Picard's Iteration: Solving dy/dx=y^2

[heman]

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,!

 

[Hurkyl]

That one's actually pretty easy: just use separation of variables.

 

[heman]

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

 

[heman]

Is this really penetrating question!

 

[HallsofIvy]

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).