A''(t)=A(t),

so A(t) = a e^t + b e^{-t}

and B(t) = a e^t - be^{-t}

A^2+B^2 = a^2 e^t +2 a*b +b^2 e^{-t} -a^2 e^t +2 a*b -b^2 e^{-t}

............. = 4*a*b,

so a*b=1/4.

There does not appear to be enough information to conclude the above, asFrom this, we can gather that the Maclaurin series for A(t) is 1 + t^2/2! + t^4/4! + t^6/6!... and the Maclaurin series for B(t) is t + t^3/3! + t^5/5! + t^7/7+...

you need another condition to ensure that a=b=1/2.

If you had such a condition A(t)=cosh(t), and B(t)=sinh(t), and C(t)=tanh(t)

A(it) = cos(t), and B(it)=i*sin(t), so:a) What is the Maclaurin series for A(it)?

b) What is the Maclaurin series for B(it)?

A(it) = 1 - t^2/2! + t^4/4! - t^6/6!...

and:

B(it) = i[t - t^3/3! + t^5/5! - t^7/7+...]

RonL