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-\frac43t^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-\frac43t^3`


`z_4(t)=2+\displaystyle\int_0^t4\left(2s-\frac43s^3\right)\,\mathrm ds`

`z_4(t)=2-4t^2+\frac43t^4`

One more step for good measure:


`y_5(t)=\displaystyle\int_0^t\left(\frac43s^4+4s^2+2\right)\,\mathrm ds`

`y_5(t)=2t-\frac43t^3+\frac4{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-\frac43t^3+\frac4{15}t^5+\cdots`

`x=2t-\frac8{3*2}t^3+(32)/(5*4*3*2)t^5+\cdots`

`x=\displaystyle\sum_(n=1)^\infty((-1)^(n+1)(2t)^(2n-1))/((2n-1)!)`

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