Explanation Given: Instead of the auxiliary equation m 4 + 1 = 0 , ( m 2 + 1 ) 2 − 2 m 2 = 0 is used. And it is found that m 4 + 1 = ( m 2 + 1 ) 2 − 2 m 2 . Procedure used: General solution of a higher order differential equation: In order, to solve an n th order differential equation a n y n + a n − 1 y n − 1 + ... a 2 y ″ + a 1 ′ + a 0 y = 0 , initially solve the n th degree polynomial equation a n m n + a n − 1 m n − 1 + ...... + a 0 = 0 . If all the roots of a n m n + a n − 1 m n − 1 + ...... + a 0 = 0 are real and distinct, then the general solution will be y = c 1 e m 1 x + c 2 e m 2 x + ⋯ + c n e m n x . If an auxiliary equation has k roots and all the roots of the auxiliary equation are equal, then the general solution will be c 1 e m 1 x + c 2 x e m 1 x + c 3 x 2 e m 1 x + ....... + c k x k − 1 e m 1 x . In case if higher order differential equation has repeated conjugate complex roots, then the general solution will be, y = c 1 e m 1 x + c 2 e − m 1 x + c 3 x e m 1 x + c 4 x e − m 1 x where m 1 represents an imaginary root. Rational root theorem to determine the roots of a polynomial: 1 The rational root theorem states a solution to a polynomial equation. a n x n + a n − 1 x n − 1 + ... + a 0 = 0 2 If a 0 and a n are no zero. Then each rational solution written in lowest terms p q satisfies the polynomial. Where p is an integer factor of a 0 and q is an integer factor of a n that is the leading coefficient. Therefore, ± p q gives a possible root. 3 If ( x + a ) is a factor of the function f ( x ) . Write, f ( x ) = ( x + a ) ( f ( x ) x + a ) . 4 Equate ( x + a ) ( f ( x ) x + a ) to zero. Solution to the quadratic equation: If there is a quadratic equation a x 2 + b x + c = 0 . Solution of the equation is given by the formula, x = − b ± b 2 − 4 a c 2 a . Where a , b and c are the coefficients of the degree of x . Calculation: Consider the given differential equation y ( 4 ) + y = 0 . From given statement, it is clear that m 4 + 1 = ( m 2 + 1 ) 2 − 2 m 2 . Use ( m 2 + 1 ) 2 − 2 m 2 = 0 as an auxiliary equation. Because, m 4 + 1 = 0 has four roots and it will produce four equal imaginary results which is hectic to deal with in terms of writing solution to the differential equation. Therefore, instead of using m 4 + 1 = ( m 2 + 1 ) 2 − 2 m 2 , using ( m 2 + 1 ) 2 − 2 m 2 = 0 will reduce the efforts. Solve the equation ( m 2 + 1 ) 2 − 2 m 2 = 0 as follows. ( m 2 + 1 ) 2 − 2 m 2 = 0 ( m 2 + 1 − 2 m ) ( m 2 + 1 + 2 m ) = 0 Consider the equation m 2 − 2 m + 1 = 0 . Here, the values of a = 1 , b = − 2 and c = 1 . b 2 − 4 a c = ( − 2 ) 2 − 4 ( 1 ) ( 1 ) = 2 − 4 = − 2 Hence, the solutions to the equation m 2 − 2 m + 1 = 0 are conjugative complex numbers. So, use m = − ( b ) ± b 2 − 4 a c 2 ( a ) and determine the roots

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 4.3, Problem 62E

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To solve: The differential equation y(4)+y=0.

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Chapter 4.3, Problem 61E

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Chapter 4 Solutions

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

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The differential equation y ( 4 ) + y = 0 .

Solution Summary: The author explains the general solution to the differential equation y(4)+y=0.

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