f''(x) + tan(x) * f'(x) + (cos(x))^2 * f(x) = 0
What is f(x)?

note: f'(x) and f''(x) are the first and second derivatives of f(x)
differential equations of the form
y'' + p(x)*y' + q(x)*y = 0
can be solved with substitutions
ln(y) = ln(z)  (1/2)*integral(p(x))
so we have
(1) ln(y) = ln(z) + (1/2)*ln(cos(x))
with some work we get
z'' + q(x)*z = 0
where q(x) = Q(x)  (1/2)*p'(x)  (1/4)*p(x)^2
which gives us
z=c1*exp(sqrt(cos(x)^21))/sqrt(cos(x))  c2*exp(sqrt(cos(x)^21))/(2*sqrt(cos(x)))
using e^(ix) = cos(x) + i*sin(x)
and substituting into (1) above, we get
y = c1*cos(sin(x)) + c2*sin(sin(x))

Posted by Daniel
on 20101123 17:46:20 