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

The differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 . The solution of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ . 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 + 4 ) d x + ( x − 2 y − 2 ) 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 = 4 , a 2 = 1 , b 2 = − 2 and c 2 = − 2 . Check the condition a 1 b 2 ≠ a 2 b 1 . a 1 b 2 = ( 2 ) ( − 2 ) = − 4 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 + 4 = 0 ... ( 2 ) h − 2 k − 2 = 0 ... ( 3 ) Multiply the equation (3) by − 2 and add it to the equation (2). 2 h − k + 4 = 0 − 2 h + 4 k + 4 = 0 _ 3 k + 8 = 0 k = − 8 3 Substitute k = − 8 3 in equation (3). h − 2 k − 2 = 0 h − 2 ( − 8 3 ) − 2 = 0 h + 16 3 − 2 = 0 h + 10 3 = 0 h = − 10 3 Thus, the translation of the axes becomes x = u − 10 3 and y = v − 8 3 . Then d x = d u and d y = d v . The equation can be translated as, ( 2 ( u − 10 3 ) − ( v − 8 3 ) + 4 ) d u + ( ( u − 10 3 ) − 2 ( v − 8 3 ) − 2 ) d v = 0 ( 2 ( 3 u − 10 3 ) − ( 3 v − 8 3 ) + 4 ) d u + ( ( 3 u − 10 3 ) − 2 ( 3 v − 8 3 ) − 2 ) d v = 0 ( 6 u − 20 − 3 v + 8 + 12 3 ) d u + ( 3 u − 10 − 6 v + 16 − 6 3 ) d v = 0 ( 6 u − 3 v 3 ) d u + ( 3 u − 6 v 3 ) d v = 0 Simplify further as follows. ( 2 u − v ) d u + ( u − 2 v ) d v = 0 d v d u = − 2 u − v u − 2 v d v d u = − 2 + v u 1 − 2 ( 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 − 2 z u d z d u = − 2 + z 1 − 2 z − z u d z d u = − 2 + z − z + 2 z 2 1 − 2 z Separate the variables and rewrite the differential equation as follows. u d z d u = − 2 + 2 z 2 1 − 2 z u d z d u = 2 ( z 2 − 1 ) 1 − 2 z 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u Integrate on both sides of the equation. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u ... ( 4 ) Consider the term 1 − 2 z 2 ( z + 1 ) ( z − 1 ) Split the term by using the method of partial fractions. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( A ( z + 1 ) + B ( z − 1 ) ) ... ( 5 ) Multiply 2 ( z + 1 ) ( z − 1 ) on both sides of the equation. 1 − 2 z = A ( z − 1 ) + B ( z + 1 ) Put z = 1 results B = − 1 2 and put z = − 1 results A = − 3 2 . Substitute A = − 3 2 and B = − 1 2 in the equation (5). 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( − 3 2 ( z + 1 ) − 1 2 ( z − 1 ) ) 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) Thus, the integral becomes, ∫ 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) d z = ∫ 1 u d u − 1 4 ∫ 3 ( z + 1 ) d z − 1 4 ∫ 1 ( z − 1 ) d z = ∫ 1 u d u − 3 4 ln | z + 1 | − 1 4 ln | z − 1 | = ln | u | + C − 1 4 ln | ( z + 1 ) 3 ( z − 1 ) | = ln | u | + C Simplify further as follows. ln | ( z + 1 ) 3 ( z − 1 ) | = − 4 ln | u | − 4 C 1 ln | ( z + 1 ) 3 ( z − 1 ) | = ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = e ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = C 2 u 4 Substitute back, z = v u . ( v u + 1 ) 3 ( v u − 1 ) = C 2 u 4 ( v + u u ) 3 ( v − u u ) = C 2 u 4 ( v + u ) 3 ( v − u ) = C 2 Substitute back, u = x + 10 3 and v = y + 8 3 . ( ( y + 8 3 ) + ( x + 10 3 ) ) 3 ( ( y + 8 3 ) − ( x + 10 3 ) ) = C 2 ( ( 3 y + 8 3 ) + ( 3 x + 10 3 ) ) 3 ( ( 3 y + 8 3 ) − ( 3 x + 10 3 ) ) = C 2 ( 3 x + 3 y + 18 3 ) 3 ( − 3 x + 3 y − 2 3 ) = C 2 1 81 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C 2 Simplify further as follows. ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = 81 C 2 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C ( C is an arbitrary constant ) Thus, the solutions of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ .

The differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 . The solution of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ . 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 + 4 ) d x + ( x − 2 y − 2 ) 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 = 4 , a 2 = 1 , b 2 = − 2 and c 2 = − 2 . Check the condition a 1 b 2 ≠ a 2 b 1 . a 1 b 2 = ( 2 ) ( − 2 ) = − 4 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 + 4 = 0 ... ( 2 ) h − 2 k − 2 = 0 ... ( 3 ) Multiply the equation (3) by − 2 and add it to the equation (2). 2 h − k + 4 = 0 − 2 h + 4 k + 4 = 0 _ 3 k + 8 = 0 k = − 8 3 Substitute k = − 8 3 in equation (3). h − 2 k − 2 = 0 h − 2 ( − 8 3 ) − 2 = 0 h + 16 3 − 2 = 0 h + 10 3 = 0 h = − 10 3 Thus, the translation of the axes becomes x = u − 10 3 and y = v − 8 3 . Then d x = d u and d y = d v . The equation can be translated as, ( 2 ( u − 10 3 ) − ( v − 8 3 ) + 4 ) d u + ( ( u − 10 3 ) − 2 ( v − 8 3 ) − 2 ) d v = 0 ( 2 ( 3 u − 10 3 ) − ( 3 v − 8 3 ) + 4 ) d u + ( ( 3 u − 10 3 ) − 2 ( 3 v − 8 3 ) − 2 ) d v = 0 ( 6 u − 20 − 3 v + 8 + 12 3 ) d u + ( 3 u − 10 − 6 v + 16 − 6 3 ) d v = 0 ( 6 u − 3 v 3 ) d u + ( 3 u − 6 v 3 ) d v = 0 Simplify further as follows. ( 2 u − v ) d u + ( u − 2 v ) d v = 0 d v d u = − 2 u − v u − 2 v d v d u = − 2 + v u 1 − 2 ( 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 − 2 z u d z d u = − 2 + z 1 − 2 z − z u d z d u = − 2 + z − z + 2 z 2 1 − 2 z Separate the variables and rewrite the differential equation as follows. u d z d u = − 2 + 2 z 2 1 − 2 z u d z d u = 2 ( z 2 − 1 ) 1 − 2 z 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u Integrate on both sides of the equation. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u ... ( 4 ) Consider the term 1 − 2 z 2 ( z + 1 ) ( z − 1 ) Split the term by using the method of partial fractions. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( A ( z + 1 ) + B ( z − 1 ) ) ... ( 5 ) Multiply 2 ( z + 1 ) ( z − 1 ) on both sides of the equation. 1 − 2 z = A ( z − 1 ) + B ( z + 1 ) Put z = 1 results B = − 1 2 and put z = − 1 results A = − 3 2 . Substitute A = − 3 2 and B = − 1 2 in the equation (5). 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( − 3 2 ( z + 1 ) − 1 2 ( z − 1 ) ) 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) Thus, the integral becomes, ∫ 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) d z = ∫ 1 u d u − 1 4 ∫ 3 ( z + 1 ) d z − 1 4 ∫ 1 ( z − 1 ) d z = ∫ 1 u d u − 3 4 ln | z + 1 | − 1 4 ln | z − 1 | = ln | u | + C − 1 4 ln | ( z + 1 ) 3 ( z − 1 ) | = ln | u | + C Simplify further as follows. ln | ( z + 1 ) 3 ( z − 1 ) | = − 4 ln | u | − 4 C 1 ln | ( z + 1 ) 3 ( z − 1 ) | = ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = e ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = C 2 u 4 Substitute back, z = v u . ( v u + 1 ) 3 ( v u − 1 ) = C 2 u 4 ( v + u u ) 3 ( v − u u ) = C 2 u 4 ( v + u ) 3 ( v − u ) = C 2 Substitute back, u = x + 10 3 and v = y + 8 3 . ( ( y + 8 3 ) + ( x + 10 3 ) ) 3 ( ( y + 8 3 ) − ( x + 10 3 ) ) = C 2 ( ( 3 y + 8 3 ) + ( 3 x + 10 3 ) ) 3 ( ( 3 y + 8 3 ) − ( 3 x + 10 3 ) ) = C 2 ( 3 x + 3 y + 18 3 ) 3 ( − 3 x + 3 y − 2 3 ) = C 2 1 81 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C 2 Simplify further as follows. ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = 81 C 2 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C ( C is an arbitrary constant ) Thus, the solutions of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ .

Solution Summary: The author explains the solution of the differential equation (2x-y+4)dx+ (x-2y-2)dy=0.

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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