The exact value of this solution at t 1 is P 1 60 260 e 25 393 85 b Eulers

The exact value of this solution at t 1 is p 1 60 260

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The exact value of this solution at t = 1 is P (1) = 60 + 260 e 0 . 25 393 . 85 . b. Euler’s method with stepsize of h = 0 . 25 for t [0 , 1] gives the following formula for this problem: P n +1 = P n + 0 . 25(0 . 25 P n - 15) = 1 . 0625 P n - 3 . 75 . Below is a table showing the iterations for Euler’s solution. t n P n t 0 = 0 P 0 = 320 t 1 = 0 . 25 P 1 = 1 . 0625 P 0 - 3 . 75 = 336 . 25 t 2 = 0 . 5 P 2 = 1 . 0625 P 1 - 3 . 75 = 353 . 5156 t 3 = 0 . 75 P 2 = 1 . 0625 P 2 - 3 . 75 = 371 . 8604 t 4 = 1 . 0 P 4 = 1 . 0625 P 3 - 3 . 75 = 391 . 3516 The percent error is 100 ( 391 . 35 - 393 . 85 393 . 85 ) ≈ - 0 . 633% . 5. a. The differential equation describing the radioactive decay is dR dt = - 0 . 25 R + 7 e - 0 . 05 t with R (0) = 70. The Euler’s solution with a stepsize of h = 0 . 5 for t [0 , 3]. The Euler formula for this problem is R n +1 = R n + 0 . 5 ( - 0 . 25 R n + 7 e - 0 . 05 t n ) , R n +1 = 0 . 875 R n + 3 . 5 e - 0 . 05 t n . Below is a table showing the iterations for this Euler’s solution. t n R n t 0 = 0 R 0 = 70 t 1 = 0 . 5 R 1 = 0 . 875 R 0 + 7 e - 0 . 05 t 0 = 0 . 875(70) + 7 = 64 . 75 t 2 = 1 . 0 R 2 = 0 . 875 R 1 + 7 e - 0 . 05 t 1 = 60 . 07 t 3 = 1 . 5 R 3 = 0 . 875 R 2 + 7 e - 0 . 05 t 2 = 55 . 89 t 4 = 2 . 0 R 4 = 0 . 875 R 3 + 7 e - 0 . 05 t 3 = 52 . 15 t 5 = 2 . 5 R 5 = 0 . 875 R 4 + 7 e - 0 . 05 t 4 = 48 . 80 t 6 = 3 . 0 R 6 = 0 . 875 R 5 + 7 e - 0 . 05 t 5 = 45 . 79 b. We rewrite the linear ODE as dR dt + 0 . 25 R = 7 e - 0 . 05 t , R (0) = 70 . The integrating factor is μ ( t ) = e 0 . 25 t , so the ODE becomes d dt ( e 0 . 25 t R ( t ) ) = 7 e 0 . 2 t , so e 0 . 25 t R ( t ) = 35 e 0 . 2 t + C.
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It follows that R ( t ) = 35 e - 0 . 05 t + C e - 0 . 25 t , with ICs R (0) = 70 = 35 + C or C = 35 . Thus, R ( t ) = 35 e - 0 . 05 t + 35 e - 0 . 25 t . The solution at t = 3 is R (3) = 35 e - 0 . 05 · 3 + 35 e - 0 . 25 · 3 = 46 . 6576 . The percent error satisfies 100 ( 45 . 7881 - 46 . 6576 46 . 6576 ) = - 1 . 864%. 7. a. The body temperature satisfies dT dt = - 0 . 2( T - 22) with T (0) = 36. Make the substitution z ( t ) = T ( t ) - 22, so z (0) = T (0) - 22 = 14. The translated problem becomes: dz dt = - 0 . 2 z, with z (0) = 14 , which has the solution z ( t ) = 14 e - 0 . 2 t or T ( t ) = 22 + 14 e - 0 . 2 t . The exact value of this solution at t = 2 is T (2) = 22 + 14 e - 0 . 4 31 . 384.
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