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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

To estimate: The value of y ( 0.2 ) within ± 0.0001 by numerically approximating the integrals in the variation of parameters formula. The value of y ( 0.2 ) within ± 0.0001 is 0.3784 _ . Procedure used: 1. 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. 2. Simpson’s Rule: Let the interval [ a , b ] be divided into 2 n equal parts and let x 0 , x 1 , … x 2 n be the points of the partition. That is, x k = a + k h , k = 0 , 1 , … , 2 n where h = b − a 2 n . If y k = f ( x k ) , k = 0 , 1 , … , 2 n , then the Simpson’s rule approximation I S for the value of the definite integral ∫ a b f ( x ) d x is given by, I S = h 3 ∑ k = 1 n ( y 2 k − 2 + 4 y 2 k − 1 + y 2 k ) . Calculation: The given initial value problem is, y ″ + 10 y ′ + 25 y = e t 3 , y ( 0 ) = 1 , y ′ ( 0 ) = − 5 . The auxiliary equation to the corresponding homogeneous equation is r 2 + 10 r + 25 = 0 . Simplify r 2 + 10 r + 25 = 0 as, r 2 + 5 r + 5 r + 25 = 0 r ( r + 5 ) + 5 ( r + 5 ) = 0 ( r + 5 ) ( r + 5 ) = 0 r = − 5 , − 5 That is, the repeated roots are r 1 = − 5 and r 2 = − 5 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e − 5 t + c 2 t e − 5 t . Obtain the particular solution using the method of undetermined coefficients as follows: Compare y ″ + 10 y ′ + 25 y = e t 3 with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 10 , c = 25 and f = e t 3 . Since the general solution for the homogenous equation y ″ + 10 y ′ + 25 y = 0 is y ( t ) = c 1 e − 5 t + c 2 t e − 5 t , the two independent solutions are e − 5 t and t e − 5 t . Let y 1 ( t ) = e − 5 t and y 2 ( t ) = t e − 5 t . Then, y ′ 1 ( t ) = − 5 e − 5 t and y ′ 2 ( t ) = e − 5 t ( 1 − 5 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 ) e − 5 t + v 2 ( t ) t e − 5 t . That is, a = 1 , y 1 ( t ) = e − 5 t , y 2 ( t ) = t e − 5 t , y ′ 1 ( t ) = − 5 e − 5 t , y ′ 2 ( t ) = e − 5 t ( 1 − 5 t ) and f = e t 3 . The equation 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 ) = − e t 3 t e − 5 t 1 [ e − 5 t e − 5 t ( 1 − 5 t ) − ( − 5 e − 5 t ) t e − 5 t ] = − e t 3 t e − 5 t e − 5 t e − 5 t [ 1 − 5 t + 5 t ] = − e t 3 t e − 5 t e − 5 t e − 5 t = − e t 3 t e − 5 t = − t e 5 t + t 3 Thus, v 1 ( t ) = − ∫ ( t e 5 t + t 3 ) d t . The equation 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 ) = e t 3 e − 5 t 1 [ e − 5 t e − 5 t ( 1 − 5 t ) − ( − 5 e − 5 t ) t e − 5 t ] = e t 3 e − 5 t e − 5 t e − 5 t [ 1 − 5 t + 5 t ] = e t 3 e − 5 t e − 5 t e − 5 t = e t 3 e − 5 t = e 5 t + t 3 Thus, v 2 ( t ) = ∫ ( e 5 t + t 3 ) d t . Note that, the initial