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 \(\sin2t\).
