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### Picard's Method

```
Date: 08/10/97 at 15:23:23
From: Michael Lam
Subject: Picard's Iteration method of approximating solutions to
Differential Equations

Hello,

Can you tell me about Picard's Iteration method of solving
Differential Equations?  I've looked for books in libraries but with
```

```
Date: 08/10/97 at 16:23:29
From: Doctor Anthony
Subject: Re: Picard's Iteration method of approximating solutions to
Differential Equations

Suppose you have the differential equation   dy/dx = f(x,y)  given
that y = b when x = a.

We can write   y = b + INT(a to x)[f(x,y).dx]

For a first approximation we replace the y in f(x,y) by b; for the
second approximation we replace it by the first approximation, for a
third by the second, and so on.

Example   dy/dx = x+y^2 and y = 0 when x = 0

Then   y = 0 + INT(0 to x)[x+y^2].dx

First approximation: put y = 0 in x+y^2  giving

y = INT(0 to x)[x.dx]

= 1/2 x^2

Second approximation: put y = 1/2 x^2 in  x+y^2

y = INT(0 to x)[x + 1/4 x^4].dx

=   x^2/2 + 1/20 x^5

Third approximation: put  y = x^2/2 + 1/20 x^5 for y in x+y^2

y = INT(0 to x)[x + 1/4 x^4 + 1/20 x^7 + 1/400 x^10]dx

=  x^2/2 + x^5/20 + x^8/160 + x^11/4400

and so on indefinitely.  Picard's method converts the differential
equation into an equation involving integrals, which is called an
integral equation.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/

```

```
Date: 08/10/97 at 21:16:28
From: Michael Lam
Subject: Re: Picard's Iteration method of approximating solutions to
Differential Equations

>         =  x^2/2 + x^5/20 + x^8/160 + x^11/4400

Thanks for the method!!!  Now I think I can do them...however, if I
proceed the iterations indefintely, am I gauranteed to find a
solution?
Is this solution unique?
And, if I find something which looks like a Taylor's series of some
function, can I write that function as the solution instead?

Thank you very much!!

MIchael

```

```
Date: 08/11/97 at 07:42:38
From: Doctor Anthony
Subject: Re: Picard's Iteration method of approximating solutions to
Differential Equations

As Michael Lam wrote to Dr. Math
On 08/10/97 at 21:16:28 (Eastern Time),
>>         =  x^2/2 + x^5/20 + x^8/160 + x^11/4400
>
>Thanks for the method!!!  Now I think I can do them...however, if I
>proceed the iterations indefintely, am I gauranteed to find a solution?
>Is this solution unique?
>And, if I find something which looks like a Taylor's series of some
>function, can I write that function as the solution instead?
>
>Thank you very much!!
>
>MIchael
>

Remember we are finding a series solution about some point x=a, y=b
(in
above example, about the point (0,0))  and so the number of terms you
require will depend how far from x=a you need the solution to apply.
It can
be shown that the series is convergent, and provides a solution to the
differential equation, but the number of terms you take will depend
how
rapidly the series converges, and how far from x=a you want to
operate.

Your second point, concerning a Taylor series, is perfectly acceptable
if
the  series is available.  e.g. dy/dx = x^(1/2).sin(x).  Here a Taylor
series for sin(x) is available, and could be used.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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