Explanation Formula used: The governing equation of motion is m d 2 y d t 2 + δ d y d t + k y = 0 . Theorem used: “If r 1 and r 2 are the two real and unequal roots of the auxiliary equation a r 2 + b r + c = 0 , then y = c 1 e r 1 x + c 2 e r 2 x is the general solution of a y ″ + b y ′ + c y = 0 ”. The general solution y = y ( x ) to the nonhomogeneous differential equation a y ″ + b y ′ + c y = G ( x ) has the form y = y c + y p , where the complementary solution y c is the general solution to the associated homogeneous equation a y ″ + b y ′ + c y = 0 , and y p is any particular solution to the nonhomogeneous equation a y ″ + b y ′ + c y = G ( x ) ”. Calculation: The mass ( m ) of object is 10kg and the value of spring constant( k) is k = 140 N/m . The instantaneous velocity δ = 90 N-sec/m . And the external force is f ( t ) = 5 sin t Now, the governing equation of motion is: m d 2 y d t 2 + δ d y d t + k y = f ( t ) With initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = − 1 Substitute the value of m , k , f ( t ) = 5 sin t , and δ in m d 2 y d t 2 + δ d y d t + k y = f ( t ) . 10 d 2 y d t 2 + 90 d y d t + 140 y = 5 sin t The auxiliary equation for the above differential equation is: 10 r 2 y + 90 r y + 140 y = 0 ( r 2 + 9 r + 14 ) y = 0 ( r 2 + 7 r + 2 r + 8 ) y = 0 ( r + 2 ) ( r + 7 ) y = 0 . Splitting the middle term to calculate the value of r . The roots of the above equation are ( r + 2 ) ( r + 7 ) = 0 The value of r = − 2 , − 7 . So, the general solution to the above differential equation is y c ( t ) = c 1 e − 2 t + c 2 e − 7 t To solve the nonhomogeneous solution of the form y p ( t ) = A sin t + B cos t Differentiate the equation y p = A sin t + B cos t is as follows: y ′ p ( t ) = A cos t − B sin t Again, differentiate the above equation y ′ p ( t ) = A cos t − B sin t is as follows: y ′ ′ p ( t ) = − A sin t − B cos t Substitute the derivative of y p ( t ) in the given equation 10 d 2 y d t 2 + 90 d y d t + 140 y = 5 sin t and find that A and B must satisfy the equation. 10 ( − A sin t − B cos t ) + 90 ( A cos t − B sin t ) + 140 ( A sin t + B cos t ) = 5 sin t ( 130 A − 90 B ) sin t + ( 90 A + 130 B ) cos t = 5 sin t Compare both coefficients of the above equation is as follows: 130 A − 90 B = 5 A = 13 500 To calculate the value of B is as follows: B = − 9 500 Now, substitute the value of A and B in y p ( t ) = A sin t + B cos t . y p ( t ) = 13 500 sin t − 9 500 cos t The general solution of the nonhomogeneous equation is: y ( t ) = y c ( t ) + y p ( t ) Substitute the value of y c ( t ) = c 1 e − 2 t + c 2 e − 7 t and y p ( t ) = 13 500 sin t − 9 500 cos t in y ( t ) = y c ( t ) + y p ( t )

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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43–46. Directions of change Consider the following functions f and points P. Sketch the xy-plane showing P and the level curve through P. Indicate (as in Figure 15.52) the directions of maximum increase, maximum decrease, and no change for f. ■ 45. f(x, y) = x² + xy + y² + 7; P(−3, 3)

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Chapter 17 Solutions

University Calculus: Early Transcendentals (4th Edition)

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