Method of successive approximations to solve $y'(t) = ty(t)+2 , y(0)=0$ By using the method of successive approximations, solve the problem:
$y'(t) = ty(t)+2 , y(0)=0$
I couldn't find what method of successive approximations is, or any examples on how it is used to solve differential equations in my text book. If you can help me, I will be glad. Thanks!
 A: you try the integral representation $$y(t) = \int_0^t \left(sy(s) + 2\right)\, ds  = 2t + \int_0^t s y\, ds.$$ the successive approximation defines an iterative solution $$y_{n+1} = 2t + \int_0^t sy_n\, ds, n = 0, 1, \dots\text{ with initial } y_0 = 0. $$
we get $$\begin{align} y_1 &= 2t,\\
y_2 &= 2t + \int_0^t2s^2 \,ds = 2t + \frac 23 t^3,\\
y_3 &= 2t + \int_0^ts\left(2s+ \frac 23 s^3\right)\, ds\\
&=2t+\frac 23 t^2+ \frac2{15}t^5\end{align}  $$
i will let you continue along the same line.
A: When one has a problem $y'(t)=F(y(t),t)$ with $y(0)=y_0$ given, then one may recast the ODE into the integral equation
$$y(t)=y_0+\int_0^t F(y(t'),t') dt'$$
For the zeroth approximation, we have $y^{(0)}(t)=y_0$.
The first approximation is obtained by using the zeroth approximation in the integral equation from which we write
$$y^{(1)}(t)=y_0+\int_0^t F(y^{(0)}(t'),t') dt'$$
We set up a recursive algorithm for which the $n^{th}$ approximation ($n^{th}$ iterate) is given by 
$$y^{(n)}(t)=y_0+\int_0^t F(y^{(n-1)}(t'),t') dt'$$.
For the problem of record, we see that $F(y(t),t)=ty(y)+2$.  Thus, we have for $n\ge 1$
$$y^{(n)}(t)=y_0+\int_0^t \left(t'y^{(n-1)}(t')+2\right) dt'= 2t + \int_0^t \left(t'y^{(n-1)}(t')\right) dt'$$
with $y^{(0)}=y_0=0$
