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

Whether the differential equation as ( y 2 + 2 x y ) d x − x 2 d y = 0 separable, linear, exact or having an integrating factor that is a function of either x alone or y alone. The differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 has an integrating factor that is a function of y alone _ . Definition used: The differential equation is a linear first order equation if it is of the form, a 1 ( x ) d y d x + a 0 ( x ) y = b ( x ) Theorem used: “Suppose the first partial derivatives of M ( x , y ) and N ( x , y ) are continuous in a rectangle R . Then M ( x , y ) d x + N ( x , y ) d y = 0 is an exact equation in R if and only if the compatibility condition ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R .” Result used: “A first-order equation is separable it can be written in the form d y d x = g ( x ) p ( y ) .” Special integrating factor: “If ( ∂ M ∂ y − ∂ N ∂ x ) N is continuous and depends only on x , then μ ( x ) = exp [ ∫ ( ∂ M ∂ y − ∂ N ∂ x N ) d x ] is an integrating factor for equation M ( x , y ) d x + N ( x , y ) d y = 0 . If ( ∂ N ∂ x − ∂ M ∂ y ) M is continuous and depends only on y , then μ ( y ) = exp [ ∫ ( ∂ N ∂ x − ∂ M ∂ y M ) d y ] is an integrating factor for equation M ( x , y ) d x + N ( x , y ) d y = 0 .” Calculation: The differential equation is, ( y 2 + 2 x y ) d x − x 2 d y = 0 Rewrite the equation ( y 2 + 2 x y ) d x − x 2 d y = 0 as shown below. − x 2 d y = − ( y 2 + 2 x y ) d x d y d x = ( y 2 + 2 x y ) x 2 d y d x = y 2 x 2 + 2 x y d y d x − 2 x y = y 2 x 2 Note that, the equation is not linear with respect to x and thus the equation is not linear with respect to y as dependent variable. Similarly, rewrite the equation ( y 2 + 2 x y ) d x − x 2 d y = 0 as shown below. ( y 2 + 2 x y ) d x = x 2 d y d x d y = x 2 ( y 2 + 2 x y ) d x d y = x 2 y ( y + 2 x ) It is observed that the right hand side is not linear with respect to x and thus the equation is not linear with respect to y as dependent variable. Therefore, the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 is not linear. The given differential equation can be written as follows. d x d y = x 2 ( y 2 + 2 x y ) d x d y = x 2 y ( y + 2 x ) It is observed that the right hand side cannot be represented as a product of two functions of variables x and y . That is, the given differential equation cannot be written in the form d y d x = g ( x ) p ( y ) . Thus, the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 is not separable. Compare ( y 2 + 2 x y ) d x − x 2 d y = 0 with the equation M ( x , y ) d x + N ( x , y ) d y = 0 and the following is obtained. M ( x , y ) = y 2 + 2 x y N ( x , y ) = − x 2 Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y ( x , y ) = ∂ ∂ y ( y 2 + 2 x y ) = 2 y + 2 x Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x ( x , y ) = ∂ ∂ x ( − x 2 ) = − 2 x Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 is not exact. Calculate ( ∂ M ∂ y − ∂ N ∂ x ) N as follows. ( ∂ M ∂ y − ∂ N ∂ x ) N = 2 y + 2 x + 2 x − x 2 = 2 y + 4 x − x 2 = − 2 y − 4 x x 2 = − 2 ( y + 2 x ) x 2 Note that, the function not only contains the variable x . So, calculate ( ∂ N ∂ x − ∂ M ∂ y ) M as follows. ( ∂ N ∂ x − ∂ M ∂ y ) M = − 2 x − 2 y − 2 x y 2 + 2 x y = − 4 x − 2 y y 2 + 2 x y = − 2 ( 2 x + y ) y ( y + 2 x ) = − 2 y Here, the function is continuous on ℝ − { 0 } and only depends on the variable y . Hence, the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 has an integrating factor that is a function of y alone _ .

