Originally Posted by

**Rapha** Hi!

Find the Solution of $\displaystyle y''(t) = -k^2 y(t) $

that satisfies $\displaystyle y(0) = 1, y'(0) = 0, k \not= 0$

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the equation has the obvious solution $\displaystyle y(t) = cos(kt)$

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Convert it to a system of first order

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i substituted $\displaystyle u = y' \Rightarrow u'(t) = y''(t)$

$\displaystyle \Rightarrow u'(t) = -k^2 y(t) $

and $\displaystyle y(0) = 1, u(0) = 0, k \not= 0$

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$\displaystyle y_1^{(n)}. y_2^{(n)} , n = 0,1,2,...$ are solutions found by this algorithm:

$\displaystyle y' = f(t,y), y(0) = y_0$ then

$\displaystyle y^{(0)}(t) = y_0$

$\displaystyle y^{(n+1)}(t) = y_0 + \int^t_0 f(x, y^{(n)}(x)) dx $, n=0,1,2,...

if $\displaystyle y_1^{(0)}(t) = 1$, then show that

$\displaystyle y_1^{(2n)} = \sum^n_{i=0}(-1)^i \frac{(kt)^{2i}}{(2i)!} $

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i do not to have a clue. I tried to use the algorithm and $\displaystyle y' = -k^2u_1(t) $ but i dont find the series

$\displaystyle y_1^{(2n)} = \sum^n_{i=0}(-1)^i \frac{(kt)^{2i}}{(2i)!} $

any help would be much apprectiated

Thank you!

Rapha