I'm learning advanced ODE for the first time & got this problem in my assignment. I tried to solve this problem by two method but am confused which is right. In my book the initial condition of every problem is given in terms of $t_0=a$ & $x_0=b$ but not $x_o=f(t)$. I searched for books like G.F. Simmons, Coddington-Levinson but nowhere get any clue.

Here is the differential equation: $$\dot{x}={x+t}$$ with the initial condition $x_0(t)=cost$

My solution: 1st Method: The n-th approximation of the IVP is given by $$x_n=x_0+\int_{t_0}^tf(t,x_{n-1})dt$$...(1)
Putting n=1 & initial condition we get $$x_1=x_0+\int_{t_0}^tf(t,x_0)dt$$ or, $x_1=cost+\int_{t_0}^t(cost+t)dt$
or, $x_1=cost+(sint+\frac{t^2}{2})_{t_0}^t$
or, $x_1=cost+sint+\frac{t^2}{2}-(sint_0+\frac{t_0^2}{2})$

2nd Method:Put t=0 in the initial condition to get $x_0=cos0=1$
Hence putting n=1 from (1) we get $$x_1=1+\int_0^t(1+t)dt$$
or, $x_1=1+t+\frac{t^2}{2}$
Which is the correct 1st approximation between method 1 & method 2?

  • $\begingroup$ $x_0(t)=cost$. There's another problem where $x_0(t)=e^t$ $\endgroup$ – Curious Apr 19 '17 at 16:28
  • $\begingroup$ Ok, then we can present any function $x_0(t)=f(t)$ as long as it gives $t_0=o$, $x_0=1$ $\endgroup$ – Curious Apr 19 '17 at 16:44

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