sol_assign3

# sol_assign3 - MAT1332A Calculus for the Life Sciences II...

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Unformatted text preview: MAT1332A - Calculus for the Life Sciences II - Fall 2009 Solutions to Assignment 3 (1) (a) [5 points] Check that z ( t ) = 1+ √ 1 + 2 t is a solution of the autonomous differential equation dz dt = 1 z- 1 with initial condition z (0) = 2. (b) [7 points] Estimate z (4) if z obeys the differential equation dz dt = 1 z- 1 with initial condition z (0) = 2. Use Euler’s method with Δ t = 1 for four steps. Compare with the exact answer in (a). Solution: (a) We need to verify that the function z defined by z ( t ) = 1 + √ 1 + 2 t satisfies dz dt = 1 z- 1 and the initial condition z (0) = 2. In fact, dz dt = d dt (1 + √ 1 + 2 t ) = 2 2 √ 1 + 2 t = 1 √ 1 + 2 t = 1 (1 + √ 1 + 2 t )- 1 = 1 z- 1 and z (0) = 1 + √ 1 + 2 · 0 = 2 . (b) By Euler’s method, we have z (1) ≈ ˆ z (1) = ˆ z (0 + Δ t ) = z (0) + z (0)Δ t = 2 + 1 2- 1 · 1 = 3 z (2) ≈ ˆ z (2) = ˆ z (1 + Δ t ) = z (1) + z (1)Δ t ≈ 3 + 1 3- 1 · 1 = 7 2 z (3) ≈ ˆ z (3) = ˆ z (2 + Δ t ) = z (2) + z (2)Δ t ≈ 7 2 + 1 7 2- 1 · 1 = 7 2 + 2 5 = 39 10 z (4) ≈ ˆ z (4) = ˆ z (3+Δ t ) = z (3)+ z (3)Δ t ≈ 39 10 + 1 39 10- 1 · 1 = 39 10 + 10 29 = 1231 290 ≈ 4 . 25 On the other hand, by (a), we know that the precise solution is z ( t ) = 1 + √ 1 + 2 t , that is, z (4) = 1 + √ 1 + 2 · 4 = 4 < 4 . 25 ≈ ˆ z (4). So Euler’s method gives an error of about 0 . 25....
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sol_assign3 - MAT1332A Calculus for the Life Sciences II...

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