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

The initial value problem ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 . The solution of the IVP ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 is 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = ln 2 _ . Definition used: Linear coefficients: “The equations of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) , where a i 's, b i 's, and c i 's are constants. For a 1 b 2 ≠ a 2 b 1 and b 1 ≠ a 2 , let x = u + h and y = v + k , where h and k satisfy a 1 h + b 1 k + c 1 = 0 , a 2 h + b 2 k + c 2 = 0. Then the transformed equation in the variables u and v is homogeneous. That is, d v d u = − a 1 u + b 1 v a 2 u + b 2 v = − a 1 + b 1 ( v u ) a 2 + b 2 ( v u ) .” Calculation: The differential equation is, ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 ... ( 1 ) The equation is of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) . Here, a 1 = 2 , b 1 = − 1 , c 1 = 0 , a 2 = 1 , b 2 = 1 and c 2 = − 3 . Check the condition a 1 b 2 ≠ a 2 b 1 . a 1 b 2 = ( 2 ) ( 1 ) = 2 a 2 b 1 = ( 1 ) ( − 1 ) = − 1 Thus, the condition a 1 b 2 ≠ a 2 b 1 is satisfied. Let x = u + h and y = v + k , where h and k satisfy the system, 2 h − k = 0 ... ( 2 ) h + k − 3 = 0 ... ( 3 ) Add both the equations. 2 h − k = 0 h + k − 3 = 0 _ 3 h = 3 h = 1 Substitute h = 1 in equation (2). 2 ( 1 ) − k = 0 − k = − 2 k = 2 Thus, the translation of the axes becomes x = u + 1 and y = v + 2 . Then d x = d u and d y = d v . The equation can be translated as, [ 2 ( u + 1 ) − ( v + 2 ) ] d u + [ ( u + 1 ) + ( v + 2 ) − 3 ] d v = 0 ( 2 u + 2 − v − 2 ) d u + ( u + 1 + v + 2 − 3 ) d v = 0 ( 2 u − v ) d u + ( u + v ) d v = 0 d v d u = − ( 2 u − v ) ( u + v ) d v d u = − 2 + ( v u ) 1 + ( v u ) This is homogeneous equation of the form d v d u = G ( v u ) . So, take z = v u . Then d v d u = z + u d z d u , the transformed equation is, z + u d z d u = − 2 + z 1 + z u d z d u = − 2 + z 1 + z − z u d z d u = − 2 + z − z − z 2 1 + z u d z d u = − z 2 + 2 1 + z Separate the variables and rewrite the differential equation as follows. u d z d u = − z 2 + 2 1 + z 1 + z z 2 + 2 d z = − 1 u d u Integrate on both sides of the equation. ∫ 1 + z z 2 + 2 d z = − ∫ 1 u d u ∫ 1 z 2 + 2 d z + ∫ z z 2 + 2 d z = − ∫ 1 u d u ∫ 1 z 2 + ( 2 ) 2 d z + 1 2 ∫ 2 z z 2 + 2 d z = − ∫ 1 u d u 1 2 tan − 1 ( z 2 ) + 1 2 ln | z 2 + 2 | = − ln | u | + C 1 Simplify further as follows. 2 tan − 1 ( z 2 ) + ln | z 2 + 2 | = − 2 ln | u | + 2 C 1 2 tan − 1 ( z 2 ) + ln | z 2 + 2 | + 2 ln | u | = C ( C is an arbitrary constant ) 2 tan − 1 ( z 2 ) + ln | ( z 2 + 2 ) u 2 | + ln | u 2 | = C 2 tan − 1 ( z 2 ) + ln | ( z 2 + 2 ) u 2 | = C Substitute back, z = v u . 2 tan − 1 ( v u 2 ) + ln | ( ( v u ) 2 + 2 ) u 2 | = C 2 tan − 1 ( v 2 u ) + ln | ( v 2 + 2 u 2 u 2 ) u 2 | = C 2 tan − 1 ( v 2 u ) + ln | v 2 + 2 u 2 | = C Substitute back, u = x − 1 and v = y − 2 . 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = C Use the initial condition y ( 0 ) = 2 in the solution of the differential equation as shown. 2 tan − 1 ( ( 2 − 2 ) 2 ( ( 0 ) − 1 ) ) + ln | ( 2 − 2 ) 2 + 2 ( 0 − 1 ) 2 | = C 2 tan − 1 ( 0 ) + ln | ( 0 ) 2 + 2 ( 1 ) 2 | = C ( 0 ) + ln | 2 | = C C = ln 2 Substitute C = ln 2 in 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = C as follows. 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = ln 2 Hence, the solution of the IVP ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 is 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = ln 2 _ .

