solution_2.10

solution_2.10 - t 1 = t h = 0 h = 0 1 y 1 = y hf t,y = 1 1...

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Solutions to selected problems in Section 2.10 1(a). From the given initial condition, we have t 0 = 1 ,y 0 = 0 . From the given ODE, we have f ( t,y ) = 2 t - 1 . Hence the Euler’s method iteration becomes y k +1 = y k + hf ( t k ,y k ) = y k + h (2 t k - 1) . 1(b). Using Euler’s method: t 1 = t 0 + h = 1 + h = 1 . 1 , y 1 = y 0 + hf ( t 0 ,y 0 ) = 0 + 0 . 1 * (2 * 1 - 1) = 0 . 1 , t 2 = t 1 + h = 1 . 2 , y 2 = y 1 + hf ( t 1 ,y 1 ) = 0 . 2 + 0 . 1 * (2 * 1 . 1 - 1) = 0 . 22 , t 3 = t 2 + h = 1 . 3 , y 3 = y 2 + hf ( t 2 ,y 2 ) = 0 . 32 + 0 . 1 * (2 * 1 . 2 - 1) = 0 . 36 . 1(c). In the integrating factor, we get the general solution to the given ODE y = t 2 - t + C. Applying the initial condition, we have 1 2 - 1 + C = y (1) = 0 from which we have C = 0 and the solution to the IVP is y ( t ) = t 2 - t. 1(d). Using the solution formula found in 1(c), we have y (1 . 1) = 0 . 11, y (1 . 2) = 0 . 24, y (1 . 3) = 0 . 39. Hence e 1 = y ( t 1 ) - y 1 = y (1 . 1) - y 1 = 0 . 11 - 0 . 1 = 0 . 01 , e 2 = y ( t 2 ) - y 2 = y (1 . 2) - y 2 = 0 . 24 - 0 . 22 = 0 . 02 , e 3 = y ( t 3 ) - y 3 = y (1 . 3) - y 3 = 0 . 39 - 0 . 36 = 0 . 03 . 3(a). From the given initial condition, we have t 0 = 0 ,y 0 = 1 . 1

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From the given ODE, we have f ( t,y ) = - ty. Hence the Euler’s method iteration becomes y k +1 = y k + hf ( t k ,y k ) = y k - ht k y k . 3(b). Using Euler’s method:
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Unformatted text preview: t 1 = t + h = 0 + h = 0 . 1 , y 1 = y + hf ( t ,y ) = 1-. 1 * * 1 = 1 , t 2 = t 1 + h = 0 . 2 , y 2 = y 1 + hf ( t 1 ,y 1 ) = 1-. 1 * . 1 * 1 = 0 . 99 , t 3 = t 2 + h = 0 . 3 , y 3 = y 2 + hf ( t 2 ,y 2 ) = 0 . 99-. 1 * . 2 * . 99 = 0 . 9702 . 3(c). In the integrating factor, we can compute the exact solution y ( t ) = exp(-t 2 / 2). 3(d). Using the formula for the exact solution, we have y (0 . 1) = 0 . 9950, y (0 . 2) = 0 . 9802, y (0 . 3) = 0 . 9560. Hence e 1 = y ( t 1 )-y 1 = y (0 . 1)-y 1 = 0 . 9950-1 =-. 005 , e 2 = y ( t 2 )-y 2 = y (0 . 2)-y 2 = 0 . 9802-. 99 =-. 0098 , e 3 = y ( t 3 )-y 3 = y (0 . 3)-y 3 = 0 . 9560-. 9702 =-. 0142 . 2...
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This note was uploaded on 10/16/2009 for the course MATH 2214 at Virginia Tech.

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solution_2.10 - t 1 = t h = 0 h = 0 1 y 1 = y hf t,y = 1 1...

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