Question

derive the taylor series for sin2t by applying the picard method to the first-order system corresponding to the second-order IVP x"=-4x ; x(0)=0; x'(0)=2

Answer 1

You're certainly right about reducing this second order linear ODE into a system of first order linear ODEs. I'll use different symbols to avoid any confusion with subscripts as part of the iterative process. Setting $y=x$ and $z=x'$ gives $y'=x'$ and $z'=x''$. Your initial conditions can then be written as $x(0)=y(0)=0$ and $x'(0)=z(0)=2$. The system is

$\begin{cases}y(0)=0\\z(0)=2\\y'=z=f(t,y,z)\\z'=-4y=g(t,y,z)\end{cases}$

The Picard process will involve the recurrence relation

$\begin{cases}t_0=0,~y_0(t)=0,~z_0(t)=2\\\\y_{n+1}(t)=y_0+\displaystyle\int_{t_0}^tf(s,y_n(s),z_n(s))\,\mathrm ds\\\\z_{n+1}(t)=z_0+\displaystyle\int_{t_0}^tg(s,y_n(s),z_n(s))\,\mathrm ds\end{cases}$

First step:

$y_1(t)=y_0+\displaystyle\int_{t_0}^tf(s,y_0(s),z_0(s))\,\mathrm ds$
$y_1(t)=0+\displaystyle\int_0^tz_0(s)\,\mathrm ds$
$y_1(t)=2\displaystyle\int_0^t\mathrm ds$
$y_1(t)=2t$

$z_1(t)=z_0+\displaystyle\int_{t_0}^tg(s,y_0(s),z_0(s))\,\mathrm ds$
$z_1(t)=2+\displaystyle\int_0^t-4y_0(s)\,\mathrm ds$
$z_1(t)=2+\displaystyle\int_0^t0\,\mathrm ds$
$z_1(t)=2$

Second step:

$y_2(t)=y_0+\displaystyle\int_{t_0}^tf(s,y_1(s),z_1(s))\,\mathrm ds$
$y_2(t)=\displaystyle\int_0^tz_1(s)\,\mathrm ds$
$y_2(t)=2\displaystyle\int_0^t\mathrm ds$
$y_2(t)=2t$

$z_2(t)=z_0+\displaystyle\int_{t_0}^tg(s,y_1(s),z_1(s))\,\mathrm ds$
$z_2(t)=2+\displaystyle\int_0^t-4y_1(s)\,\mathrm ds$
$z_2(t)=2-\displaystyle\int_0^t4(2s)\,\mathrm ds$
$z_2(t)=2-4t^2$

Third step:

$y_3(t)=\displaystyle\int_0^t(2-4s^2)\,\mathrm ds$
$y_3(t)=2t-\dfrac43t^3$

$z_3(t)=2+\displaystyle\int_0^t-4(2s)\,\mathrm ds$
$z_3(t)=2-4t^2$

Fourth step:

$y_4(t)=\displaystyle\int_0^t(2-4s^2)\,\mathrm ds$
$y_4(t)=2t-\dfrac43t^3$

$z_4(t)=2+\displaystyle\int_0^t4\left(2s-\dfrac43s^3\right)\,\mathrm ds$
$z_4(t)=2-4t^2+\dfrac43t^4$

One more step for good measure:

$y_5(t)=\displaystyle\int_0^t\left(\dfrac43s^4+4s^2+2\right)\,\mathrm ds$
$y_5(t)=2t-\dfrac43t^3+\dfrac4{15}t^5$

We don't actually need $z_5(t)$ unless you want to continue looking for $y_6(t)$, but I don't think we'll need to. We have enough of a pattern to find $y=\lim\limits_{n\to\infty}y_n$, which is the solution to the ODE because $x=y$.

$x=2t-\dfrac43t^3+\dfrac4{15}t^5+\cdots$
$x=2t-\dfrac8{3\times2}t^3+\dfrac{32}{5\times4\times3\times2}t^5+\cdots$
$x=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}(2t)^{2n-1}}{(2n-1)!}$

which is indeed the Taylor series for $\sin2t$.