y(x) =___?
asked May 16, 2013 in Calculus Answers by anonymous

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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)

 

answered Sep 8, 2015 by Fermat Level 10 User (61,380 points)

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