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

(a) To evaluate: The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s directly with x 0 = 0 , y 0 = 0 to rework Example 1. The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y 2 − x _ . Given: The Example 1 given in the textbook was, “Solve the differential equation d y d x = − 2 x y 2 + 1 2 x 2 y ”. The formula is F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s where x 0 = 0 , y 0 = 0 . That is, F ( x , y ) = ∫ 0 y N ( x , t ) d t + ∫ 0 x M ( s , 0 ) d s ... ( 1 ) Calculation: The differential equation is, d y d x = − 2 x y 2 + 1 2 x 2 y Rewrite the differential equation as follows. ( 2 x 2 y ) d y = − ( 2 x y 2 + 1 ) d x ( 2 x y 2 + 1 ) d x + ( 2 x 2 y ) d y = 0 Compare ( 2 x y 2 + 1 ) d x + ( 2 x 2 y ) 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 ) = 2 x y 2 + 1 ⇒ M ( s , 0 ) = 2 s ( 0 ) 2 + 1 = 1 N ( x , y ) = 2 x 2 y ⇒ N ( x , t ) = 2 x 2 t Use Formula (1) to solve the differential equation as follows. F ( x , y ) = ∫ 0 y ( N ( x , t ) ) d t + ∫ 0 x ( M ( s , 0 ) ) d s = ∫ 0 y ( 2 x 2 t ) d t + ∫ 0 x d s = [ 2 x 2 t 2 2 ] 0 y + [ s ] 0 x = x 2 y 2 − 0 + x − 0 = x 2 y 2 − x Thus, the formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y 2 − x _ by reworking Example 1. (b) To evaluate: The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s directly with x 0 = 0 , y 0 = 0 to rework Example 2. The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y + y 2 − tan x _ . Given: The Example 2 given in the textbook was, “Solve ( 2 x y − sec 2 x ) d x + ( x 2 + 2 y ) d y = 0 ”. The formula is F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s where x 0 = 0 , y 0 = 0 . That is, F ( x , y ) = ∫ 0 y N ( x , t ) d t + ∫ 0 x M ( s , 0 ) d s ... ( 1 ) Calculation: The differential equation is, ( 2 x y − sec 2 x ) d x + ( x 2 + 2 y ) d y = 0 Compare ( 2 x y − sec 2 x ) d x + ( x 2 + 2 y ) 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 ) = 2 x y − sec 2 x ⇒ M ( s , 0 ) = 2 s ( 0 ) − sec 2 s = − sec 2 s N ( x , y ) = x 2 + 2 y ⇒ N ( x , t ) = x 2 + 2 t Use Formula (1) to solve the differential equation as follows. F ( x , y ) = ∫ 0 y ( N ( x , t ) ) d t + ∫ 0 x ( M ( s , 0 ) ) d s = ∫ 0 y ( x 2 + 2 t ) d t + ∫ 0 x ( − sec 2 s ) d s = [ x 2 t + t 2 ] 0 y + [ − tan s ] 0 x = x 2 y + y 2 − tan x + tan 0 = x 2 y + y 2 − tan x Thus, the formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y + y 2 − tan x _ . (c) To evaluate: The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s directly with x 0 = 0 , y 0 = 0 to rework Example 3. The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x e x y + 2 y + x _ . Given: The Example 3 given in the textbook was, “Solve ( 1 + e x y + x e x y ) d x + ( x e x + 2 ) d y = 0 ”. The formula is F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s where x 0 = 0 , y 0 = 0 . That is, F ( x , y ) = ∫ 0 y N ( x , t ) d t + ∫ 0 x M ( s , 0 ) d s ... ( 1 ) Calculation: The differential equation is, ( 1 + e x y + x e x y ) d x + ( x e x + 2 ) d y = 0 Compare ( 1 + e x y + x e x y ) d x + ( x e 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 ) = 1 + e x y + x e x y ⇒ M ( s , 0 ) = 1 + e s ( 0 ) + s e s ( 0 ) = 1 N ( x , y ) = x e x + 2 ⇒ N ( x , t ) = x e x + 2 Use Formula (1) to solve the differential equation as follows. F ( x , y ) = ∫ 0 y ( N ( x , t ) ) d t + ∫ 0 x ( M ( s , 0 ) ) d s = ∫ 0 y ( x e x + 2 ) d t + ∫ 0 x ( 1 ) d s = [ x e x t + 2 t ] 0 y + [ s ] 0 x = x e x y + 2 y − 0 + x − 0 = x e x y + 2 y + x Thus, the formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x e x y + 2 y + x _ .

