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

To express: The solution to the initial value problem d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 using a definite integral. The solution to the initial value problem d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 can be expressed as y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r _ . The differential equation is, d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 Rewrite the differential equation of the standard form d y d x + P ( x ) y = Q ( x ) as follows. d y d t + y = 1 1 + t 2 ... ( 1 ) Here, P ( t ) = 1 and Q ( t ) = 1 1 + t 2 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( t ) = exp [ ∫ ( 1 ) d t ] = exp ( t ) = e t Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . e t ( d y d t + y ) = e t 1 1 + t 2 e t d y d t + e t y = e t 1 + t 2 d d t ( e t y ) = e t 1 + t 2 The value of t starts at 2 because the initial condition is y ( 2 ) = 3 . Also, the end value is unknown. Therefore, the limits of integration would be [ 2 , t ] . Integrate both sides of the equation with respect to r from r = 2 to r = t . [ e r y ] 2 t = ∫ 2 t e r 1 + r 2 d r e t y − e 2 y ( 2 ) = ∫ 2 t e r 1 + r 2 d r e t y − e 2 ( 3 ) = ∫ 2 t e r 1 + r 2 d r ( ∵ y ( 2 ) = 3 ) e t y = 3 e 2 + ∫ 2 t e r 1 + r 2 d r y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r Thus, the solution to the initial value problem d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 can be expressed as y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r _ . To approximate: The expression y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r at t = 3 by using numerical integration. The value of y ( 3 ) is 1.1883 _ 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: From part (a), the expression is y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r . Substitute t = 3 in the above solution to find the value of y ( 3 ) . y ( 3 ) = 3 e 2 − 3 + e − 3 ∫ 2 3 e r 1 + r 2 d r y ( 3 ) = 3 e − 1 + e − 3 ∫ 2 3 e r 1 + r 2 d r y ( 3 ) = 3 e + 1 e 3 ∫ 2 3 e r 1 + r 2 d r y ( 3 ) = 3 e + 1 e 3 I s ... ( 1 ) Approximate the integral I s = ∫ 2 3 e r 1 + r 2 d r for many intervals by using Simpson’s rule as follows. Here, f ( t ) = e t 1 + t 2 , a = 2 , and b = 3 . Four intervals: For n = 2 , the interval [ 2 , 3 ] 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 = 2 + k h , k = 0 , 1 , … , 4 . Calculate h as follows. h = 3 − 2 2 ( 2 ) = 1 4 Thus, the values of x 0 , x 1 , … , x 4 are 2 , 9 4 , 5 2 , 11 4 , 3 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 2 3 e r 1 + r 2 d r 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 ( 2 ) + 4 f ( 9 4 ) + 2 f ( 5 2 ) + 4 f ( 11 4 ) + f ( 3 ) ] = 1 12 [ e 2 1 + ( 2 ) 2 + 4 e ( 9 4 ) 1 + ( 9 4 ) 2 + 2 e ( 5 2 ) 1 + ( 5 2 ) 2 + 4 e ( 11 4 ) 1 + ( 11 4 ) 2 + e 3 1 + ( 3 ) 2 ] ≈ 1.701209099 Substitute I s ≈ 1.701209099 in equation (1). y ( 3 ) = 3 e + 1 e 3 ( 1.701209099 ) ≈ 1.18834 Thus, the value of y ( 3 ) ≈ 1.18834 for 4 intervals. Six intervals: For n = 3 , the interval [ 2 , 3 ] 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 = 2 + k h , k = 0 , 1 , … , 6 . Calculate h as follows. h = 3 − 2 2 ( 3 ) = 1 6 Thus, the values of x 0 , x 1 , … , x 6 are 2 , 13 6 , 7 3 , 5 2 , 8 3 , 17 6 , 3 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 2 3 e t 2 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 ( 2 ) + 4 f ( 13 6 ) + 2 f ( 7 3 ) + 4 f ( 5 2 ) + 2 f ( 8 3 ) + 4 f ( 17 6 ) + f ( 3 ) ] = 1 18 [ e 2 1 + ( 2 ) 2 + 4 e ( 13 6 ) 1 + ( 13 6 ) 2 + 2 e ( 7 3 ) 1 + ( 7 3 ) 2 + 4 e ( 5 2 ) 1 + ( 5 2 ) 2 + 2 e ( 8 3 ) 1 + ( 8 3 ) 2 + 4 e ( 17 6 ) 1 + ( 17 6 ) 2 + e 3 1 + ( 3 ) 2 ] ≈ 1.701204866 Substitute I s ≈ 1.701204866 in equation (1). y ( 3 ) = 3 e + 1 e 3 ( 1.701204866 ) ≈ 1.18834 Thus, the value of y ( 3 ) ≈ 1.18834 for 6 intervals. Eight intervals: For n = 4 , the interval [ 2 , 3 ] 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 = 2 + k h , k = 0 , 1 , … , 8 . Calculate h as follows. h = 3 − 2 2 ( 4 ) = 1 8 Thus, the values of x 0 , x 1 , … , x 8 are 2 , 17 8 , 9 4 , 19 8 , 5 2 , 21 8 , 11 4 , 23 8 , 3 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 2 3 e t 2 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 ( 2 ) + 4 f ( 17 8 ) + 2 f ( 9 4 ) + 4 f ( 19 8 ) + 2 f ( 5 2 ) + 4 f ( 21 8 ) + 2 f ( 11 4 ) + 4 f ( 23 8 ) + f ( 3 ) ] = 1 24 [ e 2 1 + ( 2 ) 2 + 4 e ( 17 8 ) 1 + ( 17 8 ) 2 + 2 e ( 9 4 ) 1 + ( 9 4 ) 2 + 4 e ( 19 8 ) 1 + ( 19 8 ) 2 + 2 e ( 5 2 ) 1 + ( 5 2 ) 2 + 4 e ( 21 8 ) 1 + ( 21 8 ) 2 + 2 e ( 11 4 ) 1 + ( 11 4 ) 2 + 4 e ( 23 8 ) 1 + ( 23 8 ) 2 + e 3 1 + ( 3 ) 2 ] ≈ 1.701203846 Substitute I s ≈ 1.701203846 in equation (1). y ( 3 ) = 3 e + 1 e 3 ( 1.701203846 ) ≈ 1.18834 Thus, the value of y ( 3 ) ≈ 1.18834 for 8 intervals. The values of y ( 3 ) for the intervals 4 ,6, and 8 are tabulated as shown below. Intervals y ( 3 ) 4 1.18834 6 1.18834 8 1.18834 From the above table, it is concluded that the value of y ( 3 ) is 1.1883 corrected to 3 decimal places. Thus, the value of y ( 3 ) is 1.1883 _ approximately.

