Explanation Given information: The initial value problem is ( 3 t 4 + 4 t 2 + 1 ) d x d t = 2 3 . The given initial value condition is x ( 1 ) = − π 3 4 . Formula: ∫ 1 x 2 + 1 d x = tan − 1 x + C ∫ 1 x 2 + a 2 d x = 1 a tan − 1 ( x a ) + C Calculation: Write the initial value problem as below. ( 3 t 4 + 4 t 2 + 1 ) d x d t = 2 3 Simplify it as below. d x = 2 3 d t ( 3 t 4 + 4 t 2 + 1 ) Apply integration operator as below. x = ∫ 2 3 d t ( t 2 ( 3 t 2 + 1 ) + 3 t 2 + 1 ) = ∫ 2 3 d t ( 3 t 2 + 1 ) ( t 2 + 1 ) Express the integrand as a sum of partial fractions with undetermined coefficients. 2 3 ( 3 t 2 + 1 ) ( t 2 + 1 ) = A ( 3 t 2 + 1 ) + B ( t 2 + 1 ) (1) Simplify the above equation. 2 3 ( 3 t 2 + 1 ) ( t 2 + 1 ) = A ( t 2 + 1 ) + B ( 3 t 2 + 1 ) ( 3 t 2 + 1 ) ( t 2 + 1 ) 2 3 = A ( t 2 + 1 ) + B ( 3 t 2 + 1 ) 2 3 = t 2 ( A + 3 B ) + A + B (2) Equate the constant term on both the sides of equation (2). A + B = 2 3 (3) Equate the coefficient of the t 2 term on both the sides of equation (2)

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8.4, Problem 68E

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The value of x as a function of t using the given initial value problem.

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Chapter 8.4, Problem 67E

Chapter 8.4, Problem 69E

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