PLS, some one help me find the 1st 4 iterates using Picard's iteration.

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Use Picard's iteration to find the first 4 iterates of solutions of the problem dx/dt = -tx, x(0) = 1.

If anyone can help me solve this, I'll be so grateful. I wait anxiously to get the solution.
 
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I'm finding it impossible to write the equations here, though the clue can be gotten at http://www.cse.uiuc.edu/eot/modules/ode/picard/
 
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If you have the differential equation dx/dt= f(x,t) with x(t0)= x0 then "Picard's interation", used in his proof of the existence of solutions of the initial value problem, is
x(t)= \int_{t_0}^t f(x,t)dt+ x_0
starting with x(t)= x0 and then using the resulting x(t) in the next iteration.

Here, f(x,t)= -tx and x(0)= x0= 1 so the first "iteration" is
\int_0^t -tdt+ 1[/itex]<br /> Surely you can continue from there.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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