It is easy to extend the idea of Picard iteration to higher order problems.  For illustration purposes we mention second order D.E.'s.
    
Theorem 3 (Picard Iteration for Second Order D. E.'s).  Given the second order initial value problem  
        ,   with  
(10)    
        .  
The solution to the I.V.P in (10) is found by constructing recursively a sequence of functions
        ,   and  
(11)
            for   .  
Then the solution to (10) is given by the limit:

(12)        .
Proof.

Begin by reformulating (10) as an equivalent integral equation.  If   is the solution of (10), then  

(13)          

The Fundamental Theorem of Calculus is used to integrate the left side of (13), and the result after rearranging terms is;

(14)        

Now integrate (14) one more time and get

        
    then
        

which can be further simplified to produce equivalent integral equation

(15)        .  

    If we use equation (15) and input a formula for   and its derivative    in the integrand  ,  then the function    on the left side is considered output.  
    Start the iteration with the initial function

          

then define the next function    as follows

        .  

Next    is used to construct    as follows  

        .  

The process is repeated, and once    has been obtained, the next function is given recursively by  

(16)        .

    In a fashion similar to the first order case we must take the limit as    in (16).  Assume that the limit (12) exists, then and and we write  

        .  

If there is "no problem" when taking limits on the right side then we might expect the following

            

This is the "intuitive proof"  of equation (15).                                  Q.E.D.

For More Proof.  
    
    A more rigorous proof based on of uniform convergence can be found in the article:

James Fabrey, Picard's Theorem (in Classroom Notes),The American Mathematical Monthly, Vol. 79, No. 9. (Nov., 1972), pp. 1020-1023, Jstor.  

(c) John H. Mathews 2005