Solve the initial value problem

In summary: I should have written \frac{dy}{dx}=-\frac{2y}{15}=-\frac{y}{4}\delta(t-2),where y is the unknown solution to the equation. I apologize for the inconvenience.In summary, the student is trying to solve the initial value problem y^{''} + 2y^{'} - 15y = 4\delta(t-2), y(0) = 1, y'(0) = -1. They have identified that the equation has real roots -5 and 3 and that the solution cannot involve "sine" or "cosine". They are also having difficulty with sign errors. The student has
  • #1
jegues
1,097
3

Homework Statement



Solve the following initial value problem

[tex]y^{''} + 2y^{'} - 15y = 4\delta(t-2), \quad y(0) = 1, \quad y'(0) = -1. [/tex]

Homework Equations





The Attempt at a Solution



See figure attached for my attempt at the solution.

Does anyone see any problems? Sorry if it's kinda crunched in there.
 

Attachments

  • 08Q10.jpg
    08Q10.jpg
    26 KB · Views: 484
Physics news on Phys.org
  • #2
The characteristic equation for this d.e. is [itex]r^2+ 2r- 15= 0[/itex] which has real roots r= -5 and r= 3. The solution cannot possibly involve "sine" or "cosine". I suspect you have a sign error.
 
  • #3
HallsofIvy said:
The characteristic equation for this d.e. is [itex]r^2+ 2r- 15= 0[/itex] which has real roots r= -5 and r= 3. The solution cannot possibly involve "sine" or "cosine". I suspect you have a sign error.


Hmmm... I can't seem to spot any errors. We haven't learned anything about the characteristic equations so I can't really relate to what you're telling me.

Is this a problem that should be solved with another method instead of using laplace transforms?
 
  • #4
Hi there jegues! :smile:

What is the inverse Laplace of 1 / (s + a) ?
 
  • #5
I have always disliked Laplace transform- quite frankly there is NO problem that can be solve by Laplace transform that cannot be solved by some other method (and probably simpler). What I would do with this problem is to first get the two independent solutions to the associated homogeneous equation, y''+ 2y'- 15y= 0 by solving the characteristic equation [itex]r^2+ 2r- 15= 0[/itex], then using "variation of parameters" to solve the entire equation.

As Saladsamurai suggests, you appear to be confusing the inverse Laplace transform of [itex]1/(s+ a)[/itex] with that of [itex]1/(s^2+ a^2)[/itex].
 
  • #6
HallsofIvy said:
I have always disliked Laplace transform- quite frankly there is NO problem that can be solve by Laplace transform that cannot be solved by some other method (and probably simpler). What I would do with this problem is to first get the two independent solutions to the associated homogeneous equation, y''+ 2y'- 15y= 0 by solving the characteristic equation [itex]r^2+ 2r- 15= 0[/itex], then using "variation of parameters" to solve the entire equation.

As Saladsamurai suggests, you appear to be confusing the inverse Laplace transform of [itex]1/(s+ a)[/itex] with that of [itex]1/(s^2+ a^2)[/itex].

Whoops there's my mistake!

This clears things up now. :biggrin:

Thanks again!

EDIT: I also made a mistake in my partial fractions decomposition.
 
Last edited:

FAQ: Solve the initial value problem

What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function that satisfies a given set of conditions. These conditions typically include a specific starting point, or initial value, and a relationship that describes how the function changes over time.

How do you solve an initial value problem?

To solve an initial value problem, you need to first determine the type of differential equation it represents. Then, you can use techniques such as separation of variables, integrating factors, or substitution to find a general solution. Finally, you can use the given initial value to determine the specific solution that satisfies the conditions.

What is the difference between an initial value problem and a boundary value problem?

The main difference between an initial value problem and a boundary value problem is the type of conditions that are given. In an initial value problem, the conditions are given at a single point, usually the starting point. In a boundary value problem, the conditions are given at multiple points, typically at the boundaries of the domain.

Can all initial value problems be solved analytically?

No, not all initial value problems can be solved analytically. Some problems may have complex or nonlinear relationships that do not have an explicit solution. In these cases, numerical methods can be used to approximate a solution.

What are some real-world applications of initial value problems?

Initial value problems have many applications in science and engineering, including modeling population growth, predicting the weather, and analyzing electrical circuits. They are also used in economics, biology, and other fields to understand and predict dynamic systems.

Similar threads

How do we solve the ODE for the initial value problem with Burger's equation?
Replies
5
Views
2K
Solve the initial value problem
Replies
1
Views
1K
How to get general solution via Green's function?
Replies
1
Views
1K
Solving initial value problem using Laplace transform
Replies
1
Views
737
Two Solutions for y' = 5x^3(y-1)^1/5 with Initial Condition y(0)=1
Replies
11
Views
1K
Differential Equation Initial Value Problem
Replies
15
Views
2K
Differential Equations, solve the following: y^(4) - y'' - 2y' +2y = 0
Replies
6
Views
3K
Solve Initial Value Problem and Determining Interval
Replies
1
Views
2K
Is d'Alembert's Formula Correct for Neumann Boundary Conditions in PDEs?
Replies
4
Views
2K
Initial value problem question
Replies
15
Views
2K

Hot Threads

Recent Insights

Back
Top