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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

The general solution to the differential equation y ″ + y = tan t + e 3 t − 1 . The general solution to the differential equation y ″ + y = tan t + e 3 t − 1 is y ( t ) = c 1 cos t + c 2 sin t − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 _ . Procedure used: 1. Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e r t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with: (i) s = 0 , if r is not a root of the associated auxiliary equation; (ii) s = 1 , if r is a simple root of the associated auxiliary equation; (iii) s = 2 , if r is a double root of the associated auxiliary equation. A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. 2. Method of variation of parameters: The following are the steps to obtain a particular solution to a y ″ + b y ′ + c y = f . (i) Find two linearly independent solutions { y 1 ( t ) , y 2 ( t ) } to the corresponding homogenous equation and take y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . (ii) Determine v 1 ( t ) and v 2 ( t ) by solving the below system for v ′ 1 ( t ) and v ′ 2 ( t ) and integrating. y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a This can be simplified using Cramer’s rule as, v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] and v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] (iii) Substitute v 1 ( t ) and v 2 ( t ) into the expression for y p ( t ) to obtain a particular solution. 3. Superposition principle: If y 1 is a solution to the differential equation a y ″ + b y ′ + c y = f 1 ( t ) and y 2 is a solution to a y ″ + b y ′ + c y = f 2 ( t ) , then for any constants k 1 and k 2 , the function k 1 y 1 + k 2 y 2 is a solution to the differential equation a y ″ + b y ′ + c y = k 1 f 1 ( t ) + k 2 f 2 ( t ) . Calculation: The given differential equation is, y ″ + y = tan t + e 3 t − 1 . The corresponding auxiliary equation is r 2 + 1 = 0 . Simplify r 2 + 1 = 0 as, r 2 = − 1 r = ± − 1 r = ± i ( ∵ i = − 1 ) That is, the roots are r 1 = i and r 2 = − i . Here, α = 0 and β = 1 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos t + c 2 sin t . The differential equation y ″ + y = tan t + e 3 t − 1 is split into equations, y ″ + y = tan t and y ″ + y = e 3 t − 1 . The solution to y ″ + y = tan t is obtained using variation of parameters, the solution to y ″ + y = e 3 t − 1 is obtained using undetermined coefficients and the superposition principle is applied to combine both solutions. Obtain the solution to y ″ + y = e 3 t − 1 using the method of undetermined coefficients as follows: Since the RHS of the differential equation y ″ + y = e 3 t − 1 is e 3 t − 1 . The particular solution will be of the form y p ( t ) = A e 3 t − B . The second derivative of y p ( t ) = A e 3 t − B is y ″ p ( t ) = 9 A e 3 t . Thus, y ″ + y for y p will be, 9 A e 3 t + A e 3 t − B = 10 A e 3 t − B . Set 10 A e 3 t − B = e 3 t − 1 and comparing the coefficients will results in 10 A = 1 and − B = − 1 . On simplification, A = 1 10 and B = 1 . Thus, the particular solution y p ( t ) = A e 3 t − B becomes y p ( t ) = 1 10 e 3 t − 1 . That is, the solution to y ″ + y = e 3 t − 1 using the method of undetermined coefficients is y p ( t ) = 1 10 e 3 t − 1 . Obtain the solution to y ″ + y = tan t is obtained using variation of parameters as follows: Since the general solution for the homogenous equation y ″ + y = 0 is y ( t ) = c 1 cos t + c 2 sin t , the two independent solutions are cos t and sin t . Let y 1 ( t ) = cos t and y 2 ( t ) = sin t . Then, y ′ 1 ( t ) = − sin t and y ′ 2 ( t ) = cos t . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as, y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t . Obtain the Wronskian y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) as, W = | cos t sin t − sin t cos t | = cos 2 t + sin 2 t = 1 Compare y ″ + y = tan t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 1 and f = tan t . The value of v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 1 ( t ) = ( − tan t ) sin t 1 [ 1 ] = − tan t sin t The value of v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 2 ( t ) = ( tan t ) ( cos t ) 1 [ 1 ] = ( sin t cos t ) ( cos t ) = sin t Integrating v ′ 1 ( t ) = − tan t sin t gives, v 1 ( t ) = ∫ v 1 ′ ( t ) d t = − ∫ tan t sin t d t = − ∫ sin 2 t cos t d t = − ∫ 1 − cos 2 t cos t d t Simplify further as, v 1 ( t ) = − ∫ 1 cos t d t + ∫ cos t d t = − ∫ sec t d t + ∫ cos t d t = − ln | sec t + tan t | + sin t + C 1 Integrating v ′ 2 ( t ) = sin t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) = ∫ sin t d t = − cos t + C 2 Since particular solution is to be obtained, take both C 1 and C 2 zero for simplicity. That is, v 1 ( t ) = − ln | sec t + tan t | + sin t and v 2 ( t ) = − cos t . Thus, the particular solution y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t becomes, y p ( t ) = [ − ln | sec t + tan t | + sin t ] cos t + ( − cos t ) sin t = − cos t ln | sec t + tan t | + cos t sin t − cos t sin t = − cos t ln | sec t + tan t | That is, the solution to y ″ + y = tan t using the method of variation of parameters is y p ( t ) = − cos t ln | sec t + tan t | . By superposition principle, the particular solution to y ″ + y = tan t + e 3 t − 1 will be, y p ( t ) = − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 . Combine the general solution y ( t ) = c 1 cos t + c 2 sin t of the corresponding homogenous equation and the particular solution y p ( t ) = − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 to obtain y ( t ) = c 1 cos t + c 2 sin t − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 . Therefore, the general solution to the differential equation y ″ + y = tan t + e 3 t − 1 is y ( t ) = c 1 cos t + c 2 sin t − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 _ .

