Program Explanation Given information: The differential equation is written below. 3 y ( 3 ) + 2 y ″ = 0 ........ (1) Initial conditions are written below. y ( 0 ) = − 1 y ′ ( 0 ) = 0 y ″ ( 0 ) = 1 Explanation: Rewrite the differential equation (1) as follows: 3 y ( 3 ) + 2 y ″ = 0 3 d 3 d x 3 ( y ) + 2 d 2 d x 2 ( y ) = 0 ( 3 d 3 d x 3 + 2 d 2 d x 2 ) y = 0 So, the characteristic equation for the differential equation is written as, 3 m 3 + 2 m 2 = 0 Solve the characteristic equation and obtain the possible roots as, m 2 ( 3 m + 2 ) = 0 m 2 = 0 or 3 m + 2 = 0 m 1 , 2 = 0 or m 3 = − 2 3 So, the general solution of the differential equation is written as, y ( x ) = C 1 e m 1 x + C 2 x e m 2 x + C 3 e m 3 x y ( x ) = C 1 e ( 0 ) x + C 2 x e ( 0 ) x + C 3 e ( − 2 3 ) x y ( x ) = C 1 + C 2 x + C 3 e − 2 3 x Substitute, the initial condition y ( 0 ) = − 1 in the general solution as, y ( 0 ) = C 1 + C 2 ( 0 ) + C 3 e − 2 3 ( 0 ) − 1 = C 1 + C 3 C 1 = − C 3 − 1 Differentiate the general solution and substitute the second initial condition y ′ ( 0 ) = 0 . y ( x ) = C 1 + C 2 x + C 3 e − 2 3 x y ′ ( x ) = C 2 − 2 3 C 3 e − 2 3 x y ′ ( 0 ) = C 2 − 2 3 C 3 e − 2 3 ( 0 ) 0 = C 2 − 2 3 C 3 Further, solve the above equation as follows: C 2 = 2 3 C 3 Differentiate the general solution twice and substitute the third initial condition y ″ ( 0 ) = 1

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Author: Calvis

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Chapter 3.3, Problem 25P

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Program Description: Purpose of the problem is to obtain the particular solution of the differential equation 3y(3)+2y″=0;y(0)=−1,y′(0)=0,y″(0)=1 .

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Chapter 3.3, Problem 24P

Chapter 3.3, Problem 26P

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Chapter 3 Solutions

EBK DIFFERENTIAL EQUATIONS

Ch. 3.1 - In Problems 1 through 16, a homogeneous...Ch. 3.1 - Prob. 2P Ch. 3.1 - Prob. 3P Ch. 3.1 - Prob. 4P Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 7P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P

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Program Description: Purpose of the problem is to obtain the particular solution of the differential equation 3 y ( 3 ) + 2 y ″ = 0 ; y ( 0 ) = − 1 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 1 .

Solution Summary: The author explains that the problem is to obtain the particular solution of the differential equation.

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Program Description: Purpose of the problem is to obtain the particular solution of the differential equation 3y(3)+2y″=0;y(0)=−1,y′(0)=0,y″(0)=1 .

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Chapter 3 Solutions