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

(a) To show: That the substitution y = u + 1 v reduces equation (18), d y d x = P ( x ) y 2 + Q ( x ) y + R ( x ) to a linear equation in v if u ( x ) is one solution of (18). The Riccati equation (18) is, d y d x = P ( x ) y 2 + Q ( x ) y + R ( x ) Take the substitution y = u + 1 v . Then d y d x = d u d x − v − 2 d v d x , the transformed equation is d u d x − v − 2 d v d x = P ( x ) ( u + 1 v ) 2 + Q ( x ) ( u + 1 v ) + R ( x ) d u d x − 1 v 2 d v d x = P ( x ) ( u 2 + 1 v 2 + 2 u v ) + Q ( x ) ( u + 1 v ) + R ( x ) d u d x − 1 v 2 d v d x = u 2 P ( x ) + 1 v 2 P ( x ) + 2 u v P ( x ) + u Q ( x ) + 1 v Q ( x ) + R ( x ) d u d x − 1 v 2 d v d x = u 2 P ( x ) + P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v + u Q ( x ) + R ( x ) ... ( 1 ) Given that u ( x ) is one of the solution of equation (18) and so it must satisfy the equation (18). That is, d u ( x ) d x = P ( x ) ( u ( x ) ) 2 + Q ( x ) ( u ( x ) ) + R ( x ) d u d x = P ( x ) u 2 + Q ( x ) u + R ( x ) ... ( 2 ) Substitute equation (2) in equation (1). P ( x ) u 2 + Q ( x ) u + R ( x ) − 1 v 2 d v d x = { u 2 P ( x ) + P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v + u Q ( x ) + R ( x ) } − 1 v 2 d v d x = { u 2 P ( x ) + P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v + u Q ( x ) + R ( x ) − u 2 P ( x ) − u Q ( x ) − R ( x ) } − 1 v 2 d v d x = P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v d v d x = − P ( x ) − ( 2 u P ( x ) + Q ( x ) ) v d v d x + ( 2 u P ( x ) + Q ( x ) ) v = − P ( x ) This is a linear differential equation. Hence the result is true for the solution u ( x ) and the substitution y = u + 1 v . (b) To find: All the other solutions to the equation d y d x = x 3 ( y − x ) 2 + y x if u ( x ) = x is one solution of d y d x = x 3 ( y − x ) 2 + y x . The another solution of the differential equation d y d x = x 3 ( y − x ) 2 + y x is y = x + 5 x C − x 5 _ . The given equation is, d y d x = x 3 ( y − x ) 2 + y x Rewrite the equation as follows. d y d x = x 3 ( y 2 + x 2 − 2 x y ) + y x d y d x = x 3 y 2 + x 5 − 2 x 4 y + y x d y d x = x 3 y 2 + x 5 + ( 1 x − 2 x 4 ) y Compare the above equation with the Riccati equation (18) d y d x = P ( x ) y 2 + Q ( x ) y + R ( x ) and the following is obtained. P ( x ) = x 3 , Q ( x ) = 1 x − 2 x 4 and R ( x ) = x 5 From part (a), the reduced form of Riccati equation is d v d x + ( 2 u P ( x ) + Q ( x ) ) v + P ( x ) = 0 and it is true for the solution u ( x ) = x and the substitution y = u + 1 v . Substitute the values of P ( x ) = x 3 , Q ( x ) = 1 x − 2 x 4 , R ( x ) = x 5 and u ( x ) = x in d v d x + ( 2 u P ( x ) + Q ( x ) ) v + P ( x ) = 0 . d v d x + ( 2 x x 3 + 1 x − 2 x 4 ) v + x 3 = 0 d v d x + 1 x v = − x 3 ... ( 3 ) This is a linear differential equation of the form d y d x + P ( x ) y = Q ( x ) . Here, P ( x ) = 1 x and Q ( x ) = − x 3 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ ( 1 x ) d x ] = e ln x = x Multiply the equation (3) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . x ( d v d x + 1 x v ) = − x 3 ⋅ x x d v d x + v = − x 4 d d x ( x v ) = − x 4 To obtain the general solution y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] , integrate both sides of the equation d d x ( x v ) = − x 4 and solve for v ( x ) by dividing μ ( x ) . x v = ∫ ( − x 4 ) d x x v = − ∫ x 4 d x x v = − x 5 5 + C v = − x 4 5 + C x Substitute back, y = u + 1 v that is v = 1 y − u . 1 y − u = − x 4 5 + C 1 x y − x = 1 − x 4 5 + C 1 x ( ∵ u = x ) y = x + 5 x 5 C 1 − x 5 y = x + 5 x C − x 5 ( C is an arbitrary constant ) Thus, the another solution of the differential equation d y d x = x 3 ( y − x ) 2 + y x is y = x + 5 x C − x 5 _ .

