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

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.

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

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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