Solution to Differential Equation with Limit Boundary Condition

[a1234]

Homework Statement

A second order differential equation was found to have a solution, y_2 (provided below). Apply the boundary conditions y(1) = 1 and lim of y as y approaches 0 = 0 to find the unique solution.

Relevant Equations

Original differential equation and obtained solution provided in the box below.

The original differential equation is:

My solution is below, where C and D are constants. I have verified that it satisfies the original DE.

When I apply the first boundary condition, I obtain that

, but I'm unsure where to go from there to apply the second boundary condition. I know that I should try to choose C such that the undefined contributions from both terms cancel out, but haven't found anything that does this.

 

[pasmith]

I would use $$\exp(\pm ikx^2/2) = \cos(kx^2/2) \pm i\sin(kx^2/2)$$ to write the solution in the form $$y(x) = \frac{A\cos(kx^2/2) + B\sin(kx^2/2)}{\sqrt{x}}.$$

 

[Mark44]

Homework Statement:: A second order differential equation was found to have a solution, y_2 (provided below). Apply the boundary conditions y(1) = 1 and lim of y as y approaches 0 = 0 to find the unique solution.
Relevant Equations:: Original differential equation and obtained solution provided in the box below.

The original differential equation is:

View attachment 314495
My solution is below, where C and D are constants. I have verified that it satisfies the original DE.
View attachment 314494

When I apply the first boundary condition, I obtain that View attachment 314496, but I'm unsure where to go from there to apply the second boundary condition. I know that I should try to choose C such that the undefined contributions from both terms cancel out, but haven't found anything that does this.

The right side of the equation for D simplifies to $$D = e^{-ik/2}(1 + \frac C {2ik})$$
I would substitute the above for D into the solution you found, and take the limit as x approaches 0.

 

[a1234]

I was able to figure out this problem.