Solve the initial value problem?

Question

Solve the initial value problem y'=2t(1+y), y(0)=0 by the method of successive approximations.

Answer 1

A differential equation with initial conditions is equivalent to a second (Volterra) type of integral equation with the variable upper limit of integration, that is

                                          y = 2∫₀^t t(1+y)dt  = 2∫₀^t tydt + t² 

In order to use successive approximations you have to start with the first one (y₁) which can be found by putting into the equation your zero approximation, y(0) = 0. It gives you y₁ = ^_t2.

The second approximation (y₂) can be found by plugging y₁ into the equation. Thus, we have

                                     y₂ = 2∫₀ ^t ty₁dt + t²  = t⁴/2 + t²

The same way the next (3rd) approximation

                                     y₃   = 2∫₀ ^t ty₂dt + t² = 2∫₀ ^t t(t⁴/2  + t²)dt +t² = t⁶/6 + t⁴/2 + t²

If you continue the same way to get your next approximation you wil see this is a Tylor expansion of the function

                                                       y(t) = exp(t²) - 1

This is a solution of your equation