conditions are at t = 0 . That is, the range is from 1 to t . That is, v 1 ( t ) = − ∫ 0 t ( u e 5 u + u 3 ) d u and v 2 ( t ) = ∫ 0 t ( e 5 u + u 3 ) d u . The particular solution y p ( t ) = v 1 ( t ) e − 5 t + v 2 ( t ) t e − 5 t becomes, y p ( t ) = e − 5 t [ − ∫ 0 t ( u e 5 u + u 3 ) d u ] + t e − 5 t [ ∫ 0 t ( e 5 u + u 3 ) d u ] = − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u Combine the general solution y ( t ) = c 1 e − 5 t + c 2 t e − 5 t of the corresponding equation and the particular solution y p ( t ) = − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u to obtain y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u . Obtain the value of the constants using the initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = − 5 as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u , y ( 0 ) = c 1 e − 5 × 0 + c 2 ( 0 ) e − 5 × 0 − e − 5 × 0 ∫ 0 0 ( u e 5 u + u 3 ) d u + ( 0 ) e − 5 × 0 ∫ 0 0 ( e 5 u + u 3 ) d u 1 = c 1 e 0 + c 2 ( 0 ) − 1 ( 0 ) + 0 ( ∵ ∫ 0 0 f ( u ) d u = 0 ) c 1 = 1 Use the result d d t ∫ a t f ( u ) d u = f ( t ) and differentiate y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u as, y ′ ( t ) = [ − 5 c 1 e − 5 t + c 2 ( − 5 t e − 5 t + e − 5 t ) − ( e − 5 t ( t e 5 t + t 3 ) − 5 e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u ) + ( t e − 5 t ( e 5 t + t 3 ) + ( − 5 t e − 5 t + e − 5 t ) ∫ 0 t ( e 5 u + u 3 ) d u ) ] = [ − 5 c 1 e − 5 t + c 2 ( e − 5 t − 5 t e − 5 t ) − t e t 3 + 5 t − 5 t + 5 e − 5 t ∫ 0 t u e 5 u + u 3 d u + t e t 3 + 5 t − 5 t + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t e 5 u + u 3 d u ] Simplify further as, y ′ ( t ) = − 5 c 1 e − 5 t + c 2 ( e − 5 t − 5 t e − 5 t ) − t e t 3 + 5 e − 5 t ∫ 0 t u e 5 u + u 3 d u + t e t 3 + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t e 5 u + u 3 d u = − 5 c 1 e − 5 t − 5 c 2 t e − 5 t + c 2 e − 5 t + 5 e − 5 t ∫ 0 t u e 5 u + u 3 d u + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t e 5 u + u 3 d u = ( c 2 − 5 c 1 ) e − 5 t + 5 e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t ( e 5 u + u 3 ) d u Substitute t = 0 and y ′ ( 0 ) = − 5 in y ′ ( t ) = ( c 2 − 5 c 1 ) e − 5 t + 5 e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t ( e 5 u + u 3 ) d u , y ′ ( 0 ) = ( c 2 − 5 c 1 ) e − 5 × 0 + 5 e − 5 × 0 ∫ 0 0 ( u e 5 u + u 3 ) d u + ( e − 5 × 0 − 5 ( 0 ) e − 5 × 0 ) ∫ 0 0 ( e 5 u + u 3 ) d u − 5 = ( c 2 − 5 c 1 ) + 5 ( 0 ) + ( 1 − 5 ( 0 ) ) ( 0 ) ( ∵ ∫ 0 0 f ( u ) d u = 0 ) c 2 = − 5 + 5 c 1 Use c 2 = 1 and simplify further as, c 2 = − 5 + 5 ( 1 ) = 0 That is, c 2 = 0 and c 1 = 1 . The solution y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u