Explanation Procedure used: Solution to a second order differential equation. Begin with considering the case of a second order differential equation a y ″ + b y ′ + c y = 0 , where a , b and c are constants. If we try to find a solution of the form y = e m x . Then substitution of y ′ = m e m x and y ″ = m 2 e m x into the equation a y ″ + b y ′ + c y = 0 becomes a m 2 e m x + b m e m x + c e m x = 0 or e m x ( a m 2 + b m + c ) = 0. Since, for a real x e m x ≠ 0 , it is apparent that the only way y = e m x satisfy the differential equation a y ″ + b y ′ + c y = 0 is when m is chosen as a root of the quadratic equation a m 2 + b m + c = 0 . Case I: Distinct real roots The general solution of a y ″ + b y ′ + c y = 0 on this interval ( − ∞ , ∞ ) is y = c 1 e m 1 x + c 2 e m 2 x . Case II: Repeated real roots The general solution is y = c 1 e m 1 x + c 2 x e m 1 x . Case III: Conjugate Complex Roots If m 1 and m 2 and are complex, then m 1 = α + i β and m 2 = α − i β , where α and β > 0 are real and i 2 = − 1 . The general solution is y = e a x ( c 1 cos β x + c 2 sin β x ) . Calculation: Consider the given differential equation y ″ + y ′ + 2 y = 0 . Consider y = e m x as the solution of the differential equation y ″ + y ′ + 2 y = 0 . Substitute y = e m x , y ″ = m 2 e m x and y ′ = m e m x into y ″ + y ′ + 2 y = 0 to obtain the auxiliary equation. Therefore, m 2 e m x + m e m x + 2 e m x = 0 e m x ( m 2 + m + 2 ) = 0 For real x, e m x ≠ 0 . Therefore, m 2 + m + 2 = 0 . Here, a = 1 , b = 1 and c = 2 . Use m = − ( b ) ± b 2 − 4 a c 2 ( a ) to determine the roots of auxiliary equation. m 1 = − ( b ) + b 2 − 4 a c 2 ( a ) = − ( 1 ) + − 7 2 ( 1 ) = − 1 + i 7 2 = − 1 2 + i 7 2 The other root is, m 2 = − ( b ) − b 2 − 4 a c 2 ( a ) = − ( 1 ) − − 7 2 ( 1 ) = − 1 − i 7 2 = − 1 2 − i 7 2 Hence, roots of the auxiliary equation are ( − 1 2 + i 7 2 , − 1 2 − i 7 2 )

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.3, Problem 33E

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To solve: The initial value problem y″+y′+2y=0,y(0)=y′(0)=0.

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Chapter 4.3, Problem 32E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - Given that y=c1+c2x2 is a two-parameter family of...Ch. 4.1 - Find two members of the family of solutions in...Ch. 4.1 - Prob. 7E Ch. 4.1 - Use the general solution of x + 2x = 0 given in...Ch. 4.1 - In Problems 9 and 10 find an interval centered...Ch. 4.1 - In Problems 9 and 10 find an interval centered...

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The initial value problem y ″ + y ′ + 2 y = 0 , y ( 0 ) = y ′ ( 0 ) = 0 .

Solution Summary: The author explains how to solve a second order differential equation.

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To solve: The initial value problem y″+y′+2y=0,y(0)=y′(0)=0.

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