Explanation Formula used: The auxiliary equation of a linear homogeneous differential equation of the form a 0 D 2 y + a 1 D y + a 2 y = 0 with constant coefficients is given by a 0 m 2 + a 1 m + 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 m is given by y = e m x ( c 1 + c 2 x ) . The general solution of the auxiliary equation with two distinct roots m 1 and m 2 , is given by y = c 1 e m 1 x + c 2 e m 2 x . The solution of a linear non-homogeneous differential equation of the form a 0 D 2 y + a 1 D y + a 2 y = b where the a ’s are constants and b is a function of x or is a constant, is given by y = y c + y p . The y c is called the complementary solution obtained by solving the corresponding homogeneous equation and where y p is the particular solution. The different forms of b and their corresponding particular solution to be used is shown below: b y p ( x ) a e β x A e β x a cos ( β x ) A cos ( β x ) + B sin ( β x ) b sin ( β x ) A cos ( β x ) + B sin ( β x ) a cos ( β x ) + b sin ( β x ) A cos ( β x ) + B sin ( β x ) n th degree polynomial A n x n + A n − 1 x n − 1 + A 1 x + A 0 The solution of the quadratic equation a x 2 + b x + c = 0 is calculated as x = − b ± b 2 − 4 a c 2 a . Theorem used: If r 1 = α + j β and r 1 = α − j β are the two complex roots of the auxiliary equation a r 2 + b r + c = 0 , then y = e α x ( c 1 cos β x + c 2 sin β x ) is the general solution of a y ″ + b y ′ + c y = 0 . Calculation: The given differential equation is: d 2 v d t 2 + d v d t + 4 v = 0 The auxiliary equation is: m 2 + m + 4 = 0 Compare the above equation with a x 2 + b x + c = 0 . Here a = 1 , b = 1 , c = 4 . Substitute the value of a = 1 , b = 1 , c = 4 in the quadratic formula x = − b ± b 2 − 4 a c 2 a . m = − 1 ± 1 − 16 2 = − 1 ± − 15 2 = − 1 2 ± 15 2 j Therefore, the general solution to the above differential equation is: v = e − t 2 ( c 1 sin 15 2 t + c 2 cos 15 2 t ) Obtain the derivatives as shown below

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Differential Equations

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Chapter 31, Problem 45RE

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The particular solution of the differential equation by using the initial conditions.

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Chapter 31, Problem 44RE

Chapter 31, Problem 46RE

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Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

Ch. 31.1 - Show that is a solution of . Is it the general...Ch. 31.1 - Prob. 1E Ch. 31.1 - In Exercises 1 and 2, show that the indicated...Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 4E Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 6E Ch. 31.1 - In Exercises 7–10, show that each function y =...Ch. 31.1 - Prob. 8E Ch. 31.1 - In Exercises 7–10, show that each function y =...

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