# 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!

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.

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 $y^{(0)}=y_0=0$