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

To approximate: The solution to the initial value problem d y d x + sin 2 x 2 ( 1 + sin 2 x ) y = 1 , y ( 0 ) = 0 at x = 1 by using Simpson’s rule with the accuracy of three decimal places. The value of the solution y ( x ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t at x = 1 is 0.860 _ approximately. Method used: 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 [ y 0 + 4 y 1 + 2 y 2 + 4 y 3 + ⋯ + 2 y 2 n − 2 + 4 y 2 n − 1 + y 2 n ] = h 3 ∑ k = 1 n ( y 2 k − 2 + 4 y 2 k − 1 + y 2 k ) . .” Calculation: The differential equation is, d y d x + sin 2 x 2 ( 1 + sin 2 x ) y = 1 , y ( 0 ) = 0 Rewrite the differential equation of the standard form d y d x + P ( x ) y = Q ( x ) as follows. d y d x + sin 2 x 2 ( 1 + sin 2 x ) y = 1 ... ( 1 ) Here, P ( x ) = sin 2 x 2 ( 1 + sin 2 x ) and Q ( x ) = 1 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ sin 2 x 2 ( 1 + sin 2 x ) d x ] = exp [ 1 2 ∫ sin 2 x ( 1 + sin 2 x ) d x ] = exp [ 1 2 ∫ 2 cos x sin x ( 1 + sin 2 x ) d x ] ( ∵ sin 2 x = 2 cos x sin x ) = exp [ ∫ cos x sin x ( 1 + sin 2 x ) d x ] Integrate by the method of U-substitution Take u = sin x , then d u = cos x d x , d x = 1 cos x d u . Thus, the integral becomes, μ ( x ) = exp [ ∫ cos x u ( 1 + u 2 ) 1 cos x d u ] μ ( x ) = exp [ ∫ u ( 1 + u 2 ) d u ] Again, integrate by the method of U-substitution Take v = 1 + u 2 , then d v = 2 u d u , d u = 1 2 u d v . Thus, the integral becomes, μ ( x ) = exp [ ∫ u v 1 2 u d v ] = exp [ ∫ 1 2 v d v ] = exp [ 1 2 ∫ 1 v d v ] = exp [ 1 2 ln | v | ] Simplify further, μ ( x ) = exp [ 1 2 ln | 1 + u 2 | ] ( Substitute back, v = 1 + u 2 ) = exp [ 1 2 ln | 1 + sin 2 x | ] ( Substitute back, u = sin x ) = e ln ( | 1 + sin 2 x | ) 1 2 = 1 + sin 2 x Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . 1 + sin 2 x d y d x + 1 + sin 2 x sin 2 x 2 ( 1 + sin 2 x ) y = 1 + sin 2 x d d x [ 1 + sin 2 x y ] = 1 + sin 2 x The value of x starts at 0 because the initial condition is y ( 0 ) = 0 . Also, the end value is unknown. Therefore, the limits of integration would be [ 0 , x ] . Integrate both sides of the equation with respect to t from t = 0 to t = x . [ 1 + sin 2 t y ] 0 x = ∫ 0 x 1 + sin 2 t d t 1 + sin 2 x y − 1 + sin 2 ( 0 ) ⋅ y ( 0 ) = ∫ 0 x 1 + sin 2 t d t 1 + sin 2 x y − 1 ⋅ ( 0 ) = ∫ 0 x 1 + sin 2 t d t y = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t Thus, the solution is y ( x ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t . Substitute x = 1 in the above solution to find the value of y ( 1 ) . y ( 1 ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 1 1 + sin 2 t d t y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ∫ 0 1 1 + sin 2 t d t y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( I s ) ... ( 1 ) Approximate the integral I s = ∫ 0 1 1 + sin 2 t d t for many intervals by using Simpson’s rule as follows. Here, f ( x ) = 1 + sin 2 x , a = 0 , and b = 1 . Four intervals: For n = 2 , the interval [ 0 , 1 ] can be divided into 4 equal parts