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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9 Existence and Uniqueness (cont’d) Example. We give an example of the proof of the theorem. Consider the initial value problem dy = 2 t (1 + y ), y (0) = 0. dt First we ﬁnd an exact solution y = y (t), then we ﬁnd a sequence of approximate solutions yn = yn (t) via Picard’s Method. To ﬁnd the exact solution, note that this diﬀerential equation is a separable equation: dy = 2 t (1 + y ) dt 1 dy = 2t y + 1 dt ￿ ￿ d ln |y + 1| = 2 t dt ln |y + 1| = t2 + C =⇒ for some constant C1 = ±eC . For the initial condition, set t = 0: C1 = 1 =⇒ 2 y + 1 = C1 et 2 y (t) = et − 1. For a sequence of approximate solutions, deﬁne the following: ￿t ￿ ￿ yn+1 = G τ , yn (τ ) dτ where G(t, y ) = 2 t (1 + y ). 0 The initial function is a constant, so we have y0 (t) = 0; ￿t y1 (t) = 2 τ (1 + 0) dτ = t2 ; 0 ￿t ￿ ￿ t4 y2 (t) = 2 τ 1 + τ 2 d τ = t2 + ; 2 0 ￿ ￿ ￿t 4 τ t4 t6 y3 (t) = 2 τ 1 + τ2 + d τ = t2 + + . 2 2 6 0 It is easy to show in general that n yn (t) = t2 + ￿ t2 k t4 t6 t2 n + + ··· + = . 2! 3! n! k! k=0 Now we take the limit as n increases without bound: ￿∞ ￿ ∞ ￿ t2 k ￿ xk lim yn (t) = = −1 n→∞ k! k! k=1 where x = t2 . k=0 Recall the Taylor Series expansion for the exponential function: ex = ∞ ￿ xk k=0 k! =⇒ 2 lim yn (t) = et − 1 = y (t). n→∞ Hence the functions yn = yn (t) do indeed give a good approximation of the exact solution y = y (t). 1 ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.

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