# Picard's Iteration

## Main Question or Discussion Point

Sorry for making another thread,but this problem is really penetrating for me!

dy/dx=y^2 with initial condition y(0)=1

I have reached upto
5/9 + 4/9((1+x) + (1+x)^4/4 + (1+x)^7/7 + ........)
And ahead of that i have no clue,!!

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Hurkyl
Staff Emeritus
Gold Member
That one's actually pretty easy: just use separation of variables.

the question has to be done by Picard's iteration,not by seperation of variables!

Is this really penetrating question!

HallsofIvy
Homework Helper
WHY do you have it in terms of (1+ x) ? Your initial value is given at x= 0.

For those of you who don't know, Picard's iteration is this:

Given the intial value problem, y'= f(x,y), y(x0)= y0, imagine that we know y as a function of x and integrate both sides:
$$y(x)= y_0+ \int_{x_0}^x f(t,y(t))dt$$
The initial value problem has a solution if and only if that integral equation has a solution. The integral equation can be thought of as a "fixed value" problem and, since Banach's fixed value theorem holds (see thread on "existance and uniqueness"), we can do it by iteration. Let Y be any function. The constant y(x)= y_0 works nicely. Plug that into the righthand side and integrate. Use the value of y(x) you get to repeat.

In this case, the initial value problem is y'= y2, y(0)= 1. That converts to the integral equation $y(x)= 1+ \int_0^x (y(t))^2 dt$.

Taking y(t)= 1 we get the new solution
$$y(x)= 1+ \int_0^x (1)^2 dt= 1+ x$$.
Taking y(t)= 1+ t, we get
$$y(x)= 1+ \int_0^x(1+ t)^2 dt= 1+ x+ x^2+ (1/3)x^3$$.
Continue until you think you see a pattern (or until you are exhausted).

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