The general solution to the differential equation y ″ + y = tan t + e 3 t − 1 . The general solution to the differential equation y ″ + y = tan t + e 3 t − 1 is y ( t ) = c 1 cos t + c 2 sin t − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 _ . Procedure used: 1. Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e r t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with: (i) s = 0 , if r is not a root of the associated auxiliary equation; (ii) s = 1 , if r is a simple root of the associated auxiliary equation; (iii) s = 2 , if r is a double root of the associated auxiliary equation. A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. 2. Method of variation of parameters: The following are the steps to obtain a particular solution to a y ″ + b y ′ + c y = f . (i) Find two linearly independent solutions { y 1 ( t ) , y 2 ( t ) } to the corresponding homogenous equation and take y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . (ii) Determine v 1 ( t ) and v 2 ( t ) by solving the below system for v ′ 1 ( t ) and v ′ 2 ( t ) and integrating. y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a This can be simplified using Cramer’s rule as, v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] and v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] (iii) Substitute v 1 ( t ) and v 2 ( t ) into the expression for y p ( t ) to obtain a particular solution. 3. Superposition principle: If y 1 is a solution to the differential equation a y ″ + b y ′ + c y = f 1 ( t ) and y 2 is a solution to a y ″ + b y ′ + c y = f 2 ( t ) , then for any constants k 1 and k 2 , the function k 1 y 1 + k 2 y 2 is a solution to the differential equation a y ″ + b y ′ + c y = k 1 f 1 ( t ) + k 2 f 2 ( t ) . Calculation: The given differential equation is, y ″ + y = tan t + e 3 t − 1 . The corresponding auxiliary equation is r 2 + 1 = 0 . Simplify r 2 + 1 = 0 as, r 2 = − 1 r = ± − 1 r = ± i ( ∵ i = − 1 ) That is, the roots are r 1 = i and r 2 = − i . Here, α = 0 and β = 1 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos t + c 2 sin t . The differential equation y ″ + y = tan t + e 3 t − 1 is split into equations, y ″ + y = tan t and y ″ + y = e 3 t − 1 . The solution to y ″ + y = tan t is obtained using variation of parameters, the solution to y ″ + y = e 3 t − 1 is obtained using undetermined coefficients and the superposition principle is applied to combine both solutions. Obtain the solution to y ″ + y = e 3 t − 1 using the method of undetermined coefficients as follows: Since the RHS of the differential equation y ″ + y = e 3 t − 1 is e 3 t − 1 . The particular solution will be of the form y p ( t ) = A e 3 t − B . The second derivative of y p ( t ) = A e 3 t − B is y ″ p ( t ) = 9 A e 3 t . Thus, y ″ + y for y p will be, 9 A e 3 t + A e 3 t − B = 10 A e 3 t − B . Set 10 A e 3 t − B = e 3 t − 1 and comparing the coefficients will results in 10 A = 1 and − B = − 1 . On simplification, A = 1 10 and B = 1 . Thus, the particular solution y p ( t ) = A e 3 t − B becomes y p ( t ) = 1 10 e 3 t − 1 . That