(a) To evaluate: The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s directly with x 0 = 0 , y 0 = 0 to rework Example 1. The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y 2 − x _ . Given: The Example 1 given in the textbook was, “Solve the differential equation d y d x = − 2 x y 2 + 1 2 x 2 y ”. The formula is F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s where x 0 = 0 , y 0 = 0 . That is, F ( x , y ) = ∫ 0 y N ( x , t ) d t + ∫ 0 x M ( s , 0 ) d s ... ( 1 ) Calculation: The differential equation is, d y d x = − 2 x y 2 + 1 2 x 2 y Rewrite the differential equation as follows. ( 2 x 2 y ) d y = − ( 2 x y 2 + 1 ) d x ( 2 x y 2 + 1 ) d x + ( 2 x 2 y ) d y = 0 Compare ( 2 x y 2 + 1 ) d x + ( 2 x 2 y ) 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 ) = 2 x y 2 + 1 ⇒ M ( s , 0 ) = 2 s ( 0 ) 2 + 1 = 1 N ( x , y ) = 2 x 2 y ⇒ N ( x , t ) = 2 x 2 t Use Formula (1) to solve the differential equation as follows. F ( x , y ) = ∫ 0 y ( N ( x , t ) ) d t + ∫ 0 x ( M ( s , 0 ) ) d s = ∫ 0 y ( 2 x 2 t ) d t + ∫ 0 x d s = [ 2 x 2 t 2 2 ] 0 y + [ s ] 0 x = x 2 y 2 − 0 + x − 0 = x 2 y 2 − x Thus, the formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y 2 − x _ by reworking Example 1. (b) To evaluate: The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s directly with x 0 = 0 , y 0 = 0 to rework Example 2. The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y + y 2 − tan x _ . Given: The Example 2 given in the textbook was, “Solve ( 2 x y − sec 2 x ) d x + ( x 2 + 2 y ) d y = 0 ”. The formula is F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s where x 0 = 0 , y 0 = 0 . That is, F ( x , y ) = ∫ 0 y N ( x , t ) d t + ∫ 0 x M ( s , 0 ) d s ... ( 1 ) Calculation: The differential equation is, ( 2 x y − sec 2 x ) d x + ( x 2 + 2 y ) d y = 0 Compare ( 2 x y − sec 2 x ) d x + ( x 2 + 2 y ) 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 ) = 2 x y − sec 2 x ⇒ M ( s , 0 ) = 2 s ( 0 ) − sec 2 s = − sec 2 s N ( x , y ) = x 2 + 2 y ⇒ N ( x , t ) = x 2 + 2 t Use Formula (1) to solve the differential equation as follows. F ( x , y ) = ∫ 0 y ( N ( x , t ) ) d t + ∫ 0 x ( M ( s , 0 ) ) d s = ∫ 0 y ( x 2 + 2 t ) d t + ∫ 0 x ( − sec 2 s ) d s = [ x 2 t + t 2 ] 0 y + [ − tan s ] 0 x = x 2 y + y 2 − tan x + tan 0 = x 2 y + y 2 − tan x Thus, the formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x 2 y + y 2 − tan x _ . (c) To evaluate: The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s directly with x 0 = 0 , y 0 = 0 to rework Example 3. The formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x e x y + 2 y + x _ . Given: The Example 3 given in the textbook was, “Solve ( 1 + e x y + x e x y ) d x + ( x e x + 2 ) d y = 0 ”. The formula is F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s where x 0 = 0 , y 0 = 0 . That is, F ( x , y ) = ∫ 0 y N ( x , t ) d t + ∫ 0 x M ( s , 0 ) d s ... ( 1 ) Calculation: The differential equation is, ( 1 + e x y + x e x y ) d x + ( x e x + 2 ) d y = 0 Compare ( 1 + e x y + x e x y ) d x + ( x e 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 ) = 1 + e x y + x e x y ⇒ M ( s , 0 ) = 1 + e s ( 0 ) + s e s ( 0 ) = 1 N ( x , y ) = x e x + 2 ⇒ N ( x , t ) = x e x + 2 Use Formula (1) to solve the differential equation as follows. F ( x , y ) = ∫ 0 y ( N ( x , t ) ) d t + ∫ 0 x ( M ( s , 0 ) ) d s = ∫ 0 y ( x e x + 2 ) d t + ∫ 0 x ( 1 ) d s = [ x e x t + 2 t ] 0 y + [ s ] 0 x = x e x y + 2 y − 0 + x − 0 = x e x y + 2 y + x Thus, the formula F ( x , y ) = ∫ y 0 y N ( x , t ) d t + ∫ x 0 x M ( s , y 0 ) d s can be evaluated as F = x e x y + 2 y + x _ .

Solution Summary: The author evaluates the formula F(x,y) as underset_F=x2y

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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