Picard's method of successive approximations

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Prove that, applying Picard's method to the equation y′=yy' = y, with a=0a=0, b=1b=1, we obtain:

y=1+x+x22!+…+xn−1(n−1)!+xnn!+… y = 1 + x + \frac{x^{2}}{2!} + \ldots + \frac{x^{n-1}}{(n-1)!} + \frac{x^{n}}{n!} + \ldots
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Remark.
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Picard's method of finding the integral of the equation y′=φ(x,y)y' = \varphi (x,y), which equals bb at x=ax=a, consists in successively constructing a sequence of functions y0,y1,y2,…y_{0}, y_{1}, y_{2}, \ldots by the formula yn=b+∫axφ(x,yn−1) dxy_{n} = b + \displaystyle \int_{a}^{x} \varphi (x,y_{n-1})\, dx; y0=by_{0}=b. If yn→yy_{n} \to y as n→∞n \to \infty, then yy is the desired integral.

Prove that in the preceding example (problem 5351), yn→yy_{n} \to y as n→∞n \to \infty for an arbitrary choice of the function y0y_{0}.

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Prove that, replacing the equation

y′′+p(x)y′+q(x)y=0 y'' + p(x)y' + q(x)y = 0

by the system of equations

y′=u;u′=−p(x)u−q(x)y y' = u; \qquad u' = -p(x)u - q(x)y

and applying Picard's method to solve these equations, we obtain a convergent process if the radius vector from the origin to the point xx does not pass, in the complex plane of the variable xx, through any singular point of the functions p(x)p(x) and q(x)q(x).

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Solve the equation y′−y=axyy' - y = axy by the method of successive approximations, setting y0=0y_{0}=0 and computing the quantities yny_{n} by the formula yn′−yn=axyn−1y_{n}' - y_{n} = axy_{n-1} (at x=0x=0 we must have y=Cy=C).

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In the same way (see problem 5354), solve the equation y′′+a2y=σxyy'' + a^{2}y = \sigma xy, where σ\sigma is a small parameter. Here y=0y=0, y′=ay'=a at x=0x=0. It is useful to apply the theorem: if y′′+a2y=f(x)y''+a^{2}y=f(x), then

y=C1sin⁡ax+C2cos⁡ax+1a∫0xf(t)sin⁡a(x−t) dt. y = C_{1}\sin ax + C_{2}\cos ax + \frac{1}{a}\int _{0}^{x} f(t)\sin a(x-t)\, dt.
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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.