Hello guys here is my question thank you for all the help.
I need to determine
On which integral the Picard's Method is applicable for y'=xy^2, $y(0)=0$
and need to calculate the first two iterates.
Hello guys here is my question thank you for all the help.
I need to determine
On which integral the Picard's Method is applicable for y'=xy^2, $y(0)=0$
and need to calculate the first two iterates.
I assume you want to apply the following:
Suppose $f$ and $f_y$ are continuous on some disc $D$ containing $(x_0,y_0)$. Then the initial value problem
\tag{1}\biggl\{ {y'=f(x,y) \atop y(x_0)=y_0 } has a unique solution in some interval $[x_0-\delta, x_0+\delta$] contained in $D$. Moreover, the sequence of iterates with first term $a_1(x)$ and $\tag{2} a_{n+1} (x)=y_0+\int_{x_0}^x f(s, a_n(s) ) \,ds $ converges to this solution.
Here, you make an initial guess $y=a_1(x)$ for the solution to (1), and use (2) to find a "refined" guess: plug $a_1(x)$ into the right hand side of (2), and calculate the integral. Then you have your refined guess $y=a_2(x)$. Then repeat...
In your problem $f(x,y)=xy^2$ and $y(0)=0$.
So the iteration scheme (2) becomes
$\tag{3} a_{n+1} (x)= 0+\int_{0}^x s ( a_n(s) )^2 \,ds $ The first guess should satisfy the initial condition. Let's try $a_1(x)=x$. This satisfies $a(0)=0$.
To find $a_2(x)$, use (3): $ a_2(x) = \int_0^x s (a_1(s))^2\, ds =\int_0^x s s^2\, ds ={s^4 \over4 }\biggl|_0^x={x^4\over4}. $
To find the next iteration, use (3) with $n=2$, and $a_2(s)= {x^4\over4}$. I'll leave this for you...
I'm not sure what you mean by "on which integral". If you meant "interval", I think the best you can do, without further analysis, is say "the iteration scheme gives a solution on some interval centered at 0".
Note that the iterates above are always powers of $x$ over a constant. The iterates in fact converge to 0 pointwise on any interval of the form $(-a,a)\subset[-1,1]$. In fact, it is easily seen that $y=0$ is the solution to your initial value problem.
(Note, also, we could have used $a_1=0$, and evaluating the integrals would be much easier...)