Question

Solve by picard's method ( until to 3-th aproximation) , y'=x+cos(y) , y(0)=1

Answer 1

We're generating a sequence of functions
`Y_n` with

`\begin{cases}Y_0(x)=y(0)\\\\Y_(n+1)(x)=y(0)+\displaystyle\int_(t=0)^(t=x)f(t,Y_n(t))\,\mathrm dt&\text{for }n\ge0\end{cases}`

where
`y'=f(x,y)`, so that the sequence
`Y_n` converges to
`y` as
`n\to\infty`.

`n=0` :

`Y_0(x)=1`

`n=1` :

`\displaystyle Y_1(x)=1+\int_0^x(t+\cos1)\,\mathrm dt=1+(\cos 1)x+\frac{x^2}2`

`n=2` :

`\displaystyle Y_2(x)=1+\int_0^x\left(t+\cos\left(1+(\cos 1)t+\frac{t^2}2\right)\right)\,\mathrm dt`

Unless you're familiar with Fresnel integrals, you won't be able to simplify this any further.

`n=3` :

`\displaystyle Y_3(x)=1+\int_0^x(t+\cos(Y_2(t)))\,\mathrm dt`

My computer takes a really long time to compute
`Y_3(x)`, and even longer to plot it, so I've ultimately omitted
`Y_3` from the plot. (I wonder now if by "until to 3-th aproximation" it's intended that you only go up to
`n=2`...)

I've attached a plot of the approximations (dashed and colored) along with a more precisely computed numerical solution (black)