gives, y ( t ) = e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u . Therefore, solution to the initial value problem y ″ + 10 y ′ + 25 y = e t 3 , y ( 0 ) = 1 , y ′ ( 0 ) = − 5 is y ( t ) = e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u _ . The aim is to obtain an approximation for y ( 0.2 ) within ± 0.0001 . Substitute t = 0.2 in y ( t ) = e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u , y ( 0.2 ) = e − 5 × 0.2 − e − 5 × 0.2 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + ( 0.2 ) e − 5 × 0.2 ∫ 0 0.2 ( e 5 u + u 3 ) d u = e − 1 − e − 1 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + 0.2 e − 1 ∫ 0 0.2 ( e 5 u + u 3 ) d u = 0.3678 − 0.3678 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + 0.07356 ∫ 0 0.2 ( e 5 u + u 3 ) d u The approximations ∫ 0 0.2 ( u e 5 u + u 3 ) d u and ∫ 0 0.2 ( e 5 u + u 3 ) d u is obtained using the Simpson’s rule approximation for [ a , b ] = [ 0 , 0.2 ] as follows: Assume n = 1 . Then, h = 0.2 − 0 2 = 0.1 and x k = 0 + 0.1 k , k = 0 , 1 , 2 . That is, x 0 = 0 , x 1 = 0.1 and x 2 = 0.2 . The approximation I S = h 3 ∑ k = 1 n ( y 2 k − 2 + 4 y 2 k − 1 + y 2 k ) becomes, I S = h 3 ( y 0 + 4 y 1 + y 2 ) . Consider ∫ 0 0.2 ( u e 5 u + u 3 ) d u , where y k = f ( x ) = x e 5 x + x 3 . The Simpson’s rule approximation for ∫ 0 0.2 ( u e 5 u + u 3 ) d u will be, I S = h 3 ( y 0 + 4 y 1 + y 2 ) = h 3 ( y ( x 0 ) + 4 y ( x 1 ) + y ( x 2 ) ) = h 3 ( x 0 e 5 x 0 + x 0 3 + 4 x 1 e 5 x 1 + x 1 3 + x 2 e 5 x 2 + x 2 3 ) Substitute h = 0.1 , x 0 = 0 , x 1 = 0.1 and x 2 = 0.2 and simplify as, I S = 0.1 3 ( ( 0 ) e 5 × 0 + 0 3 + 4 ( 0.1 ) e 5 × 0.1 + 0.1 3 + 0.2 e 5 × 0.2 + 0.2 3 ) = 0.1 3 ( 0.4 e 0.501 + 0.2 e 1.008 ) = 0.1 3 ( 0.4 ( 1.65 ) + 0.2 ( 2.74 ) ) = 0.1 3 ( 1.208 ) ≈ 0.0403 That is, ∫ 0 0.2 ( u e 5 u + u 3 ) d u ≈ 0.0403 . Similarly, the Simpson’s rule approximation for ∫ 0 0.2 ( e 5 u + u 3 ) d u will be, I S = h 3 ( y 0 + 4 y 1 + y 2 ) = h 3 ( y ( x 0 ) + 4 y ( x 1 ) + y ( x 2 ) ) = h 3 ( e 5 x 0 + x 0 3 + 4 e 5 x 1 + x 1 3 + e 5 x 2 + x 2 3 ) Substitute h = 0.1 , x 0 = 0 , x 1 = 0.1 and x 2 = 0.2 and simplify as, I S = 0.1 3 ( e 5 × 0 + 0 3 + 4 e 5 × 0.1 + 0.1 3 + e 5 × 0.2 + 0.2 3 ) = 0.1 3 ( e 0 + 4 e 0.501 + e 1.008 ) = 0.1 3 ( 1 + 4 ( 1.65 ) + 2.74 ) = 0.1 3 ( 10.34 ) ≈ 0.345 That is, ∫ 0 0.2 ( e 5 u + u 3 ) d u ≈ 0.345 . Thus, y ( 0.2 ) = 0.3678 − 0.3678 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + 0.07356 ∫ 0 0.2 ( e 5 u + u 3 ) d u becomes, y ( 0.2 ) = 0.3678 − 0.3678 ( 0.0403 ) + 0.07356 ( 0.345 ) = 0.3678 − 0.0148 + 0.0254 = 0.3784 Therefore, the value of y ( 0.2 ) within ± 0.0001 is 0.3784 _ .

