How to Solve a Differential Equation with Laplace Transform?

In summary: Please explain why you are using Laplace transforms.In summary, the conversation discusses solving a differential equation using Laplace transforms. The equation is given as y'' + 4y = x for 0<=x<π and y'' + 4y = πe^-x for π<=x, with initial conditions of y(0)=0 and y'(0)=1. The use of a Heaviside step function is mentioned, but there is disagreement on whether or not it is necessary.
  • #1
Loadme
5
0
I have a differential equation that has to be solved with Laplace. I wish someone can provide a full answer

y'' + 4y = x , 0<=x<π
y'' + 4y = πe^-x , π<=x

Initial Conditions:
y(0)=0 y'(0)=1
 
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  • #2
Sorry we do not do that here.

Start by applying Laplace tranforms to each of the equations.
 
  • #3
Learn this: http://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx
and you will know what to do :)

also your ODE can be written using the step (Heaviside) function:

y'' + 4y = x + H[x-π](πe^(-x) - x)

H[x-π] = 0 at x<π; H[x-π] = 1 at x>=π; Good Luck.
 
  • #4
Loadme said:
I have a differential equation that has to be solved with Laplace. I wish someone can provide a full answer

y'' + 4y = x , 0<=x<π
y'' + 4y = πe^-x , π<=x

Initial Conditions:
y(0)=0 y'(0)=1
No "full answer" but:
1) Solve y"+ 4y= x, 0<= x< [itex]\pi[/itex]
with initial conditions y(0)= 0, y'(0)= 1.

Evaluate the function, [itex]y_1(x)[/itex], satifying those conditions and its derivative at [itex]x= \pi[/itex] and solve
2) [itex]y''[/itex][itex]+ 4y= [/itex][itex]\pi e^{-x}[/itex]
with initial conditions [itex]y(\pi)= y_1(\pi)[/itex], [itex]y'(\pi)= y_1'(\pi)[/itex].
 
  • #5
Where I am stuck is how to transform the right part as to write it for the proper laplace transform
How I would do it(and correct me where I am wrong)
y'' + 4y = x[u(x-0)-u(x-π)] + πe^(-x)*u(x-π)

How do you apply the Heavyside? Can you explain me your technique?
 
  • #6
Why would you want to use a Heaviside step function? Please advise.

You need to transform the equation. What have you got so far?
 
  • #7
I can see no good reason to use "Laplace transform". The problem is close to trivial with regular methods (find the general solution to the associated homogeneous equation and add a particular solution found by "undetermined coefficients".
 

1. What is a Laplace differential equation?

A Laplace differential equation is a type of second-order differential equation that includes a term involving the second derivative of the dependent variable. It is commonly used in physics and engineering to model systems with damping or oscillations.

2. How is a Laplace differential equation solved?

A Laplace differential equation can be solved by taking the Laplace transform of both sides of the equation, which transforms the differential equation into an algebraic equation. The solution can then be found by applying inverse Laplace transform to the resulting equation.

3. What is the significance of the Laplace transform in solving differential equations?

The Laplace transform is a powerful tool in solving differential equations because it converts a difficult problem involving derivatives into a simpler algebraic problem. This allows for the use of algebraic techniques to solve the equation, which may be easier and more efficient than traditional methods.

4. What are some real-world applications of Laplace differential equations?

Laplace differential equations have numerous applications in physics and engineering, such as in modeling circuits with resistors, capacitors, and inductors, analyzing mechanical systems with damping, and studying heat transfer and diffusion processes. They are also used in signal processing and control systems.

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

While Laplace differential equations are useful in solving many problems, they do have some limitations. They are most effective for linear systems and may not be suitable for non-linear systems. Additionally, the transformation process can be complex and may not always result in a closed-form solution, making it difficult to interpret the results.

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