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