Whether the differential equation as ( y 2 + 2 x y ) d x − x 2 d y = 0 separable, linear, exact or having an integrating factor that is a function of either x alone or y alone. The differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 has an integrating factor that is a function of y alone _ . Definition used: The differential equation is a linear first order equation if it is of the form, a 1 ( x ) d y d x + a 0 ( x ) y = b ( x ) Theorem used: “Suppose the first partial derivatives of M ( x , y ) and N ( x , y ) are continuous in a rectangle R . Then M ( x , y ) d x + N ( x , y ) d y = 0 is an exact equation in R if and only if the compatibility condition ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R .” Result used: “A first-order equation is separable it can be written in the form d y d x = g ( x ) p ( y ) .” Special integrating factor: “If ( ∂ M ∂ y − ∂ N ∂ x ) N is continuous and depends only on x , then μ ( x ) = exp [ ∫ ( ∂ M ∂ y − ∂ N ∂ x N ) d x ] is an integrating factor for equation M ( x , y ) d x + N ( x , y ) d y = 0 . If ( ∂ N ∂ x − ∂ M ∂ y ) M is continuous and depends only on y , then μ ( y ) = exp [ ∫ ( ∂ N ∂ x − ∂ M ∂ y M ) d y ] is an integrating factor for equation M ( x , y ) d x + N ( x , y ) d y = 0 .” Calculation: The differential equation is, ( y 2 + 2 x y ) d x − x 2 d y = 0 Rewrite the equation ( y 2 + 2 x y ) d x − x 2 d y = 0 as shown below. − x 2 d y = − ( y 2 + 2 x y ) d x d y d x = ( y 2 + 2 x y ) x 2 d y d x = y 2 x 2 + 2 x y d y d x − 2 x y = y 2 x 2 Note that, the equation is not linear with respect to x and thus the equation is not linear with respect to y as dependent variable. Similarly, rewrite the equation ( y 2 + 2 x y ) d x − x 2 d y = 0 as shown below. ( y 2 + 2 x y ) d x = x 2 d y d x d y = x 2 ( y 2 + 2 x y ) d x d y = x 2 y ( y + 2 x ) It is observed that the right hand side is not linear with respect to x and thus the equation is not linear with respect to y as dependent variable. Therefore, the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 is not linear. The given differential equation can be written as follows. d x d y = x 2 ( y 2 + 2 x y ) d x d y = x 2 y ( y + 2 x ) It is observed that the right hand side cannot be represented as a product of two functions of variables x and y . That is, the given differential equation cannot be written in the form d y d x = g ( x ) p ( y ) . Thus, the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 is not separable. Compare ( y 2 + 2 x y ) d x − x 2 d y = 0 with the equation M ( x , y ) d x + N ( x , y ) d y = 0 and the following is obtained. M ( x , y ) = y 2 + 2 x y N ( x , y ) = − x 2 Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y ( x , y ) = ∂ ∂ y ( y 2 + 2 x y ) = 2 y + 2 x Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x ( x , y ) = ∂ ∂ x ( − x 2 ) = − 2 x Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 is not exact. Calculate ( ∂ M ∂ y − ∂ N ∂ x ) N as follows. ( ∂ M ∂ y − ∂ N ∂ x ) N = 2 y + 2 x + 2 x − x 2 = 2 y + 4 x − x 2 = − 2 y − 4 x x 2 = − 2 ( y + 2 x ) x 2 Note that, the function not only contains the variable x . So, calculate ( ∂ N ∂ x − ∂ M ∂ y ) M as follows. ( ∂ N ∂ x − ∂ M ∂ y ) M = − 2 x − 2 y − 2 x y 2 + 2 x y = − 4 x − 2 y y 2 + 2 x y = − 2 ( 2 x + y ) y ( y + 2 x ) = − 2 y Here, the function is continuous on ℝ − { 0 } and only depends on the variable y . Hence, the differential equation ( y 2 + 2 x y ) d x − x 2 d y = 0 has an integrating factor that is a function of y alone _ .

Solution Summary: The author analyzes whether the differential equation is separable, linear, exact or having an integrating factor.

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. 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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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