To estimate: The value of y ( 0.2 ) within ± 0.0001 by numerically approximating the integrals in the variation of parameters formula. The value of y ( 0.2 ) within ± 0.0001 is 0.3784 _ . Procedure used: 1. 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. 2. Simpson’s Rule: Let the interval [ a , b ] be divided into 2 n equal parts and let x 0 , x 1 , … x 2 n be the points of the partition. That is, x k = a + k h , k = 0 , 1 , … , 2 n where h = b − a 2 n . If y k = f ( x k ) , k = 0 , 1 , … , 2 n , then the Simpson’s rule approximation I S for the value of the definite integral ∫ a b f ( x ) d x is given by, I S = h 3 ∑ k = 1 n ( y 2 k − 2 + 4 y 2 k − 1 + y 2 k ) . Calculation: The given initial value problem is, y ″ + 10 y ′ + 25 y = e t 3 , y ( 0 ) = 1 , y ′ ( 0 ) = − 5 . The auxiliary equation to the corresponding homogeneous equation is r 2 + 10 r + 25 = 0 . Simplify r 2 + 10 r + 25 = 0 as, r 2 + 5 r + 5 r + 25 = 0 r ( r + 5 ) + 5 ( r + 5 ) = 0 ( r + 5 ) ( r + 5 ) = 0 r = − 5 , − 5 That is, the repeated roots are r 1 = − 5 and r 2 = − 5 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e − 5 t + c 2 t e − 5 t . Obtain the particular solution using the method of undetermined coefficients as follows: Compare y ″ + 10 y ′ + 25 y = e t 3 with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 10 , c = 25 and f = e t 3 . Since the general solution for the homogenous equation y ″ + 10 y ′ + 25 y = 0 is y ( t ) = c 1 e − 5 t + c 2 t e − 5 t , the two independent solutions are e − 5 t and t e − 5 t . Let y 1 ( t ) = e − 5 t and y 2 ( t ) = t e − 5 t . Then, y ′ 1 ( t ) = − 5 e − 5 t and y ′ 2 ( t ) = e − 5 t ( 1 − 5 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 ) e − 5 t + v 2 ( t ) t e − 5 t . That is, a = 1 , y 1 ( t ) = e − 5 t , y 2 ( t ) = t e − 5 t , y ′ 1 ( t ) = − 5 e − 5 t , y ′ 2 ( t ) = e − 5 t ( 1 − 5 t ) and f = e t 3 . The equation 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 ) = − e t 3 t e − 5 t 1 [ e − 5 t e − 5 t ( 1 − 5 t ) − ( − 5 e − 5 t ) t e − 5 t ] = − e t 3 t e − 5 t e − 5 t e − 5 t [ 1 − 5 t + 5 t ] = − e t 3 t e − 5 t e − 5 t e − 5 t = − e t 3 t e − 5 t = − t e 5 t + t 3 Thus, v 1 ( t ) = − ∫ ( t e 5 t + t 3 ) d t . The equation 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 ) = e t 3 e − 5 t 1 [ e − 5 t e − 5 t ( 1 − 5 t ) − ( − 5 e − 5 t ) t e − 5 t ] = e t 3 e − 5 t e − 5 t e − 5 t [ 1 − 5 t + 5 t ] = e t 3 e − 5 t e − 5 t e − 5 t = e t 3 e − 5 t = e 5 t + t 3 Thus, v 2 ( t ) = ∫ ( e 5 t + t 3 ) d t . Note that, the initial conditions are at t = 0 . That is, the range is from 1 to t . That is, v 1 ( t ) = − ∫ 0 t ( u e 5 u + u 3 ) d u and v 2 ( t ) = ∫ 0 t ( e 5 u + u 3 ) d u . The particular solution y p ( t ) = v 1 ( t ) e − 5 t + v 2 ( t ) t e − 5 t becomes, y p ( t ) = e − 5 t [ − ∫ 0 t ( u e 5 u + u 3 ) d u ] + t e − 5 t [ ∫ 0 t ( e 5 u + u 3 ) d u ] = − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u Combine the general solution y ( t ) = c 1 e − 5 t + c 2 t e − 5 t of the corresponding equation and the particular