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Math Advanced Math Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

The fraction of the normal year T that Earth would take to splash into the sun. The fraction of the normal year T that Earth would take to splash into the sun is 1 4 2 _ . Theorem used: The Energy Integral Lemma: Let y ( t ) be a solution to the differential equation y ″ = f ( y ) , where f ( y ) is a continuous function that does not depend on y ′ or the independent variable t . Let F ( y ) be an indefinite integral of f ( y ) . That is, f ( y ) = d d y F ( y ) . Then, the quantity E ( t ) = 1 2 y ′ ( t ) 2 − F ( y ( t ) ) is constant. That is, d d t E ( t ) = 0 . The equation y ″ = f ( y ) can be reduced to 1 2 ( y ′ ) 2 − F ( y ) = K , for some constant K , which is equivalent to the separable first-order equation d y d t = ± 2 [ F ( y ) + K ] having the implicit two-parameter solution t = ± ∫ d y 2 [ F ( y ) + K ] + c . Calculation: By the Newon’s second law, m d 2 r d t 2 = − G M m r 2 . Rewrite the equation as, d 2 r d t 2 = − G M r 2 . Here, f ( r ) = − G M r 2 . The indefinite integral of f ( r ) will be, F ( r ) = ∫ f ( r ) d r = ∫ ( − G M r 2 ) d r = − G M ∫ r − 2 d r = − G M ( − r − 1 ) + C = G M r + C Assuming the constant to be zero, implies F ( r ) = G M r . That is, f ( r ) = d d r ( G M r ) . Thus, the equation E ( t ) = 1 2 y ′ ( t ) 2 − F ( y ( t ) ) for r becomes, E ( r ) = 1 2 ( r ′ ( t ) ) 2 − F ( r ( t ) ) = 1 2 ( r ′ ( t ) ) 2 − G M r Given that, Earth revolves around the sun in an approximately circular orbit of radius r = a . That is, r ( 0 ) = a and r ′ ( 0 ) = 0 . Substitute r ( 0 ) = a and r ′ ( 0 ) = 0 in E ( r ) = 1 2 ( r ′ ( t ) ) 2 − G M r as, E ( r ) = 1 2 ( 0 ) 2 − G M a = − G M a Thus, the equation E ( r ) = 1 2 ( r ′ ( t ) ) 2 − G M r becomes, − G M a = 1 2 ( r ′ ( t ) ) 2 − G M r 1 2 ( r ′ ( t ) ) 2 = G M r − G M a ( r ′ ( t ) ) 2 = 2 G M ( 1 r − 1 a ) r ′ ( t ) = ± 2 G M ( 1 r − 1 a ) Here, the radius with respect to time is a decreasing function. Thus, d r d t = − 2 G M ( 1 r − 1 a ) d r 2 G M ( 1 r − 1 a ) = − d t 1 2 G M × d r 1 r − 1 a = − d t d r 1 r − 1 a = − 2 G M d t Integrate both sides as, ∫ d r 1 r − 1 a = − 2 G M ∫ d t . Obtain the value of ∫ d r 1 r − 1 a as follows: Let u = 1 r − 1 a . Then, u 2 = 1 r − 1 a 1 r = u 2 + 1 a 1 r = a u 2 + 1 a r = a a u 2 + 1 The derivative of 1 r = u 2 + 1 a is, − 1 r 2 d r = 2 u d u d r = − 2 u r 2 d u d r = − 2 u ( a a u 2 + 1 ) 2 d u ( ∵ r = a a u 2 + 1 ) d r = − 2 a 2 u ( a u 2 + 1 ) 2 d u Thus, ∫ d r 1 r − 1 a in terms of u will be, ∫ d r 1 r − 1 a = ∫ 1 u ( − 2 a 2 u ( a u 2 + 1 ) 2 d u ) = − 2 a 2 ∫ 1 ( a u 2 + 