(a) To show: That the substitution y = u + 1 v reduces equation (18), d y d x = P ( x ) y 2 + Q ( x ) y + R ( x ) to a linear equation in v if u ( x ) is one solution of (18). The Riccati equation (18) is, d y d x = P ( x ) y 2 + Q ( x ) y + R ( x ) Take the substitution y = u + 1 v . Then d y d x = d u d x − v − 2 d v d x , the transformed equation is d u d x − v − 2 d v d x = P ( x ) ( u + 1 v ) 2 + Q ( x ) ( u + 1 v ) + R ( x ) d u d x − 1 v 2 d v d x = P ( x ) ( u 2 + 1 v 2 + 2 u v ) + Q ( x ) ( u + 1 v ) + R ( x ) d u d x − 1 v 2 d v d x = u 2 P ( x ) + 1 v 2 P ( x ) + 2 u v P ( x ) + u Q ( x ) + 1 v Q ( x ) + R ( x ) d u d x − 1 v 2 d v d x = u 2 P ( x ) + P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v + u Q ( x ) + R ( x ) ... ( 1 ) Given that u ( x ) is one of the solution of equation (18) and so it must satisfy the equation (18). That is, d u ( x ) d x = P ( x ) ( u ( x ) ) 2 + Q ( x ) ( u ( x ) ) + R ( x ) d u d x = P ( x ) u 2 + Q ( x ) u + R ( x ) ... ( 2 ) Substitute equation (2) in equation (1). P ( x ) u 2 + Q ( x ) u + R ( x ) − 1 v 2 d v d x = { u 2 P ( x ) + P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v + u Q ( x ) + R ( x ) } − 1 v 2 d v d x = { u 2 P ( x ) + P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v + u Q ( x ) + R ( x ) − u 2 P ( x ) − u Q ( x ) − R ( x ) } − 1 v 2 d v d x = P ( x ) v 2 + 2 u P ( x ) + Q ( x ) v d v d x = − P ( x ) − ( 2 u P ( x ) + Q ( x ) ) v d v d x + ( 2 u P ( x ) + Q ( x ) ) v = − P ( x ) This is a linear differential equation. Hence the result is true for the solution u ( x ) and the substitution y = u + 1 v . (b) To find: All the other solutions to the equation d y d x = x 3 ( y − x ) 2 + y x if u ( x ) = x is one solution of d y d x = x 3 ( y − x ) 2 + y x . The another solution of the differential equation d y d x = x 3 ( y − x ) 2 + y x is y = x + 5 x C − x 5 _ . The given equation is, d y d x = x 3 ( y − x ) 2 + y x Rewrite the equation as follows. d y d x = x 3 ( y 2 + x 2 − 2 x y ) + y x d y d x = x 3 y 2 + x 5 − 2 x 4 y + y x d y d x = x 3 y 2 + x 5 + ( 1 x − 2 x 4 ) y Compare the above equation with the Riccati equation (18) d y d x = P ( x ) y 2 + Q ( x ) y + R ( x ) and the following is obtained. P ( x ) = x 3 , Q ( x ) = 1 x − 2 x 4 and R ( x ) = x 5 From part (a), the reduced form of Riccati equation is d v d x + ( 2 u P ( x ) + Q ( x ) ) v + P ( x ) = 0 and it is true for the solution u ( x ) = x and the substitution y = u + 1 v . Substitute the values of P ( x ) = x 3 , Q ( x ) = 1 x − 2 x 4 , R ( x ) = x 5 and u ( x ) = x in d v d x + ( 2 u P ( x ) + Q ( x ) ) v + P ( x ) = 0 . d v d x + ( 2 x x 3 + 1 x − 2 x 4 ) v + x 3 = 0 d v d x + 1 x v = − x 3 ... ( 3 ) This is a linear differential equation of the form d y d x + P ( x ) y = Q ( x ) . Here, P ( x ) = 1 x and Q ( x ) = − x 3 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ ( 1 x ) d x ] = e ln x = x Multiply the equation (3) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . x ( d v d x + 1 x v ) = − x 3 ⋅ x x d v d x + v = − x 4 d d x ( x v ) = − x 4 To obtain the general solution y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] , integrate both sides of the equation d d x ( x v ) = − x 4 and solve for v ( x ) by dividing μ ( x ) . x v = ∫ ( − x 4 ) d x x v = − ∫ x 4 d x x v = − x 5 5 + C v = − x 4 5 + C x Substitute back, y = u + 1 v that is v = 1 y − u . 1 y − u = − x 4 5 + C 1 x y − x = 1 − x 4 5 + C 1 x ( ∵ u = x ) y = x + 5 x 5 C 1 − x 5 y = x + 5 x C − x 5 ( C is an arbitrary constant ) Thus, the another solution of the differential equation d y d x = x 3 ( y − x ) 2 + y x is y = x + 5 x C − x 5 _ .

Solution Summary: The author explains that the substitution y=u+1v reduces equation (18).

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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