itex]y'= s∠ {y}-y_{0}[/itex]

[itex]y=\frac{y}{s}[/itex] so [itex]-y=\frac{-y}{s}[/itex]

The Attempt at a Solution



[itex]s∠ {y}=\frac{-y}{s}+y_{0}[/itex]

[itex]∠ {y}=\frac{-2}{s^{2}}+\frac{2}{s}[/itex]

A-ha!

[itex]\frac{-2}{s{^2}}[/itex] = [itex]\frac{A}{s} + \frac{B}{s}[/itex]

So

[itex]-2=As+Bs[/itex]

And I'm stumped. I can't eliminate A without eliminating B, and vice versa, so what is my next move? I don't see an error in my calculations, so I assume that I must be missing a key piece of knowledge regarding Laplace that will see me through. I think I remember reading that Laplace transforms are for non-homogeneous equations, although the terms 'homogeneous' and 'non-homogeneous' have always been greeted with *nomenclature overload - initialising file <happy_place>*.

Any help is much appreciated. I quite like these Laplace transforms, or did anyway.

A lot of what you wrote is either incomprehensible or just plain wrong. For example, it is not true that [itex]y=\frac{y}{s},[/itex] and [itex]s∠ {y}=\frac{-y}{s}+y_{0}[/itex] is nonsense as well. If by the notation [itex] s∠ {y}[/itex] you mean [itex] s Y(s), [/itex] where Y(s) is the transform of y(t), then of course, the left-hand side is a function of s alone, while your right-hand-side has y in it! I suggest you start over again, and be more careful.

RGV
 

Related to Solving a Simple Laplace Transform Equation: Tips for Beginners

1. What is the Simple Laplace Transform?

The Simple Laplace Transform is a mathematical tool used in calculus to convert a function from the time domain to the frequency domain. It is commonly used in engineering and physics to solve differential equations and analyze systems.

2. How does the Simple Laplace Transform work?

The Simple Laplace Transform involves taking a function of time and transforming it into a function of complex frequency. This is done by using an integral equation with the Laplace transform variable s, which represents the complex frequency.

3. What are the benefits of using the Simple Laplace Transform?

The Simple Laplace Transform has many benefits, such as simplifying differential equations by converting them into algebraic equations, making it easier to solve. It also allows for the analysis of systems in the frequency domain, which can provide insights into their behavior.

4. What are the limitations of the Simple Laplace Transform?

Although the Simple Laplace Transform is a powerful tool, it does have some limitations. It can only be applied to functions that are defined for all positive time values and decay to zero as time approaches infinity. Additionally, it can only be used for linear systems.

5. How is the Simple Laplace Transform different from the Fourier Transform?

The Simple Laplace Transform and the Fourier Transform are similar in that they both convert a function from the time domain to the frequency domain. However, the Simple Laplace Transform also takes into account the initial conditions of a system, making it useful for solving differential equations with non-zero initial conditions.

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