and let x 0 , x 1 , … , x 4 be the points of the partition, that is, x k = 0 + k h , k = 0 , 1 , … , 4 . Calculate h as follows. h = 1 − 0 2 ( 2 ) = 1 4 Thus, the values of x 0 , x 1 , … , x 4 are 0 , 1 4 , 1 2 , 3 4 , 1 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 0 1 1 + sin 2 t d t as follows. I s = 1 4 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + f ( x 4 ) ] = 1 12 [ f ( 0 ) + 4 f ( 1 4 ) + 2 f ( 1 2 ) + 4 f ( 3 4 ) + f ( 1 ) ] = 1 12 [ 1 + sin 2 ( 0 ) + 4 1 + sin 2 ( 1 4 ) + 2 1 + sin 2 ( 1 2 ) + 4 1 + sin 2 ( 3 4 ) + 1 + sin 2 ( 1 ) ] ≈ 1.12387 Substitute I s ≈ 1.12387 in equation (1). y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( 1.12387 ) ≈ 0.860 Thus, the value of y ( 1 ) ≈ 0.860 for 4 intervals. Six intervals: For n = 3 , the interval [ 0 , 1 ] can be divided into 6 equal parts and let x 0 , x 1 , … , x 6 be the points of the partition, that is, x k = 0 + k h , k = 0 , 1 , … , 6 . calculate h as follows. h = 1 − 0 2 ( 3 ) = 1 6 Thus, the values of x 0 , x 1 , … , x 6 are 0 , 1 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 0 1 1 + sin 2 t d t as follows. I s = 1 6 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + 4 f ( x 5 ) + f ( x 6 ) ] = 1 18 [ f ( 0 ) + 4 f ( 1 6 ) + 2 f ( 1 3 ) + 4 f ( 1 2 ) + 2 f ( 2 3 ) + 4 f ( 5 6 ) + f ( 1 ) ] = 1 18 [ 1 + sin 2 ( 0 ) + 4 1 + sin 2 ( 1 6 ) + 2 1 + sin 2 ( 1 3 ) + 4 1 + sin 2 ( 1 2 ) + 2 1 + sin 2 ( 2 3 ) + 4 1 + sin 2 ( 5 6 ) + 1 + sin 2 ( 1 ) ] ≈ 1.12389 Substitute I s ≈ 1.12389 in equation (1). y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( 1.12389 ) ≈ 0.860 Thus, the value of y ( 1 ) ≈ 0.860 for 6 intervals. Eight intervals: For n = 4 , the interval [ 0 , 1 ] can be divided into 8 equal parts and let x 0 , x 1 , … , x 8 be the points of the partition, that is, x k = 0 + k h , k = 0 , 1 , … , 8 . calculate h as follows. h = 1 − 0 2 ( 4 ) = 1 8 Thus, the values of x 0 , x 1 , … , x 8 are 0 , 1 8 , 1 4 , 3 8 , 1 2 , 5 8 , 3 4 , 7 8 , 1 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 0 1 1 + sin 2 t d t as follows. I s = 1 8 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + 4 f ( x 5 ) + 2 f ( x 6 ) + 4 f ( x 7 ) + f ( x 8 ) ] = 1 24 [ f ( 0 ) + 4 f ( 1 8 ) + 2 f ( 1 4 ) + 4 f ( 3 8 ) + 2 f ( 1 2 ) + 4 f ( 5 8 ) + 2 f ( 3 4 ) + 4 f ( 7 8 ) + f ( 1 ) ] = 1 24 [ 1 + sin 2 ( 0 ) + 4 1 + sin 2 ( 1 8 ) + 2 1 + sin 2 ( 1 4 ) + 4 1 + sin 2 ( 3 8 ) + 2 1 + sin 2 ( 1 2 ) + 4 1 + sin 2 ( 5 8 ) + 2 1 + sin 2 ( 3 4 ) + 4 1 + sin 2 ( 7 8 ) + 1 + sin 2 ( 1 ) ] ≈ 1.12389 Substitute I s ≈ 1.12389 in equation (1). y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( 1.12389 ) ≈ 0.860 Thus, the value of y ( 1 ) ≈ 0.860 for 8 intervals. The values of y ( 1 ) for the intervals 4, 6, and 8 are tabulated as below. Intervals y ( 1 ) 4 0.860 6 0.860 8 0.860 From the above table, it is concluded that the value of y ( 1 ) is 0.860 corrected to 3 decimal places. Thus, the value of the solution y ( x ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t at x = 1 is 0.860 _ approximately.

