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Unformatted text preview: Exercises 4.7 (b) Separating variables and integrating from t to t yields dW W = pdt t Z t dW W = t Z t p ( ) d (4.4) ln W ( t ) W ( t ) = t Z t p ( ) d  W ( t )  =  W ( t )  exp  t Z t p ( ) d . Since the integral on the righthand side is continuous (even differentiable) on ( a,b ), the exponential function does not vanish on ( a,b ). Therefore, W ( t ) has a constant sign on ( a,b ) (by the intermediate value theorem), and so we can drop the absolute value signs and obtain W ( t ) = W ( t ) exp  t Z t p ( ) d . (4.5) The constant C := W ( t ) = y 1 ( t ) y 2 ( t ) y 1 ( t ) y 2 ( t ) depends on y 1 and y 2 (and t ). Thus, the Abels formula is proved. (c) If, at some point t in ( a,b ), W ( t ) = 0, then (4.5) implies that W ( t ) 0. 34. Using the superposition principle (see Problem 30), we conclude the following. (a) y 1 ( t ) = t 2 t and y 2 ( t ) = t 3 t are solutions to the corresponding homogeneous equation. These two functions are linearly independent on any interval because their nontrivial linear combination c 1 y 1 + c 2 y 2 = c 2 t 3 + c 1 t 2 ( c 1 + c 2 ) t is a nonzero polynomial of degree at most three, which cannot have more than three zeros. (b) A general solution to the given equation is a sum of a general solution y h to the cor responding homogeneous equation and a particular solution to the nonhomogeneous equation, say, t . Hence, y = t + c 1 ( t 2 t ) + c 2 ( t 3 t ) 151 Chapter 4 y = 1 + c 1 (2 t 1) + c 2 ( 3 t 2 1 ) ....
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 Spring '08
 MAZMANI
 Equations

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