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2 MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6 Example. Consider the initial value problem dy dt = 3 + e t 1 2 y, y (0) = 1 . 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 di ff erential equation is a linear equation: dy dt + 1 2 y = 3 + e t e t/ 2 dy dt + 1 2 e t/ 2 y = e t/ 2 3 + e t d dt e t/ 2 y = 3 e t/ 2 + e t/ 2 e t/ 2 y ( t ) = 6 e t/ 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 t n = n h and y n +1 = y n + G ( t n , y n ) h, y 0 = 1 where G ( t, y ) = 3 + e t 1 2 y. We list some values in the following table: n t n Approximate: y n Slope: G ( t n , y n ) Exact: y ( t n ) Error: | y ( t n ) y n | 0 0.0 1.0000 3.5000 1.0000 0.0000 1 0.1 1.3500 3.2298 1.3366 0.0134 2 0.2 1.6730 2.9822 1.6480 0.0250 3 0.3 1.9712 2.7552 1.9362 0.0350 4 0.4 2.2467 2.5470 2.2032 0.0435 Note that y n y ( t n ) – even though the error gets larger as
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