is, the solution to y ″ + y = e 3 t − 1 using the method of undetermined coefficients is y p ( t ) = 1 10 e 3 t − 1 . Obtain the solution to y ″ + y = tan t is obtained using variation of parameters as follows: Since the general solution for the homogenous equation y ″ + y = 0 is y ( t ) = c 1 cos t + c 2 sin t , the two independent solutions are cos t and sin t . Let y 1 ( t ) = cos t and y 2 ( t ) = sin t . Then, y ′ 1 ( t ) = − sin t and y ′ 2 ( t ) = cos t . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as, y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t . Obtain the Wronskian y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) as, W = | cos t sin t − sin t cos t | = cos 2 t + sin 2 t = 1 Compare y ″ + y = tan t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 1 and f = tan t . The value of v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 1 ( t ) = ( − tan t ) sin t 1 [ 1 ] = − tan t sin t The value of v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] becomes, v ′ 2 ( t ) = ( tan t ) ( cos t ) 1 [ 1 ] = ( sin t cos t ) ( cos t ) = sin t Integrating v ′ 1 ( t ) = − tan t sin t gives, v 1 ( t ) = ∫ v 1 ′ ( t ) d t = − ∫ tan t sin t d t = − ∫ sin 2 t cos t d t = − ∫ 1 − cos 2 t cos t d t Simplify further as, v 1 ( t ) = − ∫ 1 cos t d t + ∫ cos t d t = − ∫ sec t d t + ∫ cos t d t = − ln | sec t + tan t | + sin t + C 1 Integrating v ′ 2 ( t ) = sin t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) = ∫ sin t d t = − cos t + C 2 Since particular solution is to be obtained, take both C 1 and C 2 zero for simplicity. That is, v 1 ( t ) = − ln | sec t + tan t | + sin t and v 2 ( t ) = − cos t . Thus, the particular solution y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t becomes, y p ( t ) = [ − ln | sec t + tan t | + sin t ] cos t + ( − cos t ) sin t = − cos t ln | sec t + tan t | + cos t sin t − cos t sin t = − cos t ln | sec t + tan t | That is, the solution to y ″ + y = tan t using the method of variation of parameters is y p ( t ) = − cos t ln | sec t + tan t | . By superposition principle, the particular solution to y ″ + y = tan t + e 3 t − 1 will be, y p ( t ) = − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 . Combine the general solution y ( t ) = c 1 cos t + c 2 sin t of the corresponding homogenous equation and the particular solution y p ( t ) = − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 to obtain y ( t ) = c 1 cos t + c 2 sin t − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 . Therefore, the general solution to the differential equation y ″ + y = tan t + e 3 t − 1 is y ( t ) = c 1 cos t + c 2 sin t − cos t ln | sec t + tan t | + 1 10 e 3 t − 1 _ .

Solution Summary: The author explains the general solution to the differential equation y′′+y=mathrmtan

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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Chapter 4 Solutions

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

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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