Solution to Differential Equation with Limit Boundary Condition

a1234
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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:

1663934426962.png

My solution is below, where C and D are constants. I have verified that it satisfies the original DE.
1663934166996.png


When I apply the first boundary condition, I obtain that
1663934588145.png
, 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.
 
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I would use <br /> \exp(\pm ikx^2/2) = \cos(kx^2/2) \pm i\sin(kx^2/2) to write the solution in the form <br /> y(x) = \frac{A\cos(kx^2/2) + B\sin(kx^2/2)}{\sqrt{x}}.
 
a1234 said:
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.
 
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I was able to figure out this problem.
 
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