# Thread: solution of a differential equation

1. ## solution of a differential equation

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

2. Originally Posted by Rapha Hi!

Find the Solution of $\displaystyle y''(t) = -k^2 y(t)$
that satisfies $\displaystyle y(0) = 1, y'(0) = 0, k \not= 0$
--------------

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

--------------

Convert it to a system of first order

--------------

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

$\displaystyle \Rightarrow u'(t) = -k^2 y(t)$
and $\displaystyle y(0) = 1, u(0) = 0, k \not= 0$

--------------

$\displaystyle y_1^{(n)}. y_2^{(n)} , n = 0,1,2,...$ 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

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

--------------

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
What, exactly, is the question? You have the solution. Do you want to solve it by that specific method, $\displaystyle y^{(n+1)}(t) = y_0 + \int^t_0 f(x, y^{(n)}(x)) dx$?

Here your "Y" is really the pair (y, u) so you will need to apply that to both terms. You are told that $\displaystyle y_0= 1$ and y'(0)= u(0)= 0 so Y (0)= (1, 0) and $\displaystyle f(x,t)= (u, -k^2y)$.

$\displaystyle (y1, u1)= (1, 0)+ (\int_0^t (0)dt, \int_0^t -k^2(1)dt)= (1, -k^2t)$

$\displaystyle (y2, u2)= (1, 0)+ (\int_0^t (-k^2 t)dt, \int_0^t(-k^2)dt)= (1- k^2t, -k^2 t)$

etc. You should be able to see pretty quickly that the first, y, component is the first n terms of the Taylor's series for cos(kt).

3. Hello HallsofIvy,

thank you very much, this really helps a lot.

Rapha

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