To express: The solution to the initial value problem d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 using a definite integral. The solution to the initial value problem d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 can be expressed as y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r _ . The differential equation is, d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 Rewrite the differential equation of the standard form d y d x + P ( x ) y = Q ( x ) as follows. d y d t + y = 1 1 + t 2 ... ( 1 ) Here, P ( t ) = 1 and Q ( t ) = 1 1 + t 2 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( t ) = exp [ ∫ ( 1 ) d t ] = exp ( t ) = e t Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . e t ( d y d t + y ) = e t 1 1 + t 2 e t d y d t + e t y = e t 1 + t 2 d d t ( e t y ) = e t 1 + t 2 The value of t starts at 2 because the initial condition is y ( 2 ) = 3 . Also, the end value is unknown. Therefore, the limits of integration would be [ 2 , t ] . Integrate both sides of the equation with respect to r from r = 2 to r = t . [ e r y ] 2 t = ∫ 2 t e r 1 + r 2 d r e t y − e 2 y ( 2 ) = ∫ 2 t e r 1 + r 2 d r e t y − e 2 ( 3 ) = ∫ 2 t e r 1 + r 2 d r ( ∵ y ( 2 ) = 3 ) e t y = 3 e 2 + ∫ 2 t e r 1 + r 2 d r y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r Thus, the solution to the initial value problem d y d t = 1 1 + t 2 − y , y ( 2 ) = 3 can be expressed as y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r _ . To approximate: The expression y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r at t = 3 by using numerical integration. The value of y ( 3 ) is 1.1883 _ 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: From part (a), the expression is y = 3 e 2 − t + e − t ∫ 2 t e r 1 + r 2 d r . Substitute t = 3 in the above solution to find the value of y ( 3 ) . y ( 3 ) = 3 e 2 − 3 + e − 3 ∫ 2 3 e r 1 + r 2 d r y ( 3 ) = 3 e − 1 + e − 3 ∫ 2 3 e r 1 + r 2 d r y ( 3 ) = 3 e + 1 e 3 ∫ 2 3 e r 1 + r 2 d r y ( 3 ) = 3 e + 1 e 3 I s ... ( 1 ) Approximate the integral I s = ∫ 2 3 e r 1 + r 2 d r for many intervals by using Simpson’s rule as follows. Here, f ( t ) = e t 1 + t 2 , a = 2 , and b = 3 . Four intervals: For n = 2 , the interval [ 2 , 3 ] 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 = 2 + k h , k = 0 , 1 , … , 4 . Calculate h as follows. h = 3 − 2 2 ( 2 ) = 1 4 Thus, the values of x 0 , x 1 , … , x 4 are 2 , 9 4 , 5 2 , 11 4 , 3 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 2 3 e r 1 + r 2 d r 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 ( 2 ) + 4 f ( 9 4 ) + 2 f ( 5 2 ) + 4 f ( 11 4 ) + f ( 3 ) ] = 1 12 [ e 2 1 + ( 2 ) 2 + 4 e ( 9 4 ) 1 + ( 9 4 ) 2 + 2 e ( 5 2 ) 1 + ( 5 2 ) 2 + 4 