Solve: y^n=k^2y y=(1)=y(-1)=A
Assuming that y^n is the nth differential coefft of y(x), i.e. y^n = y^(n) = d^n(y)/dx^n, then what we have is
y^(n) = k^2y
Assume a solution of the form, y = C.e^(mx), then
y^(n) = C.(m^n).e^(mx)
The auxiliary eqn then is,
m^n – k^2 = 0
m^n = k^2
m = k^(2/n) (n odd), or m = +/- k^(2/n) (n even)
Therefore,
y = P.e^(øx) (n odd), or y = Q.e^(øx) + R.e^(-øx) (n even), using ø = k^(2/n)
Using the initial values y(1) = y(-1) = A,
n = odd
y(1) = A = P.e^(ø)
y(-1) = A = P.e^(-ø)
Since we cannot have e^(ø) = e^(-ø), then n = odd is not a solution.
n = even
y(1) = A = Q.e^(ø) + R.e^(-ø)
y(-1) = A = Q.e^(-ø) + R.e^(ø)
The above eqns require that Q = R
Our final solution will then be of the form,
y = B.e^(øx) + B.e^(-øx)
i.e. y = 2B.cosh(øx)
Using the initial condition,
A = 2B.cosh(ø)
B = A/2cosh(ø)
Finally,
y(x) = A. cosh(øx)/ cosh(ø), ø = k^(2/n), n = 2m, (m e N)