Solve equation $y'=1-3x+y+x^2+xy$ with another initial value $y(0)=1$.
Solve: We solve this by using Newton's extraordinary method.We assume that the solution is analytic,which means it can be expanded in Taylor series.$y(0)=1$ means that
$$ y'(0)=2 $$ So $$y=1+2x+cdots$$.
So
$$y'=1-3x+(1+2x+cdots)+x^2+x(1+2x+cdots)=2+0cdot x+cdots$$
So we have
$$y=1+2x+0cdot x^{2}+cdots$$
So we have
$$ y'=1-3x+(1+2x+0cdot x^{2}+cdots)+x^2+x(1+2x+0cdot x^{2}+cdots)=2+0cdot x+3x^2+cdots $$
So we have $$ y=1+2x+0cdot x^2+x^3+cdots $$
$$ vdots $$