The initial value problem ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 . The solution of the IVP ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 is 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = ln 2 _ . Definition used: Linear coefficients: “The equations of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) , where a i 's, b i 's, and c i 's are constants. For a 1 b 2 ≠ a 2 b 1 and b 1 ≠ a 2 , let x = u + h and y = v + k , where h and k satisfy a 1 h + b 1 k + c 1 = 0 , a 2 h + b 2 k + c 2 = 0. Then the transformed equation in the variables u and v is homogeneous. That is, d v d u = − a 1 u + b 1 v a 2 u + b 2 v = − a 1 + b 1 ( v u ) a 2 + b 2 ( v u ) .” Calculation: The differential equation is, ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 ... ( 1 ) The equation is of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) . Here, a 1 = 2 , b 1 = − 1 , c 1 = 0 , a 2 = 1 , b 2 = 1 and c 2 = − 3 . Check the condition a 1 b 2 ≠ a 2 b 1 . a 1 b 2 = ( 2 ) ( 1 ) = 2 a 2 b 1 = ( 1 ) ( − 1 ) = − 1 Thus, the condition a 1 b 2 ≠ a 2 b 1 is satisfied. Let x = u + h and y = v + k , where h and k satisfy the system, 2 h − k = 0 ... ( 2 ) h + k − 3 = 0 ... ( 3 ) Add both the equations. 2 h − k = 0 h + k − 3 = 0 _ 3 h = 3 h = 1 Substitute h = 1 in equation (2). 2 ( 1 ) − k = 0 − k = − 2 k = 2 Thus, the translation of the axes becomes x = u + 1 and y = v + 2 . Then d x = d u and d y = d v . The equation can be translated as, [ 2 ( u + 1 ) − ( v + 2 ) ] d u + [ ( u + 1 ) + ( v + 2 ) − 3 ] d v = 0 ( 2 u + 2 − v − 2 ) d u + ( u + 1 + v + 2 − 3 ) d v = 0 ( 2 u − v ) d u + ( u + v ) d v = 0 d v d u = − ( 2 u − v ) ( u + v ) d v d u = − 2 + ( v u ) 1 + ( v u ) This is homogeneous equation of the form d v d u = G ( v u ) . So, take z = v u . Then d v d u = z + u d z d u , the transformed equation is, z + u d z d u = − 2 + z 1 + z u d z d u = − 2 + z 1 + z − z u d z d u = − 2 + z − z − z 2 1 + z u d z d u = − z 2 + 2 1 + z Separate the variables and rewrite the differential equation as follows. u d z d u = − z 2 + 2 1 + z 1 + z z 2 + 2 d z = − 1 u d u Integrate on both sides of the equation. ∫ 1 + z z 2 + 2 d z = − ∫ 1 u d u ∫ 1 z 2 + 2 d z + ∫ z z 2 + 2 d z = − ∫ 1 u d u ∫ 1 z 2 + ( 2 ) 2 d z + 1 2 ∫ 2 z z 2 + 2 d z = − ∫ 1 u d u 1 2 tan − 1 ( z 2 ) + 1 2 ln | z 2 + 2 | = − ln | u | + C 1 Simplify further as follows. 2 tan − 1 ( z 2 ) + ln | z 2 + 2 | = − 2 ln | u | + 2 C 1 2 tan − 1 ( z 2 ) + ln | z 2 + 2 | + 2 ln | u | = C ( C is an arbitrary constant ) 2 tan − 1 ( z 2 ) + ln | ( z 2 + 2 ) u 2 | + ln | u 2 | = C 2 tan − 1 ( z 2 ) + ln | ( z 2 + 2 ) u 2 | = C Substitute back, z = v u . 2 tan − 1 ( v u 2 ) + ln | ( ( v u ) 2 + 2 ) u 2 | = C 2 tan − 1 ( v 2 u ) + ln | ( v 2 + 2 u 2 u 2 ) u 2 | = C 2 tan − 1 ( v 2 u ) + ln | v 2 + 2 u 2 | = C Substitute back, u = x − 1 and v = y − 2 . 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = C Use the initial condition y ( 0 ) = 2 in the solution of the differential equation as shown. 2 tan − 1 ( ( 2 − 2 ) 2 ( ( 0 ) − 1 ) ) + ln | ( 2 − 2 ) 2 + 2 ( 0 − 1 ) 2 | = C 2 tan − 1 ( 0 ) + ln | ( 0 ) 2 + 2 ( 1 ) 2 | = C ( 0 ) + ln | 2 | = C C = ln 2 Substitute C = ln 2 in 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = C as follows. 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = ln 2 Hence, the solution of the IVP ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 is 2 tan − 1 ( ( y − 2 ) 2 ( x − 1 ) ) + ln | ( y − 2 ) 2 + 2 ( x − 1 ) 2 | = ln 2 _ .

Solution Summary: The author explains that the solution of the IVP (2x-y)dx+ (x +y) is underset_sqrt2

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)

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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. 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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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