How do you solve
d^2y/dx^2 = k^2 y
for boundary conditions:
y = y0 when x =0
y = 0 when x = infinity
Any help would be much appreciated.
Dear dbbwgr54,
Use the trial solution $\displaystyle y = Ae^{mx}$.
$\displaystyle \Rightarrow\frac{dy}{dx}=Ame^{mx}$
$\displaystyle \Rightarrow\frac{d^{2}y}{dx^2}=Am^{2}e^{mx}$
Therefore the auxiliary equation will be,
$\displaystyle m^2-k^2=0$
$\displaystyle \Rightarrow m = \pm k$
Hope you can continue.
Not quite. You actually have to include both solutions, at first, and then use your boundary conditions to determine the two arbitrary constants. Remember: the solution to a second-order DE should always two arbitrary constants. So what will your general solution, with two arbitrary constants, look like?
The DE can be written as $\displaystyle \frac{d^2 y}{dx^2} - k^2 y = 0$. It is obviously homogeneous.
Now refer to your class notes and textbook on how to solve second order DE's with constant coefficients.
Alternatively, you can substitute $\displaystyle \frac{d^2 y}{dx^2} = \frac{d \left( \frac{v^2}{2}\right)}{dy}$ where $\displaystyle v = \frac{dy}{dx}$ and then integrate directly with respect to y and then solve the resulting 1st order DE.
As always, the techniques, applications and examples are found in the class notes and textbook.