solution y p ( t ) = − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u to obtain y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u . Obtain the value of the constants using the initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = − 5 as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u , y ( 0 ) = c 1 e − 5 × 0 + c 2 ( 0 ) e − 5 × 0 − e − 5 × 0 ∫ 0 0 ( u e 5 u + u 3 ) d u + ( 0 ) e − 5 × 0 ∫ 0 0 ( e 5 u + u 3 ) d u 1 = c 1 e 0 + c 2 ( 0 ) − 1 ( 0 ) + 0 ( ∵ ∫ 0 0 f ( u ) d u = 0 ) c 1 = 1 Use the result d d t ∫ a t f ( u ) d u = f ( t ) and differentiate y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u as, y ′ ( t ) = [ − 5 c 1 e − 5 t + c 2 ( − 5 t e − 5 t + e − 5 t ) − ( e − 5 t ( t e 5 t + t 3 ) − 5 e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u ) + ( t e − 5 t ( e 5 t + t 3 ) + ( − 5 t e − 5 t + e − 5 t ) ∫ 0 t ( e 5 u + u 3 ) d u ) ] = [ − 5 c 1 e − 5 t + c 2 ( e − 5 t − 5 t e − 5 t ) − t e t 3 + 5 t − 5 t + 5 e − 5 t ∫ 0 t u e 5 u + u 3 d u + t e t 3 + 5 t − 5 t + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t e 5 u + u 3 d u ] Simplify further as, y ′ ( t ) = − 5 c 1 e − 5 t + c 2 ( e − 5 t − 5 t e − 5 t ) − t e t 3 + 5 e − 5 t ∫ 0 t u e 5 u + u 3 d u + t e t 3 + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t e 5 u + u 3 d u = − 5 c 1 e − 5 t − 5 c 2 t e − 5 t + c 2 e − 5 t + 5 e − 5 t ∫ 0 t u e 5 u + u 3 d u + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t e 5 u + u 3 d u = ( c 2 − 5 c 1 ) e − 5 t + 5 e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t ( e 5 u + u 3 ) d u Substitute t = 0 and y ′ ( 0 ) = − 5 in y ′ ( t ) = ( c 2 − 5 c 1 ) e − 5 t + 5 e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + ( e − 5 t − 5 t e − 5 t ) ∫ 0 t ( e 5 u + u 3 ) d u , y ′ ( 0 ) = ( c 2 − 5 c 1 ) e − 5 × 0 + 5 e − 5 × 0 ∫ 0 0 ( u e 5 u + u 3 ) d u + ( e − 5 × 0 − 5 ( 0 ) e − 5 × 0 ) ∫ 0 0 ( e 5 u + u 3 ) d u − 5 = ( c 2 − 5 c 1 ) + 5 ( 0 ) + ( 1 − 5 ( 0 ) ) ( 0 ) ( ∵ ∫ 0 0 f ( u ) d u = 0 ) c 2 = − 5 + 5 c 1 Use c 2 = 1 and simplify further as, c 2 = − 5 + 5 ( 1 ) = 0 That is, c 2 = 0 and c 1 = 1 . The solution y ( t ) = c 1 e − 5 t + c 2 t e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u gives, y ( t ) = e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u . Therefore, solution to the initial value problem y ″ + 10 y ′ + 25 y = e t 3 , y ( 0 ) = 1 , y ′ ( 0 ) = − 5 is y ( t ) = e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u _ . The aim is to obtain an approximation for y ( 0.2 ) within ± 0.0001 . Substitute t = 0.2 in y ( t ) = e − 5 t − e − 5 t ∫ 0 t ( u e 5 u + u 3 ) d u + t e − 5 t ∫ 0 t ( e 5 u + u 3 ) d u , y ( 0.2 ) = e − 5 × 0.2 − e − 5 × 0.2 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + ( 0.2 ) e − 5 × 0.2 ∫ 0 0.2 ( e 5 u + u 3 ) d u = e − 1 − e − 1 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + 0.2 e − 1 ∫ 0 0.2 ( e 5 u + u 3 ) d u = 0.3678 − 0.3678 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + 0.07356 ∫ 0 0.2 ( e 5 u + u 3 ) d u The approximations ∫ 0 0.2 ( u e 5 u + u 3 ) d u and ∫ 0 0.2 ( e 5 u + u 3 ) d u is obtained using the Simpson’s rule approximation for [ a , b ] = [ 0 , 0.2 ] as follows: Assume n = 1 . Then, h = 0.2 − 0 2 = 0.1 and x k = 0 + 0.1 k , k = 0 , 1 , 2 . That is, x 0 = 0 , x 1 = 0.1 and x 2 = 0.2 . The approximation I S = h 3 ∑ k = 1 n ( y 2 k − 2 + 4 y 2 k − 1 + y 2 k ) becomes, I S = h 3 ( y 0 + 4 y 1 + y 2 ) . Consider ∫ 0 0.2 ( u e 5 u + u 3 ) d u , where y k = f ( x ) = x e 5 x + x 3 . The Simpson’s rule approximation for ∫ 0 0.2 ( u e 5 u + u 3 ) d u will be, I S = h 3 ( y 0 + 4 y 1 + y 2 ) = h 3 ( y ( x 0 ) + 4 y ( x 1 ) + y ( x 2 ) ) = h 3 ( x 0 e 5 x 0 + x 0 3 + 4 x 1 e 5 x 1 + x 1 3 + x 2 e 5 x 2 + x 2 3 ) Substitute h = 0.1 , x 0 = 0 , x 1 = 0.1 and x 2 = 0.2 and simplify as, I S = 0.1 3 ( ( 0 ) e 5 × 0 + 0 3 + 4 ( 0.1 ) e 5 × 0.1 + 0.1 3 + 0.2 e 5 × 0.2 + 0.2 3 ) = 0.1 3 ( 0.4 e 0.501 + 0.2 e 1.008 ) = 0.1 3 ( 0.4 ( 1.65 ) + 0.2 ( 2.74 ) ) = 0.1 3 ( 1.208 ) ≈ 0.0403 That is, ∫ 0 0.2 ( u e 5 u + u 3 ) d u ≈ 0.0403 . Similarly, the Simpson’s rule approximation for ∫ 0 0.2 ( e 5 u + u 3 ) d u will be, I S = h 3 ( y 0 + 4 y 1 + y 2 ) = h 3 ( y ( x 0 ) + 4 y ( x 1 ) + y ( x 2 ) ) = h 3 ( e 5 x 0 + x 0 3 + 4 e 5 x 1 + x 1 3 + e 5 x 2 + x 2 3 ) Substitute h = 0.1 , x 0 = 0 , x 1 = 0.1 and x 2 = 0.2 and simplify as, I S = 0.1 3 ( e 5 × 0 + 0 3 + 4 e 5 × 0.1 + 0.1 3 + e 5 × 0.2 + 0.2 3 ) = 0.1 3 ( e 0 + 4 e 0.501 + e 1.008 ) = 0.1 3 ( 1 + 4 ( 1.65 ) + 2.74 ) = 0.1 3 ( 10.34 ) ≈ 0.345 That is, ∫ 0 0.2 ( e 5 u + u 3 ) d u ≈ 0.345 . Thus, y ( 0.2 ) = 0.3678 − 0.3678 ∫ 0 0.2 ( u e 5 u + u 3 ) d u + 0.07356 ∫ 0 0.2 ( e 5 u + u 3 ) d u becomes, y ( 0.2 ) = 0.3678 − 0.3678 ( 0.0403 ) + 0.07356 ( 0.345 ) = 0.3678 − 0.0148 + 0.0254 = 0.3784 Therefore, the value of y ( 0.2 ) within ± 0.0001 is 0.3784 _ .

Solution Summary: The author explains how to estimate the value of y(0.2) by numerically approximating the integrals in the variation of parameters formula.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4.6, Problem 20E

Chapter 4.6, Problem 22E

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The double-angle trigonometric identity proof can be found on page 40 of the shared file. Prove the identity step-by-step using known trigonometric properties. Link: [https://drive.google.com/file/d/1RQ20 ZK-LSxp RyejKEMg1t2q15dbpVLCS/view? usp=sharing]

Refer to page 28 of the following file for a trigonometric identity verification problem. Use appropriate trigonometric transformations and properties to prove the result. Link: [https://drive.google.com/file/d/1RQ 20 ZK-LSxp RyejKEMg1t2q15dbpVLCS/view? usp=sharing]

The differential equation problem is on page 30 of the file below. Solve the equation step-by- step using an appropriate method such as substitution or integrating factors. Link: [https://drive.google.com/file/d/1RQ20 ZK-LSxp RyejKEMg 1t2q15dbpVLCS/view? usp=sharing]

Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

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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