Explanation Given: The second order differential equation y ″ + 4 y ′ + 3 y = 0 and t 0 = 1 . Formula used: The auxiliary equation of a linear homogeneous differential equation of the form a 0 y ″ + a 1 y ′ + a 2 y = 0 with constant coefficients is given by a 0 r 2 + a 1 r + a 2 = 0 . The general solution of the auxiliary equation with complex roots is given by y = e α x ( c 1 sin β x + c 2 cos β x ) . The general solution of the auxiliary equation with double roots r is given by y = e r x ( c 1 + c 2 x ) . The general solution of the auxiliary equation with two distinct roots r 1 and r 2 , is given by y = c 1 e r 1 x + c 2 e r 2 x . Calculation: The given differential equation is y ″ + 4 y ′ + 3 y = 0 . Compare the given equation with a 0 y ″ + a 1 y ′ + a 2 y = 0 and obtain the values of a 0 , a 1 , a 2 as shown below: a 0 = 1 , a 1 = 1 , a 2 = − 2 . Write the auxiliary equation of the homogeneous equation y ″ + 4 y ′ + 3 y = 0 as, r 2 + 4 r + 3 = 0 Solve the auxiliary equation and obtain the roots as follows: r 2 + 4 r + 3 = 0 ( r − 1 ) ( r + 2 ) = 0 r = − 3 , − 1 Thus, the auxiliary equation has three distinct roots r = − 3 , − 1 . Using the above formula, obtain the general solution of the homogeneous equation y ″ + 4 y ′ + 3 y = 0 as, y = c 1 e r 1 t + c 3 e r 2 t y = c 1 e − t + c 2 e − 3 t Therefore, the general solution of the differential equation is y ″ + 4 y ′ + 3 y = 0 y = c 1 e − t + c 2 e − 3 t . Differentiate the above solution and obtain the value of y ′ as shown below: y = c 1 e − t + c 2 e − 3 t y ′ = − c 1 e − t − 3 c 2 e − 3 t Let y 1 ( 0 ) = 1 , y ′ 1 ( 0 ) = 0 and y 2 ( 0 ) = 1 , y ′ 2 ( 0 ) = 0 be the two initial conditions. Substitute y = 1 , t = 1 in the above equation as, y ( 1 ) = 1 c 1 e − 1 + c 2 e − 3 ( 1 ) = 1 c 1 e − 1 + c 2 e − 3 = 1 … … ( 1 ) Substitute y ′ = 0 , t = 1 in y ′ = − c 1 e − t − 3 c 2 e − 3 t as, y ′ ( 1 ) = 1 − c 1 e − 1 − 3 c 2 e − 3 ( 1 ) = 0 − c 1 e − 1 − 3 c 2 e − 3 = 0 … … ( 2 ) Solve equations (1) and (2) as shown below: c 1 e − 1 + c 2 e − 3 + ( − c 1 e − 1 − 3 c 2 e − 3 ) = 0 − 1 2 c 2 e − 3 = − 1 c 2 = − e 3 2 Thus, c 2 = − e 3 2 . Substitute c 2 = − e 3 2 in equation (2) and obtain the value of c 1 as follows: c 1 e − 1 + c 2 e − 3 = 1 c 1 e − 1 + ( − e 3 2 ) e − 3 = 1 c 1 e − 1 + e 1 2 = 1 c 1 = 3 2 e 1 Thus, c 1 = 3 2 e 1

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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The fundamental set of solution for the given differential equation at the initial point t0=0.

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The fundamental set of solution for the given differential equation at the initial point t 0 = 0 .

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