1 ) 2 d u Let v = a u . That is, v 2 = a u 2 . The derivative of v = a u is, d v = a d u d u = d v a Thus, − 2 a 2 ∫ 1 ( a u 2 + 1 ) 2 d u in terms of v will be, − 2 a 2 ∫ 1 ( a u 2 + 1 ) 2 d u = − 2 a 2 ∫ 1 ( v 2 + 1 ) 2 ( d v a ) = − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v Use the trigonometric substitution w = tan − 1 v . That is, v = tan w . Then, d v = sec 2 w d w . Thus, − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v becomes, − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v = − 2 a a ∫ 1 ( tan 2 w + 1 ) 2 ( sec 2 w d w ) = − 2 a a ∫ 1 ( sec 2 w ) 2 ( sec 2 w d w ) = − 2 a a ∫ 1 sec 2 w d w = − 2 a a ∫ cos 2 w d w Simplify further as, − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v = − 2 a a ∫ 1 + cos 2 w 2 d w = − a a ∫ d w − a a ∫ cos 2 w d w = − a a w − a a ( sin 2 w 2 ) = − a a ( w + sin 2 w 2 ) = − a a 2 ( 2 w + sin 2 w ) That is, ∫ d r 1 r − 1 a = − a a 2 ( 2 w + sin 2 w ) . Using the substitutions w = tan − 1 v , v = a u and u = 1 r − 1 a obtain w = tan − 1 ( a 1 r − 1 a ) . Thus, ∫ d r 1 r − 1 a = − a a 2 ( 2 w + sin 2 w ) becomes, ∫ d r 1 r − 1 a = − a a 2 ( 2 tan − 1 ( a 1 r − 1 a ) + sin 2 ( tan − 1 ( a 1 r − 1 a ) ) ) = − a 3 2 [ tan − 1 ( a a − r a r ) − 1 2 sin 2 ( tan − 1 ( a a − r a r ) ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − 1 2 sin 2 ( tan − 1 ( a − r a r ) ) ] Use the identity sin ( 2 tan − 1 x ) = 2 x 1 + x 2 and simplify as, ∫ d r 1 r − 1 a = − a 3 2 [ tan − 1 ( a − r a r ) − 1 2 ( 2 ( a − r a r ) 1 + ( a − r a r ) 2 ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − ( a − r a r ) ( 1 + a − r a 2 r ) ] Simplify as, ∫ d r 1 r − 1 a = − a 3 2 [ tan − 1 ( a − r a r ) − ( a − r a r ) ( a 2 r + a − r a 2 r ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − ( a − r a r × a 2 r a 2 r + a − r ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − a r ( a − r ) a 2 r + a − r ] Thus, ∫ d r 1 r − 1 a = − 2 G M ∫ d t becomes, − a 3 2 [ tan − 1 ( a − r a r ) − a r ( a − r ) a 2 r + a − r ] = − 2 G M t . Since the fraction of the normal year T that Earth would take to splash into the sun is to be obtained, r = 0 . When r = 0 , the equation − a 3 2 [ tan − 1 ( a − r a r ) − a r ( a − r ) a 2 r + a − r ] = − 2 G M t becomes, − a 3 2 [ tan − 1 ( a − 0 a × 0 ) − a × 0 a ] = − 2 G M t − a 3 2 tan − 1 ( a 0 ) = − 2 G M t − a 3 2 tan − 1 ( ∞ ) = − 2 G M t t = a 3 2 2 G M tan − 1 ( ∞ ) t = a 3 2 2 G M ( π 2 ) Given that, T = 2 π a 3 G M . Thus, the ratio of t to T will be, t T = ( a 3 2 π 2 2 G M ) ( 2 π a 3 G M ) = a 3 2 π 2 2 G M × 1 2 π G M a 3 2 = 1 4 2 Therefore, the fraction of the normal year T that Earth would take to splash into the sun is 1 4 2 _ .

