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nagle_differential_equations_ISM_Part6

# nagle_differential_equations_ISM_Part6 - Exercises 1.2(b We...

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Exercises 1.2 (b) We have ( e mx ) + 3 ( e mx ) + 2 ( e mx ) = 0 e mx ( m 3 + 3 m 2 + 2 m ) = 0 m ( m 2 + 3 m + 2) = 0 m = 0 , - 1 , - 2 . 22. We find φ ( x ) = c 1 e x - 2 c 2 e - 2 x , φ ( x ) = c 1 e x + 4 c 2 e - 2 x . Substitution yields φ + φ - 2 φ = ( c 1 e x + 4 c 2 e - 2 x ) + ( c 1 e x - 2 c 2 e - 2 x ) - 2 ( c 1 e x + c 2 e - 2 x ) = ( c 1 + c 1 - 2 c 1 ) e x + (4 c 2 - 2 c 2 - 2 c 2 ) e - 2 x = 0 . Thus, with any choice of constants c 1 and c 2 , φ ( x ) is a solution to the given equation. (a) Constants c 1 and c 2 must satisfy the system 2 = φ (0) = c 1 + c 2 1 = φ (0) = c 1 - 2 c 2 . Subtracting the second equation from the first one yields 3 c 2 = 1 c 2 = 1 / 3 c 1 = 2 - c 2 = 5 / 3 . (b) Similarly to the part (a), we obtain the system 1 = φ (1) = c 1 e + c 2 e - 2 0 = φ (1) = c 1 e - 2 c 2 e - 2 which has the solution c 1 = (2 / 3) e - 1 , c 2 = (1 / 3) e 2 . 24. In this problem, the independent variable is t , the dependent variable is y . Writing the equation in the form dy dt = ty + sin 2 t , we conclude that f ( t, y ) = ty + sin 2 t , ∂f ( t, y ) /∂y = t . Both functions, f and ∂f/∂y , are continuous on the whole ty -plane. So, Theorem 1 applies for any initial condition, in particular, for y ( π ) = 5. 21

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Chapter 1 26. With the independent variable t and the dependent variable x , we have f ( t, x ) = sin t - cos x, ∂f ( t, x ) ∂x = sin x , which are continuous on tx -plane. So, Theorem 1 applies for any initial condition. 28. Here, f ( x, y ) = 3 x - 3 y - 1 and ∂f ( x, y ) ∂y = ∂y 3 x - ( y - 1) 1 / 3 ) = - 1 3 3 ( y - 1) 2 . The function f is continuous at any point ( x, y ) while ∂f/∂y is defined and continuous at any point ( x, y ) with y = 1 i.e., on the xy -plane excluding the horizontal line y = 1.
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nagle_differential_equations_ISM_Part6 - Exercises 1.2(b We...

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