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

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.

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 4.6, 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

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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Limits and Continuity; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=9brk313DjV8;License: Standard YouTube License, CC-BY