DiffEq1Solutions(2).pdf

# DiffEq1Solutions(2).pdf - CHEN20041 Solutions 1 1 a...

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CHEN20041 Solutions 1 1. a) Auxiliary equation: m 2 + m - 20 = ( m + 5)( m - 4) = 0, so m 1 = - 5 and m 2 = 4. Solutions are real and distinct, so y = Ae - 5 x + Be 4 x . b) Auxiliary equation: 7 m 2 + 17 m - 12 = (7 m - 4)( m + 3) = 0, so m 1 = 4 7 and m 2 = - 3. Solutions are real and distinct, so y = Ae 4 x/ 7 + Be - 3 x . c) Auxiliary equation: m 2 - 6 m + 9 = ( m - 3)( m - 3) = 0, so m = 3. Solutions real and equal, so s = ( at + b ) e 3 t . d) Auxiliary equation: m 2 + 4 = ( m - 2 i )( m + 2 i ) = 0, so m 1 = 0 ± 2 i . Solution complex, so z = e 0 ( a cos(2 x ) + b sin(2 x )), hence z = a cos(2 x ) + b sin(2 x ) . e) Auxiliary equation: m 2 - 2 m - 10 = 0. By the quadratic formula, m = 2 ± p ( - 2) 2 - 4 × ( - 10) 2 which gives m = 1 ± 11. Solutions real and distinct, so y = ae (1+ 11) t + be (1 - 11) t . 2. a) Use solution from 1a). When x = 0, y = 3, so 3 = Ae - 5 × 0 + Be 4 × 0 = A + B . Also, dy dx = - 5 Ae - 5 x + 4 Be 4 x , and since when x = 0, dy dx = - 33 we have - 33 = - 5 Ae - 5 × 0 + 4 Be 4 × 0 = - 5 A + 4 B . So we have A + B = 3 (1) - 5 A + 4 B = - 33 (2)

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Solving this system gives A = 5 and B = - 2, so the particular solution is y = 5 e - 5 x - 2 e 4 x . b) Use solution from 1c). When t = 1, s = - 2, so - 2 = ( a + b ) e 3 . Also, ds dt = ae 3 t + 3( at + b ) e 3 t , and since when t = 1, ds dt = 5 we get 5 = ae 3 + 3( a + b ) e 3 . So

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