To approximate: The solution to the initial value problem d y d x + sin 2 x 2 ( 1 + sin 2 x ) y = 1 , y ( 0 ) = 0 at x = 1 by using Simpson’s rule with the accuracy of three decimal places. The value of the solution y ( x ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t at x = 1 is 0.860 _ approximately. Method used: 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 [ y 0 + 4 y 1 + 2 y 2 + 4 y 3 + ⋯ + 2 y 2 n − 2 + 4 y 2 n − 1 + y 2 n ] = h 3 ∑ k = 1 n ( y 2 k − 2 + 4 y 2 k − 1 + y 2 k ) . .” Calculation: The differential equation is, d y d x + sin 2 x 2 ( 1 + sin 2 x ) y = 1 , y ( 0 ) = 0 Rewrite the differential equation of the standard form d y d x + P ( x ) y = Q ( x ) as follows. d y d x + sin 2 x 2 ( 1 + sin 2 x ) y = 1 ... ( 1 ) Here, P ( x ) = sin 2 x 2 ( 1 + sin 2 x ) and Q ( x ) = 1 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ sin 2 x 2 ( 1 + sin 2 x ) d x ] = exp [ 1 2 ∫ sin 2 x ( 1 + sin 2 x ) d x ] = exp [ 1 2 ∫ 2 cos x sin x ( 1 + sin 2 x ) d x ] ( ∵ sin 2 x = 2 cos x sin x ) = exp [ ∫ cos x sin x ( 1 + sin 2 x ) d x ] Integrate by the method of U-substitution Take u = sin x , then d u = cos x d x , d x = 1 cos x d u . Thus, the integral becomes, μ ( x ) = exp [ ∫ cos x u ( 1 + u 2 ) 1 cos x d u ] μ ( x ) = exp [ ∫ u ( 1 + u 2 ) d u ] Again, integrate by the method of U-substitution Take v = 1 + u 2 , then d v = 2 u d u , d u = 1 2 u d v . Thus, the integral becomes, μ ( x ) = exp [ ∫ u v 1 2 u d v ] = exp [ ∫ 1 2 v d v ] = exp [ 1 2 ∫ 1 v d v ] = exp [ 1 2 ln | v | ] Simplify further, μ ( x ) = exp [ 1 2 ln | 1 + u 2 | ] ( Substitute back, v = 1 + u 2 ) = exp [ 1 2 ln | 1 + sin 2 x | ] ( Substitute back, u = sin x ) = e ln ( | 1 + sin 2 x | ) 1 2 = 1 + sin 2 x Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . 1 + sin 2 x d y d x + 1 + sin 2 x sin 2 x 2 ( 1 + sin 2 x ) y = 1 + sin 2 x d d x [ 1 + sin 2 x y ] = 1 + sin 2 x The value of x starts at 0 because the initial condition is y ( 0 ) = 0 . Also, the end value is unknown. Therefore, the limits of integration would be [ 0 , x ] . Integrate both sides of the equation with respect to t from t = 0 to t = x . [ 1 + sin 2 t y ] 0 x = ∫ 0 x 1 + sin 2 t d t 1 + sin 2 x y − 1 + sin 2 ( 0 ) ⋅ y ( 0 ) = ∫ 0 x 1 + sin 2 t d t 1 + sin 2 x y − 1 ⋅ ( 0 ) = ∫ 0 x 1 + sin 2 t d t y = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t Thus, the solution is y ( x ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t . Substitute x = 1 in the above solution to find the value of y ( 1 ) . y ( 1 ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 1 1 + sin 2 t d t y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ∫ 0 1 1 + sin 2 t d t y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( I s ) ... ( 1 ) Approximate the integral I s = ∫ 0 1 1 + sin 2 t d t for many intervals by using Simpson’s rule as follows. Here, f ( x ) = 1 + sin 2 x , a = 0 , and b = 1 . Four intervals: For n = 2 , the interval [ 0 , 1 ] can be divided into 4 equal parts and let x 0 , x 1 , … , x 4 be the points of the partition, that is, x k = 0 + k h , k = 0 , 1 , … , 4 . Calculate h as follows. h = 1 − 0 2 ( 2 ) = 1 4 Thus, the values of x 0 , x 1 , … , x 4 are 0 , 1 4 , 1 2 , 3 4 , 1 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 0 1 1 + sin 2 t d t as follows. I s = 1 4 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + f ( x 4 ) ] = 1 12 [ f ( 0 ) + 4 f ( 1 4 ) + 2 f ( 1 2 ) + 4 f ( 3 4 ) + f ( 1 ) ] = 1 12 [ 1 + sin 2 ( 0 ) + 4 1 + sin 2 ( 1 4 ) + 2 1 + sin 2 ( 1 2 ) + 4 1 + sin 2 ( 3 4 ) + 1 + sin 2 ( 1 ) ] ≈ 1.12387 Substitute I s ≈ 1.12387 in equation (1). y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( 1.12387 ) ≈ 0.860 Thus, the value of y ( 1 ) ≈ 0.860 for 4 intervals. Six intervals: For n = 3 , the interval [ 0 , 1 ] can be divided into 6 equal parts and let x 0 , x 1 , … , x 6 be the points of the partition, that is, x k = 0 + k h , k = 0 , 1 , … , 6 . calculate h as follows. h = 1 − 0 2 ( 3 ) = 1 6 Thus, the values of x 0 , x 1 , … , x 6 are 0 , 1 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 0 1 1 + sin 2 t d t as follows. I s = 1 6 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + 4 f ( x 5 ) + f ( x 6 ) ] = 1 18 [ f ( 0 ) + 4 f ( 1 6 ) + 2 f ( 1 3 ) + 4 f ( 1 2 ) + 2 f ( 2 3 ) + 4 f ( 5 6 ) + f ( 1 ) ] = 1 18 [ 1 + sin 2 ( 0 ) + 4 1 + sin 2 ( 1 6 ) + 2 1 + sin 2 ( 1 3 ) + 4 1 + sin 2 ( 1 2 ) + 2 1 + sin 2 ( 2 3 ) + 4 1 + sin 2 ( 5 6 ) + 1 + sin 2 ( 1 ) ] ≈ 1.12389 Substitute I s ≈ 1.12389 in equation (1). y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( 1.12389 ) ≈ 0.860 Thus, the value of y ( 1 ) ≈ 0.860 for 6 intervals. Eight intervals: For n = 4 , the interval [ 0 , 1 ] can be divided into 8 equal parts and let x 0 , x 1 , … , x 8 be the points of the partition, that is, x k = 0 + k h , k = 0 , 1 , … , 8 . calculate h as follows. h = 1 − 0 2 ( 4 ) = 1 8 Thus, the values of x 0 , x 1 , … , x 8 are 0 , 1 8 , 1 4 , 3 8 , 1 2 , 5 8 , 3 4 , 7 8 , 1 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 0 1 1 + sin 2 t d t as follows. I s = 1 8 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + 4 f ( x 5 ) + 2 f ( x 6 ) + 4 f ( x 7 ) + f ( x 8 ) ] = 1 24 [ f ( 0 ) + 4 f ( 1 8 ) + 2 f ( 1 4 ) + 4 f ( 3 8 ) + 2 f ( 1 2 ) + 4 f ( 5 8 ) + 2 f ( 3 4 ) + 4 f ( 7 8 ) + f ( 1 ) ] = 1 24 [ 1 + sin 2 ( 0 ) + 4 1 + sin 2 ( 1 8 ) + 2 1 + sin 2 ( 1 4 ) + 4 1 + sin 2 ( 3 8 ) + 2 1 + sin 2 ( 1 2 ) + 4 1 + sin 2 ( 5 8 ) + 2 1 + sin 2 ( 3 4 ) + 4 1 + sin 2 ( 7 8 ) + 1 + sin 2 ( 1 ) ] ≈ 1.12389 Substitute I s ≈ 1.12389 in equation (1). y ( 1 ) = ( 1 + sin 2 1 ) − 1 2 ( 1.12389 ) ≈ 0.860 Thus, the value of y ( 1 ) ≈ 0.860 for 8 intervals. The values of y ( 1 ) for the intervals 4, 6, and 8 are tabulated as below. Intervals y ( 1 ) 4 0.860 6 0.860 8 0.860 From the above table, it is concluded that the value of y ( 1 ) is 0.860 corrected to 3 decimal places. Thus, the value of the solution y ( x ) = ( 1 + sin 2 x ) − 1 2 ∫ 0 x 1 + sin 2 t d t at x = 1 is 0.860 _ approximately.

Solution Summary: The author explains how to approximate the solution to the initial value problem by using Simpson's rule with the accuracy of three decimal places.

