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

The differential equation ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 . The solution of the differential equation ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 is y 2 x 2 + ln | x 6 y 2 | = C _ . Definition used: “The homogeneous equation is of the form d y d x = G ( y x ) . Let v = y x . Then d y d x = v + x ( d v d x ) , and the transformed equation in the variables v and x is separable.” Calculation: The differential equation is, ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 The equation can be rewritten as, d y d x = − ( 3 x 2 − y 2 ) x y − x 3 y − 1 d y d x = y 2 − 3 x 2 x y − x 3 y − 1 d y d x = y 3 − 3 y x 2 x y 2 − x 3 ( multiply both the numerator and the denominator by y ) d y d x = y 3 x 3 − 3 y x 2 x 3 x y 2 x 3 − x 3 x 3 ( divide both the numerator and the denominator by x 3 ) d y d x = ( y x ) 3 − 3 ( y x ) ( y x ) 2 − 1 This is homogeneous equation of the form d y d x = G ( y x ) . So take v = y x . Then d y d x = v + x d v d x , the transformed equation is v + x d v d x = v 3 − 3 v v 2 − 1 x d v d x = v 3 − 3 v v 2 − 1 − v x d v d x = v 3 − 3 v − v 3 + v v 2 − 1 x d v d x = − 2 v v 2 − 1 Separate the variables and rewrite the differential equation as follows. v 2 − 1 v d v = − 2 x d x ( v − 1 v ) d v = − 2 x d x Integrate on both sides of the equation. ( v − 1 v ) d v = − 2 x d x ∫ ( v − 1 v ) d v = ∫ − 2 x d x v 2 2 − ln | v | = − 2 ln | x | + C 1 v 2 − 2 ln | v | = − 4 ln | x | + 2 C 1 v 2 − ln | v 2 | + ln | x 4 | = C ( take 2 C 1 = C , C is an arbitrary constant ) Substitute back, v = y x . v 2 − ln | v 2 | + ln | x 4 | = C ( y x ) 2 − ln | ( y x ) 2 | + ln | x 4 | = C y 2 x 2 + ln | x 2 y 2 | + ln | x 4 | = C y 2 x 2 + ln | x 6 y 2 | = C Thus, the solutions of the differential equation ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 is y 2 x 2 + ln | x 6 y 2 | = C _ .

The differential equation ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 . The solution of the differential equation ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 is y 2 x 2 + ln | x 6 y 2 | = C _ . Definition used: “The homogeneous equation is of the form d y d x = G ( y x ) . Let v = y x . Then d y d x = v + x ( d v d x ) , and the transformed equation in the variables v and x is separable.” Calculation: The differential equation is, ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 The equation can be rewritten as, d y d x = − ( 3 x 2 − y 2 ) x y − x 3 y − 1 d y d x = y 2 − 3 x 2 x y − x 3 y − 1 d y d x = y 3 − 3 y x 2 x y 2 − x 3 ( multiply both the numerator and the denominator by y ) d y d x = y 3 x 3 − 3 y x 2 x 3 x y 2 x 3 − x 3 x 3 ( divide both the numerator and the denominator by x 3 ) d y d x = ( y x ) 3 − 3 ( y x ) ( y x ) 2 − 1 This is homogeneous equation of the form d y d x = G ( y x ) . So take v = y x . Then d y d x = v + x d v d x , the transformed equation is v + x d v d x = v 3 − 3 v v 2 − 1 x d v d x = v 3 − 3 v v 2 − 1 − v x d v d x = v 3 − 3 v − v 3 + v v 2 − 1 x d v d x = − 2 v v 2 − 1 Separate the variables and rewrite the differential equation as follows. v 2 − 1 v d v = − 2 x d x ( v − 1 v ) d v = − 2 x d x Integrate on both sides of the equation. ( v − 1 v ) d v = − 2 x d x ∫ ( v − 1 v ) d v = ∫ − 2 x d x v 2 2 − ln | v | = − 2 ln | x | + C 1 v 2 − 2 ln | v | = − 4 ln | x | + 2 C 1 v 2 − ln | v 2 | + ln | x 4 | = C ( take 2 C 1 = C , C is an arbitrary constant ) Substitute back, v = y x . v 2 − ln | v 2 | + ln | x 4 | = C ( y x ) 2 − ln | ( y x ) 2 | + ln | x 4 | = C y 2 x 2 + ln | x 2 y 2 | + ln | x 4 | = C y 2 x 2 + ln | x 6 y 2 | = C Thus, the solutions of the differential equation ( 3 x 2 − y 2 ) d x + ( x y − x 3 y − 1 ) d y = 0 is y 2 x 2 + ln | x 6 y 2 | = C _ .

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)

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. 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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. 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