b Eulers method with stepsize of h 0 5 for t 0 2 gives the following formula

B eulers method with stepsize of h 0 5 for t 0 2

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b. Euler’s method with stepsize of h = 0 . 5 for t [0 , 2] gives the following formula for this problem: T n +1 = T n + 0 . 5( - 0 . 2( T n - 22)) = 0 . 9 T n + 2 . 2 . Below is a table showing the iterations for Euler’s solution. t n T n t 0 = 0 T 0 = 36 t 1 = 0 . 5 T 1 = 0 . 9 T 0 + 2 . 2 = 32 . 4 + 2 . 2 = 34 . 6 t 2 = 1 . 0 T 2 = 0 . 9 T 1 + 2 . 2 = 31 . 14 + 2 . 2 = 33 . 34 t 3 = 1 . 5 T 3 = 0 . 9 T 2 + 2 . 2 = 30 . 006 + 2 . 2 = 32 . 206 t 4 = 2 . 0 T 4 = 0 . 9 T 3 + 2 . 2 = 28 . 985 + 2 . 2 = 31 . 185 The percent error is 100 ( 31 . 185 - 31 . 384 31 . 384 ) ≈ - 0 . 634% . c. The new differential equation is dT dt = - 0 . 2( T - (22 - 0 . 5 t )) T (0) = 36 . This is a linear ODE, which can be written dT dt + 0 . 2 T = 22 5 - 0 . 1 t, and has an integrating factor μ ( t ) = e 0 . 2 t . Thus, d dt ( e 0 . 2 t T ( t ) ) = 22 5 - 0 . 1 t e 0 . 2 t . Integrating gives e 0 . 2 t T ( t ) = 22 e 0 . 2 t - 0 . 5 t e 0 . 2 t + 2 . 5 e 0 . 2 t + C or T ( t ) = 24 . 5 - 0 . 5 t + C e - 0 . 2 t .
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The initial condition T (0) = 36 = 24 . 5 + C or C = 11 . 5, so T ( t ) = 24 . 5 - 0 . 5 t + 11 . 5 e - 0 . 2 t . Euler’s method with stepsize of h = 0 . 5 for t [0 , 2] gives the following formula for this problem: T n +1 = T n + 0 . 5( - 0 . 2( T n - (22 - 0 . 5 t ))) = 0 . 9 T n + 0 . 1(22 - 0 . 5 t ) . Below is a table showing the iterations for Euler’s solution. t n T n t 0 = 0 T 0 = 36 t 1 = 0 . 5 T 1 = 0 . 9 T 0 + 0 . 1(22 - 0 . 5(0)) = 34 . 6 t 2 = 1 . 0 T 2 = 0 . 9 T 1 + 0 . 1(22 - 0 . 5(0 . 5)) = 33 . 315 t 3 = 1 . 5 T 3 = 0 . 9 T 2 + 0 . 1(22 - 0 . 5(1 . 0)) = 32 . 1335 t 4 = 2 . 0 T 4 = 0 . 9 T 3 + 0 . 1(22 - 0 . 5(1 . 5)) = 31 . 045 The actual solution gives T (2) = 23 . 5 + 11 . 5 e - 0 . 4 = 31 . 2087, which implies Euler’s method has about a - 0 . 52% percent error.
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