How to Solve an ODE with Constant Non-Homogeneous Term and Boundary Conditions?

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In summary: I think this might have been what I was trying to do. In summary, the equation has two solutions, one of which is 1.
  • #1
rugabug
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I can't seem to find anywhere how to find the exact solution to the equation when the non-homogeneous part is just 1 and not a function. I would very much appreciate it if anyone could give me some assistance. Also for the bounday counditions doesn't one of them need to be phi prime?
Thanks.
 
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  • #2
y''+y=1
or
y''+y-1=0
y(0)=y(1)=0

differentiate
(y''+y-1)'=0
y'''+y'=0
D(D^2+1)y=0
substitute your solution into
y''+y=1

otherwise change variable y=u+1

and 1 is a function f(x)=1 so you could just solve
y''+y=f(x)
 
  • #3
If you're given phi and the derivative of phi at 0, that's called initial conditions, since it describes the position and velocity of your particle (historically you're interested in this for physics reasons). If you're given phi at two points, that's called boundary conditions because you know the position at the beginning and at the end.

Either way you get two equation with which to solve your unknowns, so either way is fine (although for certain cases boundary conditions are a little bit weaker, but that shouldn't be a problem here)
 
  • #4
There is a crucial difference between "initial value" problems and "boundary value" problems. The existence and uniqueness of a solution to an initial value problem depends entirely upon the equation, not the initial conditions but the existence and uniqueness of a solution to a boundary value problem may depend upon the boundary conditions as well.

For example, y"+ y= 0 y(0)= A, y'(0)= B has a unique solution for all A and B. But y"+ y= 0, y(0)= 0, y([itex]\pi[/itex])= 1 has no solution while y"+ y= 0, y(0)= 0, y([itex]\pi[/itex])= 0 has an infinite number of solutions.

rugabug, the simplest way to solve your problem is to first find the solution to the associated homogeneous equation (I imagine you have already done that), then "try" an undetermined constant, A as a specific solution to the entire equation: If y= A, y'= 0 and y"= 0 so the equation becomes A= 1. Just add 1 to your general solution to the homogeneous equation.
 
  • #5
Thank you so very much for the help.
I first found the complementary solution then add it to the particular solution and I am now solving for the constants.
It's all starting to come back to me. It's been a handful of years since I took diff eq.
 

FAQ: How to Solve an ODE with Constant Non-Homogeneous Term and Boundary Conditions?

What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes how a function changes over time. It involves derivatives of the function with respect to one or more independent variables.

How do you solve an ODE?

To solve an ODE, you need to find a function that satisfies the equation. This can be done analytically, using mathematical methods such as separation of variables or substitution, or numerically, using computer algorithms.

What are boundary conditions?

Boundary conditions are values or constraints that are specified at the boundaries of a system, which help determine a unique solution to the ODE. They can be initial conditions, which are values given at the starting point of the system, or boundary value conditions, which are values given at the boundaries of the system.

Why are boundary conditions important in solving ODEs?

Boundary conditions are important because they help determine a unique solution to the ODE. Without them, the solution would be a general function and not specific to the system being studied. They also allow for the application of mathematical techniques to solve the ODE.

What are some common methods for solving ODEs?

Some common methods for solving ODEs include separation of variables, substitution, power series, and numerical methods such as Euler's method, Runge-Kutta methods, and finite difference methods. The choice of method depends on the type of ODE and the desired level of accuracy.

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