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Unformatted text preview: MA341 Homework 8 Solution 6.1 3. This equation is already in the standard form, hence p 1 ( x ) = 1, p 2 ( x ) = √ x 1, g ( x ) = tan x . Now p 1 ( x ) is continuous on (∞ , + ∞ ), p 2 ( x ) is continuous on [1 , + ∞ ), and g ( x ) is defined on ( π 2 + kπ, π 2 + kπ ), thus p 1 , p 2 , and g ( x ) are simultaneously continuous on ( 3 π 2 , 5 π 2 ), which contains x = 5. Therefore, Theorem 1 implies that if the initial data are specified at x = 5, then there is a unique solution on ( 3 π 2 , 5 π 2 ). 5. First, we convert this given equation into the standard form: y 000 1 x √ x + 1 y + 1 √ x + 1 y = 0 . Then, we find that p 1 ( x ) = 0, p 2 ( x ) = 1 x √ x +1 , p 3 ( x ) = 1 √ x +1 , and g ( x ) = 0. Next, we observe that p 1 ( x ) and g ( x ) are constant (and hence continuous) for all x , p 2 ( x ) is contin uous on ( 1 , 0) S (0 , + ∞ ), and p 3 ( x ) is continuous on ( 1 , + ∞ ). Consequently, the largest interval that contains x = 1 2 , on which p 1 , p 2 , p 3 , and g are simultaneously continuous, is (0 , ∞ ). According to Theorem 1, the initial value problem has a unique solution on (0 , ∞ ). 15. Let us first make sure that e 3 x , e x , and e 4 x are solutions of the given differential equation. The corresponding characteristic equation is λ 3 + 2 λ 2 11 λ 12 = 0. By direct substitution, one can easily see that λ 1 = 3, λ 2 = 1, and λ 3 = 4 are indeed its roots. Then, we compute the Wronskian: W ( x ) = e 3 x e x e 4 x 3 e 3 x e x 4 e 4 x 9 e 3 x e x 16 e 4 x = 84 e 2 x and see that it is never zero on (∞ , + ∞ ). Hence, the solutions e 3 x , e x , and e 4 x are linearly independent, and according to Theorem 3, { e 3 x , e x , e 4 x } is a fundamental set of solutions for the given differential equation. Its general solution is then given by the linear combination: c 1 e 3 x + c 2 e x + c 3 e 4 x . 17. First, we need to show that x , x 2 and x 3 are indeed solutions. This is easily done by direct substitution into the given differential equation. Then, we compute the Wronskian: W ( x ) = x x 2 x 3 1 2 x 3 x 2 2 6 x = 2 x 3 and see that it is not equal to zero on (0 , + ∞ ). Hence, the solutions x , x 2 and x 3 are linearly independent, and according to Theorem 3, { x, x 2 , x 3 } 1 MA341 Homework 8 Solution is a fundamental set of solutions for the given differential equation. Its general solution is then given by the linear combination: c 1 x + c 2 x 2 + c 3 x 3 . 21. (a) According to theorem 4, a general solution to the inhomogeneous equation can be expressed in the form: y ( x ) = y p ( x ) + C 1 y 1 ( x ) + C 2 y 2 ( x ) + C 3 y 3 ( x ) = ln x + C 1 x + C 2 x ln x + C 3 x (ln x ) 2 ....
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 Spring '08
 Schecter
 Differential Equations, Equations, characteristic equation

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