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Unformatted text preview: Exercises 1.2 (b) We have ( e mx ) 000 + 3 ( e mx ) 00 + 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 ( x ) = c 1 e x 2 c 2 e 2 x , 00 ( x ) = c 1 e x + 4 c 2 e 2 x . Substitution yields 00 +  2 = ( 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 , ( x ) is a solution to the given equation. (a) Constants c 1 and c 2 must satisfy the system 2 = (0) = c 1 + c 2 1 = (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 = (1) = c 1 e + c 2 e 2 0 = (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 typlane. So, Theorem 1 applies for any initial condition, in particular, for y ( ) = 5. 21 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 txplane. 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 p ( y 1) 2 ....
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 Spring '08
 MAZMANI
 Equations

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