So I'm trying to solve a 3rd degree polynomial $$y′=t^2y−t $$ I'm not sure about how to proceed with the second order if $t$ will just become $2t$ and $y$ will just become $2$ when you derive it.
The exact question is as follows:
Find a 3rd degree Taylor polynomial for each that will approximate the solution to a given initial value problem.
On
It doesn't say the "Taylor polynomial" is centered at $0$. Let's assume it is.
So we try to look for coefficients in
$y = a_0+a_1 t+a_2 t^2 + a_3 t^3 + O(t^4)$
near $t=0$. Using this, we get
\begin{align}
y'&=a_1+2 a_2 t + 3 a_3 t^2 + O(t^3)
\\
t^2 y - t &= - t + a_0 t^2+a_1 t^3+a_2 t^4 + a_3 t^5 + O(t^6)
\end{align}
In order for these to be identically equal, we require
\begin{align}
a_1 &= 0 ,
\\
2a_2 &= -1 ,
\\
3a_3 &= a_0 .
\end{align}
Thus, our approximation is
$$
y = a_0 -\frac{1}{2} t^2+\frac{a_0}{3} t^3 + O(t^4)
$$
If we are given an initial vaulue, we can use that for $a_0$.
If you know the initial point $t,y$, then you can compute $$ y'=t^2y-t. $$ From that you can compute the second derivative per $$ y''=t^2y'+2ty-1 $$ and subsequently the third derivative $$ y'''=t^2y''+4ty'+2y $$ With these 3 values for the derivatives you can then compose the Taylor polynomial $$ y(t+h)=y+hy'+\frac{h^2}2y''+\frac{h^3}6y''' $$