The fraction of the normal year T that Earth would take to splash into the sun. The fraction of the normal year T that Earth would take to splash into the sun is 1 4 2 _ . Theorem used: The Energy Integral Lemma: Let y ( t ) be a solution to the differential equation y ″ = f ( y ) , where f ( y ) is a continuous function that does not depend on y ′ or the independent variable t . Let F ( y ) be an indefinite integral of f ( y ) . That is, f ( y ) = d d y F ( y ) . Then, the quantity E ( t ) = 1 2 y ′ ( t ) 2 − F ( y ( t ) ) is constant. That is, d d t E ( t ) = 0 . The equation y ″ = f ( y ) can be reduced to 1 2 ( y ′ ) 2 − F ( y ) = K , for some constant K , which is equivalent to the separable first-order equation d y d t = ± 2 [ F ( y ) + K ] having the implicit two-parameter solution t = ± ∫ d y 2 [ F ( y ) + K ] + c . Calculation: By the Newon’s second law, m d 2 r d t 2 = − G M m r 2 . Rewrite the equation as, d 2 r d t 2 = − G M r 2 . Here, f ( r ) = − G M r 2 . The indefinite integral of f ( r ) will be, F ( r ) = ∫ f ( r ) d r = ∫ ( − G M r 2 ) d r = − G M ∫ r − 2 d r = − G M ( − r − 1 ) + C = G M r + C Assuming the constant to be zero, implies F ( r ) = G M r . That is, f ( r ) = d d r ( G M r ) . Thus, the equation E ( t ) = 1 2 y ′ ( t ) 2 − F ( y ( t ) ) for r becomes, E ( r ) = 1 2 ( r ′ ( t ) ) 2 − F ( r ( t ) ) = 1 2 ( r ′ ( t ) ) 2 − G M r Given that, Earth revolves around the sun in an approximately circular orbit of radius r = a . That is, r ( 0 ) = a and r ′ ( 0 ) = 0 . Substitute r ( 0 ) = a and r ′ ( 0 ) = 0 in E ( r ) = 1 2 ( r ′ ( t ) ) 2 − G M r as, E ( r ) = 1 2 ( 0 ) 2 − G M a = − G M a Thus, the equation E ( r ) = 1 2 ( r ′ ( t ) ) 2 − G M r becomes, − G M a = 1 2 ( r ′ ( t ) ) 2 − G M r 1 2 ( r ′ ( t ) ) 2 = G M r − G M a ( r ′ ( t ) ) 2 = 2 G M ( 1 r − 1 a ) r ′ ( t ) = ± 2 G M ( 1 r − 1 a ) Here, the radius with respect to time is a decreasing function. Thus, d r d t = − 2 G M ( 1 r − 1 a ) d r 2 G M ( 1 r − 1 a ) = − d t 1 2 G M × d r 1 r − 1 a = − d t d r 1 r − 1 a = − 2 G M d t Integrate both sides as, ∫ d r 1 r − 1 a = − 2 G M ∫ d t . Obtain the value of ∫ d r 1 r − 1 a as follows: Let u = 1 r − 1 a . Then, u 2 = 1 r − 1 a 1 r = u 2 + 1 a 1 r = a u 2 + 1 a r = a a u 2 + 1 The derivative of 1 r = u 2 + 1 a is, − 1 r 2 d r = 2 u d u d r = − 2 u r 2 d u d r = − 2 u ( a a u 2 + 1 ) 2 d u ( ∵ r = a a u 2 + 1 ) d r = − 2 a 2 u ( a u 2 + 1 ) 2 d u Thus, ∫ d r 1 r − 1 a in terms of u will be, ∫ d r 1 r − 1 a = ∫ 1 u ( − 2 a 2 u ( a u 2 + 1 ) 2 d u ) = − 2 a 2 ∫ 1 ( a u 2 + 1 ) 2 d u Let v = a u . That is, v 2 = a u 2 . The derivative of v = a u is, d v = a d u d u = d v a Thus, − 2 a 2 ∫ 1 ( a u 2 + 1 ) 2 d u in terms of v will be, − 2 a 2 ∫ 1 ( a u 2 + 1 ) 2 d u = − 2 a 2 ∫ 1 ( v 2 + 1 ) 2 ( d v a ) = − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v Use the trigonometric substitution w = tan − 1 v . That is, v = tan w . Then, d v = sec 2 w d w . Thus, − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v becomes, − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v = − 2 a a ∫ 1 ( tan 2 w + 1 ) 2 ( sec 2 w d w ) = − 2 a a ∫ 1 ( sec 2 w ) 2 ( sec 2 w d w ) = − 2 a a ∫ 1 sec 2 w d w = − 2 a a ∫ cos 2 w d w Simplify further as, − 2 a a ∫ 1 ( v 2 + 1 ) 2 d v = − 2 a a ∫ 1 + cos 2 w 2 d w = − a a ∫ d w − a a ∫ cos 2 w d w = − a a w − a a ( sin 2 w 2 ) = − a a ( w + sin 2 w 2 ) = − a a 2 ( 2 w + sin 2 w ) That is, ∫ d r 1 r − 1 a = − a a 2 ( 2 w + sin 2 w ) . Using the substitutions w = tan − 1 v , v = a u and u = 1 r − 1 a obtain w = tan − 1 ( a 1 r − 1 a ) . Thus, ∫ d r 1 r − 1 a = − a a 2 ( 2 w + sin 2 w ) becomes, ∫ d r 1 r − 1 a = − a a 2 ( 2 tan − 1 ( a 1 r − 1 a ) + sin 2 ( tan − 1 ( a 1 r − 1 a ) ) ) = − a 3 2 [ tan − 1 ( a a − r a r ) − 1 2 sin 2 ( tan − 1 ( a a − r a r ) ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − 1 2 sin 2 ( tan − 1 ( a − r a r ) ) ] Use the identity sin ( 2 tan − 1 x ) = 2 x 1 + x 2 and simplify as, ∫ d r 1 r − 1 a = − a 3 2 [ tan − 1 ( a − r a r ) − 1 2 ( 2 ( a − r a r ) 1 + ( a − r a r ) 2 ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − ( a − r a r ) ( 1 + a − r a 2 r ) ] Simplify as, ∫ d r 1 r − 1 a = − a 3 2 [ tan − 1 ( a − r a r ) − ( a − r a r ) ( a 2 r + a − r a 2 r ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − ( a − r a r × a 2 r a 2 r + a − r ) ] = − a 3 2 [ tan − 1 ( a − r a r ) − a r ( a − r ) a 2 r + a − r ] Thus, ∫ d r 1 r − 1 a = − 2 G M ∫ d t becomes, − a 3 2 [ tan − 1 ( a − r a r ) − a r ( a − r ) a 2 r + a − r ] = − 2 G M t . Since the fraction of the normal year T that Earth would take to splash into the sun is to be obtained, r = 0 . When r = 0 , the equation − a 3 2 [ tan − 1 ( a − r a r ) − a r ( a − r ) a 2 r + a − r ] = − 2 G M t becomes, − a 3 2 [ tan − 1 ( a − 0 a × 0 ) − a × 0 a ] = − 2 G M t − a 3 2 tan − 1 ( a 0 ) = − 2 G M t − a 3 2 tan − 1 ( ∞ ) = − 2 G M t t = a 3 2 2 G M tan − 1 ( ∞ ) t = a 3 2 2 G M ( π 2 ) Given that, T = 2 π a 3 G M . Thus, the ratio of t to T will be, t T = ( a 3 2 π 2 2 G M ) ( 2 π a 3 G M ) = a 3 2 π 2 2 G M × 1 2 π G M a 3 2 = 1 4 2 Therefore, the fraction of the normal year T that Earth would take to splash into the sun is 1 4 2 _ .

Solution Summary: The author explains the Energy Integral Lemma: Let y(t) be a solution to the differential equation.

Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

9th Edition

ISBN: 9780134768748

Author: Nagle

Publisher: PEARSON

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Chapter 4 Solutions

Fundamentals of Differential Equations plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (9th Edition) (Nagle, Saff & Snider, Fundamentals of Differential Equations)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

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