Trying to solve a boundary value problem

In summary, the conversation is about solving two boundary value problems involving second order differential equations with constant coefficients. The first problem has no solution while the second one does. The person is struggling with finding a particular integral and asks for clarification. The other person explains that the complementary solution can be found by solving y''+4y=0 and the particular integral can be found by substituting a solution for the right side and comparing coefficients.
  • #1
ugetwutugiv
3
0
Trying to solve the following boundary value problems.



y'' + 4y = cos x; y(0) = 0, y(pi) = 0
y'' + 4y = sin x; y(0) = 0, y(pi) = 0




The answer key says that there's no solution to the first part, but there is a solution to the 2nd part. I'm really lost and am not sure how to go about this. I'd greatly appreciate everyone's help on this!
 
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  • #2
Were you ever taught to solve 2nd order differential equations with constant coefficients?
 
  • #3
For the most part. If it were y'' + 4y = 0, I'd know what to do. But for some reason, I'm at a mental block with this. Or is this something completely different from what you're asking?
 
  • #4
ugetwutugiv said:
For the most part. If it were y'' + 4y = 0, I'd know what to do. But for some reason, I'm at a mental block with this. Or is this something completely different from what you're asking?

The first step is to find the complementary solution by solving y''+4y=0. What will y be equal to for this?

You then find a particular integral for the right side.
 
  • #5
y = A cos 2x + B sin 2x...that would be the complementary solution. How would I go about finding a particular integral? I was going to do an integrating factor but I think that's only for first-order ODE.
 
  • #6
ugetwutugiv said:
y = A cos 2x + B sin 2x...that would be the complementary solution. How would I go about finding a particular integral? I was going to do an integrating factor but I think that's only for first-order ODE.

If the Right side is cosax,sinax, sinax+cosax, then your particular integral is Asinax+Bcosax.

To find A and B, you need to substitute back this into the differential equation and then compare coefficients.
 

Related to Trying to solve a boundary value problem

1. What is a boundary value problem?

A boundary value problem is a type of mathematical problem that involves finding a solution to a differential equation subject to specified boundary conditions. These conditions are usually given at the endpoints of the domain where the solution is sought.

2. Why are boundary value problems important in science?

Boundary value problems are important in science because they allow us to model real-world phenomena and make predictions about their behavior. They are used in many fields, including physics, engineering, and economics, to study systems that are influenced by external factors or constraints.

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

The main difference between a boundary value problem and an initial value problem is the type of conditions that are given. In a boundary value problem, the conditions are specified at the endpoints of the domain, while in an initial value problem, the conditions are given at a single point within the domain.

4. How do you solve a boundary value problem?

The process of solving a boundary value problem involves finding a solution to the differential equation that satisfies the given boundary conditions. This can be done analytically using mathematical techniques such as separation of variables or numerically using computational methods like finite difference or finite element methods.

5. What are some applications of solving boundary value problems?

Boundary value problems have many applications in science and engineering, such as predicting the behavior of physical systems like heat flow, fluid dynamics, and electrical circuits. They are also used in economics to model supply and demand and in biology to study population dynamics. Other applications include image and signal processing, control systems, and optimization problems.

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