- All categories
- Pre-Algebra Answers (11,954)
- Algebra 1 Answers (24,236)
- Algebra 2 Answers (10,059)
- Geometry Answers (4,859)
- Trigonometry Answers (2,480)
- Calculus Answers (5,562)
- Statistics Answers (2,831)
- Word Problem Answers (9,091)
- Other Math Topics (5,330)

*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)

...