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Unformatted text preview: 2 MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6 Example. Consider the initial value problem dy 1 = 3 + e−t − y, y (0) = 1. dt 2 First we find an exact solution y = y (t), then we find an approximate solution via Euler’s Method. To find the exact solution, note that this differential equation is a linear equation: dy 1 + y = 3 + e−t dt 2 ￿ ￿ 1 t/2 t/2 dy e + e y = et/2 3 + e−t dt 2 d ￿ t/2 ￿ e y = 3 et/2 + e−t/2 dt et/2 y (t) = 6 et/2 − 2 e−t/2 + C =⇒ y (t) = 6 − 2 e−t + C e−t/2 . For the initial condition, set t = 0: 1=6−2+C =⇒ C = −3 =⇒ y (t) = 6 − 2 e−t − 3 e−t/2 . For an approximate solution, we use Euler’s Method with a step size of h = 0.1. Then we have the sequences ￿ yn+1 = yn + G(tn , yn ) h, 1 tn = n h and where G(t, y ) = 3 + e−t − y. y0 = 1 2 We list some values in the following table: n 0 1 2 3 4 tn Approximate: yn 0.0 1.0000 0.1 1.3500 0.2 1.6730 0.3 1.9712 0.4 2.2467 Slope: G(tn , yn ) Exact: y (tn ) Error: |y (tn ) − yn | 3.5000 1.0000 0.0000 3.2298 1.3366 0.0134 2.9822 1.6480 0.0250 2.7552 1.9362 0.0350 2.5470 2.2032 0.0435 Note that yn ≈ y (tn ) – even though the error gets larger as n increases. Existence and Uniqueness Picard’s Method. Consider an initial value problem dy = G(t, y ), y (t0 ) = y0 . dt Say that there exists a rectangle ￿ ￿ ￿ ￿ 2￿ R = (t, y ) ∈ R ￿ α < t < β , γ < t < δ such that i. G(t, y ) is continuous on R, ∂G ii. (t, y ) is continuous on R, and ∂y iii. (t0 , y0 ) ∈ R. then there exists a unique solution y = y (t) to the initial value problem. We sketch a proof of this result. Say that y = y (t) is indeed a solution to the initial value problem. Then we have the differential equation ￿ ￿ dy = G(t, y ) = G t, y (t) dt where we view the right-hand side as a function of time t. We may integrate both sides: ￿t ￿ ￿ y (t) = G τ , y (τ ) dτ . ...
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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 University-West Lafayette.

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