Trig problem :(

[ResidentBiscuit]

Eh I can manage. It's just a summer course I needed as a prereq, I'm not too concerned.

@ne555, I haven't the slightest idea what

A_k \exp( \lambda_k )

means. Other than I've heard lambda used in various places.

[ne555]

That's the general solution.

Suppose that L{} is the differential operator \frac{ d }{ dx }. You could write the equation as
a_k L^k{ y } = 0

So the differential equation was transformed to an algebraic one.
\lambda_k are the roots of that ``polynomial''

Reverting the transformation, a solution takes the form \exp( \lambda x ) (D'Alembert)
where \exp is the exponentiation function.

The general solution is a linear combination of all the solutions.
y(x) = A_k \exp( \lambda_k x )
where the A_k coefficients are determined by the initial conditions.


Notes:
_ I'd forgot the variable in the equation in the previous post.

_ I think that I'm abusing summation notation.
The idea is that repeated indexes mean to sum over them, for all their values.
By instance the inner product \sum_{k=0}^{n-1} v^2_k = v_k v_k
In this case just take it as a linear combination.

_ If \lambda is a multiple root, then there are solutions of the form x \exp( \lambda x )

_ The method can be applied to ODE linear, of constant coefficients

[Veltas]

Yeah but we get complex roots, and so the general formula can be re-written in terms of sin and cos without imaginary numbers, and this has a specific solution to our initial values.

But as I said before this is tedious and unnecessary to anyone who knows simple harmonic motion formulae.