Given an ODE [tex]x'=f(t,x)[/tex] with initial condition [tex]x(t_0)=x_0[/tex], the general process is to write the ODE as an integral equation,
[tex]x(t)=x_0+\displaystyle\int_{t_0}^tf(u,x(u))\,\mathrm du[/tex]
By setting [tex]x_0(t)=x_0[/tex] for all [tex]t[/tex], we get the following recurrence for [tex]n\ge1[/tex].
[tex]x_{n+1}(t)=x_0+\displaystyle\int_{t_0}^tf(u,x_n(u))\,\mathrm du[/tex]
From this we work towards finding a pattern for [tex]x_n[/tex] so that we can find a solution of the form [tex]x=\lim\limits_{n\to\infty}x_n[/tex].
[tex]\begin{cases}x'=x+2\\x(0)=2\end{cases}[/tex]
Write this as the integral equation,
[tex]x_{n+1}=x(0)+\displaystyle\int_{t_0}^t(x_n(u)+2)\,\mathrm du[/tex]
First step:
[tex]x_1=x(0)+\displaystyle\int_{t_0}^t(x_0(u)+2)\,\mathrm du[/tex]
[tex]x_1=2\displaystyle\int_0^t\mathrm du[/tex]
[tex]x_1=2t[/tex]
Second step:
[tex]x_2=x(0)+\displaystyle\int_{t_0}^t(x_1(u)+2)\,\mathrm du[/tex]
[tex]x_2=\displaystyle\int_0^t(2u+2)\,\mathrm du[/tex]
[tex]x_2=t^2+2t[/tex]
Third step:
[tex]x_3=x(0)+\displaystyle\int_{t_0}^t(x_2(u)+2)\,\mathrm du[/tex]
[tex]x_3=\displaystyle\int_0^t(t^2+2t+2)\,\mathrm du[/tex]
[tex]x_3=\dfrac13t^3+t^2+2t[/tex]
Fourth step:
[tex]x_4=x(0)+\displaystyle\int_{t_0}^t(x_3(u)+2)\,\mathrm du[/tex]
[tex]x_4=\displaystyle\int_0^t\left(\frac13u^3+u^2+2u+2\right)\,\mathrm du[/tex]
[tex]x_4=\dfrac1{4\times3}t^4+\dfrac13t^3+t^2+2t[/tex]
You should already start seeing a pattern. Recall that
[tex]e^t=\displaystyle\sum_{k=0}^\infty\frac{x^k}{k!}=1+t+\frac{t^2}2+\frac{t^3}{2\times3}+\frac{t^4}{2\times3\times4}+\cdots[/tex]
Multiplying this by 2 gives
[tex]2e^t=2+2t+t^2+\dfrac{t^3}3+\dfrac{t^4}{4\times3}+\cdots[/tex]
which matches the solution we have for [tex]x_4[/tex] except for that first term. So subtracting that, we find a solution of
[tex]x=2e^t-2[/tex]
with a domain of [tex]t\in\mathbb R[/tex].
Hopefully this gives some insight on how to approach the other two problems.
