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WrittenHW07-Solutions - Math 216(Section 50 1(4.14 Let us...

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Math 216 (Section 50) Written Homework #7 – Solutions Fall 2007 1 ( 4.1–4 ) Let us first rewrite x (3) - 2 t x + 1 t 2 x + 5 t 3 x = ln t t 3 . Let us define x 1 := x, x 2 := x = x 1 , x 3 := x = x 2 . Therefore the given differential equations are rewritten as follows: x 1 = x 2 x 2 = x 3 x 3 - 2 t x 3 + 1 t 2 x 2 + 5 t 3 x 1 = ln t t 3 = x 1 = x 2 x 2 = x 3 x 3 = 2 t x 3 - 1 t 2 x 2 - 5 t 3 x 1 + ln t t 3 . 2 ( 4.1–8 ) Let us define x 1 := x, x 2 := x = x 1 , y 1 := y, y 2 := y = y 1 . Note that then x = x 2 and y = y 2 . Therefore the given system of differential equations are rewritten as follows: x + 3 x + 4 x - 2 y = 0 y + 2 y - 3 x + y = cos t = x 1 = x 2 y 1 = y 2 x 2 + 3 x 2 + 4 x 1 - 2 y 1 = 0 x 2 + 2 y 2 - 3 x 1 + y 1 = cos t = x 1 = x 2 y 1 = y 2 x 2 = - 3 x 2 - 4 x 1 + 2 y 1 y 2 = - 2 y 2 + 3 x 1 - y 1 + cos t 3 ( 4.1–9 ) Let us define x 1 := x, x 2 := x = x 1 , y 1 := y, y 2 := y = y 1 . z 1 := z, z 2 := z = z 1 . Note that then x = x 2 , y = y 2 , and z = z 2 . Therefore the given system of differential equations are rewritten as follows: x = 3 x - y + 2 z y = x + y - 4 z z = 5 x - y - z = x 1 = x 2 y 1 = y 2 z 1 = z 2 x 2 = 3 x 1 - y 1 + 2 z 1 y 2 = x 1 + y 1 - 4 z 1 z 2 = 5 x 1 - y 1 - z 1 1
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4 ( 4.1–16 ) Given x = 8 y and y = - 2 x , x = 8 y = - 16 x = x + 16 x = 0 .
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