Skip to main content

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

See similar textbooks

Concept explainers

Numerical Integration

In a mathematical investigation, numerical integration comprises a wide group of calculations for computing the mathematical estimation of definite integral, and likewise, the term is moreover in some cases used to depict the numerical solution of differ…

Videos

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 2.3, Problem 25E

Chapter 2.3, Problem 27E

Students have asked these similar questions

2

)5. The operator T on a vector space V over Q has as its sequence of invariant factors the sequence (x-1)3(x-2)5(x-3)2, (x-1)(x-2)(x-3)2, (x-3). Find the Jordan normal form of T.

18

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

Knowledge Booster

Background pattern image

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Functions and Change: A Modeling Approach to Coll...
Algebra
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Cengage Learning
College Algebra
Algebra
ISBN:9781337282291
Author:Ron Larson
Publisher:Cengage Learning
Trigonometry (MindTap Course List)
Trigonometry
ISBN:9781337278461
Author:Ron Larson
Publisher:Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
College Algebra
Algebra
ISBN:9781938168383
Author:Jay Abramson
Publisher:OpenStax
Mathematics For Machine Technology
Advanced Math
ISBN:9781337798310
Author:Peterson, John.
Publisher:Cengage Learning,

Text book image

Functions and Change: A Modeling Approach to Coll...

Algebra

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Cengage Learning

Text book image

College Algebra

Algebra

ISBN:9781337282291

Author:Ron Larson

Publisher:Cengage Learning

Text book image

Trigonometry (MindTap Course List)

Trigonometry

ISBN:9781337278461

Author:Ron Larson

Publisher:Cengage Learning

Text book image

Algebra & Trigonometry with Analytic Geometry

Algebra

ISBN:9781133382119

Author:Swokowski

Publisher:Cengage

Text book image

College Algebra

Algebra

ISBN:9781938168383

Author:Jay Abramson

Publisher:OpenStax

Text book image

Mathematics For Machine Technology

Advanced Math

ISBN:9781337798310

Author:Peterson, John.

Publisher:Cengage Learning,

SEE MORE TEXTBOOKS

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