Weird Boundary Conditions?

[juice34]

Homework Statement

d20/de2+1=0 and the boundry condition is -d0/de(evaluated at e=+/- 1)=+/-H0(evaluated at +/-1). The final result yields 0(e)=(1/2)(1-e2)+1/H. What i don't understand is how to use this boundary condition and where the 1/H comes from.

The Attempt at a Solution

When i just have a B.C. of e=+/-1(0=0(zero)), I get 0=(1/2)(1-e2)). My teacher this semester really does not explain anything so i would appreciate your guy's help with this class. IT IS A NIGHTMARE!

 

[mighty2000]

Hello,

Thank you for your post. It appears that you are trying to solve a differential equation with a boundary condition. Let's break down the equation and boundary condition to better understand them.

The first part of the equation, d20/de2+1=0, is a second order differential equation. This means that the highest derivative in the equation is a second derivative. In order to solve this type of equation, you will need two boundary conditions.

The boundary condition given is -d0/de(evaluated at e=+/- 1)=+/-H0(evaluated at +/-1). This means that when e is equal to positive or negative 1, the first derivative of 0 should be equal to positive or negative H0, respectively. This is important because it helps to narrow down the possible solutions for the differential equation.

Now, let's look at the final result, 0(e)=(1/2)(1-e2)+1/H. This is the solution to the differential equation, where 0(e) represents the function that satisfies the equation. The term 1/H comes from the boundary condition. When you integrate the differential equation, you will get a constant of integration, which can be represented as 1/H.

I hope this helps to clarify the equation and boundary condition for you. If you need further assistance, please don't hesitate to ask. Good luck with your studies!