Hw8 - 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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.

Page1 / 7

Hw8 - 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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online