# Thread: Taylor polynomial of ODE

1. ## Taylor polynomial of ODE

dy/dt = [-pi*cos(pi*t)-2ty]/[1+t^2]

Find the fourth degree polynomial centred about 0

I know the taylor polynomial goes y(0) + hy'(0) + h^2/2y''(0) + h^3/6y'''(0) + h^4/24y''''(c)

The problem I am having is not being able to find y'''(0) I know its meant to be h^3[pi(pi^2 + 6)]/6 But i just can't seem to be able to get that answer..Can anyone help out please?

2. The first thing you're going to need to do is solve the DE...

$\frac{dy}{dt} = \frac{-\pi\cos{(\pi t)} - 2ty}{1 + t^2}$

$\frac{dy}{dt} = -\frac{\pi \cos{(\pi t)}}{1 + t^2} - \left(\frac{2t}{1 + t^2}\right)y$

$\frac{dy}{dt} + \left(\frac{2t}{1 + t^2}\right)y = -\frac{\pi \cos{(\pi t)}}{1 + t^2}$.

This is first order linear, so use the Integrating Factor method.

The integrating factor is $e^{\int{\frac{2t}{1 + t^2}\,dt}} = e^{\ln{(1 + t^2)}} = 1 + t^2$.

Multiplying through by the integrating factor gives

$(1 + t^2)\frac{dy}{dt} + 2t\,y = -\pi \cos{(\pi t)}$

$\frac{d}{dt}[(1 + t^2)y] = -\pi \cos{(\pi t)}$

$(1 + t^2)y = \int{-\pi \cos{(\pi t)}\,dt}$

$(1 + t^2)y = -\sin{(\pi t)} + C$

$y = \frac{C - \sin{(\pi t)}}{1 + t^2}$.

Now you should be able to substitute the MacLaurin series for $\sin{x}$, replacing $x$ with $\pi t$.

3. If we suppose that $y(t)$ is analytic in $t=0$ we can write...

$\displaystyle y(t) = a_{0} + a_{1}\ t + a_{2}\ t^{2} + a_{3}\ t^{3} + a_{4}\ t^{4} + ...$ (1)

... and...

$\displaystyle y^{'} (t) = a_{1} + 2\ a_{2}\ t + 3\ a_{3}\ t^{2} + 4\ a_{4}\ t^{3} + ...$ (2)

The DE becomes with a certain number of steps...

$y^{'} (t) = -\{\pi - 2\ a_{0}\ t + (\frac{\pi^{3}}{2}-2\ a_{1} + \pi)\ t^{2} + (2\ a_{0} - 2\ a_{2})\ t^{3} + ...\} \ (1-t^{2} + t^{4} - ...)$ (3)

... and now we are able from (2) and (3), setting $a_{0} = y(0)$ , to derive...

$a_{1} = -\pi$

$a_{2} = a_{0}$

$3\ a_{3} = \frac{\pi^{3}}{3} -2\ a_{1} + \pi \rightarrow a_{3} = \pi + \frac{\pi^{3}}{9}$

$4\ a_{4} = -2\ a_{2} +2\ a_{0}= 0 \rightarrow a_{4} = 0$

Kind regards

$\chi$ $\sigma$