e ( 11 4 ) 1 + ( 11 4 ) 2 + e 3 1 + ( 3 ) 2 ] ≈ 1.701209099 Substitute I s ≈ 1.701209099 in equation (1). y ( 3 ) = 3 e + 1 e 3 ( 1.701209099 ) ≈ 1.18834 Thus, the value of y ( 3 ) ≈ 1.18834 for 4 intervals. Six intervals: For n = 3 , the interval [ 2 , 3 ] 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 = 2 + k h , k = 0 , 1 , … , 6 . Calculate h as follows. h = 3 − 2 2 ( 3 ) = 1 6 Thus, the values of x 0 , x 1 , … , x 6 are 2 , 13 6 , 7 3 , 5 2 , 8 3 , 17 6 , 3 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 2 3 e t 2 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 ( 2 ) + 4 f ( 13 6 ) + 2 f ( 7 3 ) + 4 f ( 5 2 ) + 2 f ( 8 3 ) + 4 f ( 17 6 ) + f ( 3 ) ] = 1 18 [ e 2 1 + ( 2 ) 2 + 4 e ( 13 6 ) 1 + ( 13 6 ) 2 + 2 e ( 7 3 ) 1 + ( 7 3 ) 2 + 4 e ( 5 2 ) 1 + ( 5 2 ) 2 + 2 e ( 8 3 ) 1 + ( 8 3 ) 2 + 4 e ( 17 6 ) 1 + ( 17 6 ) 2 + e 3 1 + ( 3 ) 2 ] ≈ 1.701204866 Substitute I s ≈ 1.701204866 in equation (1). y ( 3 ) = 3 e + 1 e 3 ( 1.701204866 ) ≈ 1.18834 Thus, the value of y ( 3 ) ≈ 1.18834 for 6 intervals. Eight intervals: For n = 4 , the interval [ 2 , 3 ] 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 = 2 + k h , k = 0 , 1 , … , 8 . Calculate h as follows. h = 3 − 2 2 ( 4 ) = 1 8 Thus, the values of x 0 , x 1 , … , x 8 are 2 , 17 8 , 9 4 , 19 8 , 5 2 , 21 8 , 11 4 , 23 8 , 3 respectively. Use Simpson’s rule to approximate the integral I s = ∫ 2 3 e t 2 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 ( 2 ) + 4 f ( 17 8 ) + 2 f ( 9 4 ) + 4 f ( 19 8 ) + 2 f ( 5 2 ) + 4 f ( 21 8 ) + 2 f ( 11 4 ) + 4 f ( 23 8 ) + f ( 3 ) ] = 1 24 [ e 2 1 + ( 2 ) 2 + 4 e ( 17 8 ) 1 + ( 17 8 ) 2 + 2 e ( 9 4 ) 1 + ( 9 4 ) 2 + 4 e ( 19 8 ) 1 + ( 19 8 ) 2 + 2 e ( 5 2 ) 1 + ( 5 2 ) 2 + 4 e ( 21 8 ) 1 + ( 21 8 ) 2 + 2 e ( 11 4 ) 1 + ( 11 4 ) 2 + 4 e ( 23 8 ) 1 + ( 23 8 ) 2 + e 3 1 + ( 3 ) 2 ] ≈ 1.701203846 Substitute I s ≈ 1.701203846 in equation (1). y ( 3 ) = 3 e + 1 e 3 ( 1.701203846 ) ≈ 1.18834 Thus, the value of y ( 3 ) ≈ 1.18834 for 8 intervals. The values of y ( 3 ) for the intervals 4 ,6, and 8 are tabulated as shown below. Intervals y ( 3 ) 4 1.18834 6 1.18834 8 1.18834 From the above table, it is concluded that the value of y ( 3 ) is 1.1883 corrected to 3 decimal places. Thus, the value of y ( 3 ) is 1.1883 _ approximately.

Solution Summary: The author explains how to express the solution to the initial value problem dy-dt using a definite integral.

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 2 Solutions

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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ISBN:9781337278461

Author:Ron Larson

Publisher:Cengage Learning

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