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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Fundamentals of Differential Equations (9th Edition)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

Ch. 2.2 - In Problems 716, solve the equation. 11....Ch. 2.2 - In Problems 716, solve the equation. 12....Ch. 2.2 - In Problems 716, solve the equation. 13....Ch. 2.2 - In Problems 716, solve the equation. 14. dxdtx3=x Ch. 2.2 - In Problems 716, solve the equation. 15....Ch. 2.2 - In Problems 716, solve the equation. 16. y1 dy +...Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 23E Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 27E Ch. 2.2 - Sketch the solution to the initial value problem...Ch. 2.2 - Uniqueness Questions. In Chapter 1 we indicated...Ch. 2.2 - As stated in this section, the separation of...Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Mixing. Suppose a brine containing 0.3 kilogram...Ch. 2.2 - Newtons Law of Cooling. According to Newtons law...Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Compound Interest. If P(t) is the amount of...Ch. 2.2 - Free Fall. In Section 2.1, we discussed a model...Ch. 2.2 - Grand Prix Race. Driver A had been leading...Ch. 2.2 - Prob. 40E Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - Radioactive Decay. In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. The equation (18) dydx+2y=xy2...Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 9E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 15E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 17E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 25E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 27E Ch. 2.4 - For each of the following equations, find the most...Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Orthogonal Trajectories. A geometric problem...Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.5 - Prob. 1E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Verify that when the linear differential equation...Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 8E Ch. 2.6 - Use the method discussed under Homogeneous...Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Use the method discussed under Bernoulli Equations...Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Use the method discussed under Equations with...Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - In Problems 3340, solve the equation given in: 36....Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Show that equation (13) reduces to an equation of...Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2 - In Problems 130, solve the equation. 1....Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - In Problems 130, solve the equation. 6. 2xy3 dx ...Ch. 2 - In Problems 130, solve the equation. 7. t3y2 dt +...Ch. 2 - Prob. 8RP Ch. 2 - In Problems 130, solve the equation. 9. (x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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Numerical Integration Introduction l Trapezoidal Rule Simpson's 1/3 Rule l Simpson's 3/8 l GATE 2021; Author: GATE Lectures by Dishank;https://www.youtube.com/watch?v=zadUB3NwFtQ;License: Standard YouTube License, CC-BY