- #1

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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,!!

- Thread starter heman
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- #1

- 349

- 0

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,!!

- #2

Hurkyl

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

- #3

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

- #4

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Is this really penetrating question!

- #5

HallsofIvy

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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(x_{0})= y_{0}, imagine that we know y as a function of x and integrate both sides:

[tex]y(x)= y_0+ \int_{x_0}^x f(t,y(t))dt[/tex]

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'= y^{2}, y(0)= 1. That converts to the integral equation [itex]y(x)= 1+ \int_0^x (y(t))^2 dt[/itex].

Taking y(t)= 1 we get the new solution

[tex]y(x)= 1+ \int_0^x (1)^2 dt= 1+ x[/tex].

Taking y(t)= 1+ t, we get

[tex]y(x)= 1+ \int_0^x(1+ t)^2 dt= 1+ x+ x^2+ (1/3)x^3[/tex].

Continue until you think you see a pattern (or until you are exhausted).

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

Given the intial value problem, y'= f(x,y), y(x

[tex]y(x)= y_0+ \int_{x_0}^x f(t,y(t))dt[/tex]

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'= y

Taking y(t)= 1 we get the new solution

[tex]y(x)= 1+ \int_0^x (1)^2 dt= 1+ x[/tex].

Taking y(t)= 1+ t, we get

[tex]y(x)= 1+ \int_0^x(1+ t)^2 dt= 1+ x+ x^2+ (1/3)x^3[/tex].

Continue until you think you see a pattern (or until you are exhausted).

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