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

The differential equation y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 . The solution of the differential equation y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 is x 2 2 y 2 − x y − 2 x y 2 = C and y ≡ 0 _ . Theorem used: Exact equation: “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 .” 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 .” Procedure used: “a) If M ( x , y ) d x + N ( x , y ) d y = 0 is exact, then ∂ F ∂ x = M . Integrate the last equation with respect to x to get F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) b) To determine g ( y ) , take the partial derivative with respect to y of both sides of equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) and substitute N for ∂ F ∂ y . We can now solve for g ′ ( y ) . c) Integrate g ′ ( y ) to obtain g ( y ) up to a numerical constant. Substituting g ( y ) into equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) gives F ( x , y ) . d) The solution to M d x + N d y = 0 is given implicitly by F ( x , y ) = C .” Calculation: The differential equation is, y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 ... ( 1 ) Compare y ( x − y − 2 ) d x + x ( y − x + 4 ) 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 ( x − y − 2 ) N ( x , y ) = x ( y − x + 4 ) Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( y ( x − y − 2 ) ) = ∂ ∂ y ( x y − y 2 − 2 y ) = x − 2 y − 2 Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( x ( y − x + 4 ) ) = ∂ ∂ x ( x y − x 2 + 4 x ) = y − 2 x + 4 Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the given differential equation is not exact, and it is also neither separable nor linear. So, use the method of special integrating factor to solve the equation. Calculate ( ∂ M ∂ y − ∂ N ∂ x ) N as follows. ( ∂ M ∂ y − ∂ N ∂ x ) N = x − 2 y − 2 − ( y − 2 x + 4 ) x ( y − x + 4 ) = 3 x − 3 y − 6 x y − x 2 + 4 x Note that, the function not only contains the variable x . So, calculate ( ∂ N ∂ x − ∂ M ∂ y ) M as follows. ( ∂ N ∂ x − ∂ M ∂ y ) M = y − 2 x + 4 − ( x − 2 y − 2 ) y ( x − y − 2 ) = 3 y − 3 x + 6 x y − y 2 − 2 y = − 3 ( − y + x − 2 ) y ( x − y − 2 ) = − 3 y Here, the function is continuous on ℝ − { 0 } and only depends on the variable y . Calculate μ ( y ) as follows. μ ( y ) = exp [ ∫ ( − 3 y ) d y ] = exp ( − 3 ln y ) = e ln y − 3 = y − 3 = 1 y 3 Multiply the equation (1) by μ ( y ) , to make the equation is exact. 1 y 3 [ y ( x − y − 2 ) d x + x ( y − x + 4 ) d y ] = 0 ( x − y − 2 ) y 2 d x + x ( y − x + 4 ) y 3 d y = 0 ( x y 2 − 1 y − 2 y 2 ) d x + ( x y 2 − x 2 y 3 + 4 x y 3 ) d y = 0 ... ( 2 ) Check whether the equation (2) is exact or not. M y ( x , y ) = ∂ ∂ y ( x y 2 − 1 y − 2 y 2 ) = − 2 x y 3 + 1 y 2 + 4 y 3 N x ( x , y ) = ∂ ∂ x ( x y 2 − x 2 y 3 + 4 x y 3 ) = 1 y 2 − 2 x y 3 + 4 y 3 Therefore, the differential equation ( x y 2 − 1 y − 2 y 2 ) d x + ( x y 2 − x 2 y 3 + 4 x y 3 ) d y = 0 is exact. Integrate M ( x , y ) with respect to x and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( x y 2 − 1 y − 2 y 2 ) d x + g ( y ) = x 2 2 y 2 − x y − 2 x y 2 + g ( y ) Differentiate F ( x , y ) = x 2 2 y 2 − x y − 2 x y 2 + g ( y ) partially with respect to y and substitute x y 2 − x 2 y 3 + 4 x y 3 for N as shown below. ∂ F ∂ y = N ∂ ∂ y ( x 2 2 y 2 − x y − 2 x y 2 + g ( y ) ) = x y 2 − x 2 y 3 + 4 x y 3 − 2 x 2 2 y 3 + x y 2 + 4 x y 3 + g ′ ( y ) = x y 2 − x 2 y 3 + 4 x y 3 − x 2 y 3 + x y 2 + 4 x y 3 + g ′ ( y ) = − x 2 y 3 + x y 2 + 4 x y 3 Thus, it is observed that g ′ ( y ) = 0 . Note that as the choice of constant of integration is not important obtain g ( y ) as follows. g ( y ) = ∫ ( 0 ) d y = 0 Substitute g ( y ) = 0 in F ( x , y ) = x 2 2 y 2 − x y − 2 x y 2 + g ( y ) and the following is obtained. F ( x , y ) = x 2 2 y 2 − x y − 2 x y 2 Obtain the solution of the differential equation ( x y 2 − 1 y − 2 y 2 ) d x + ( x y 2 − x 2 y 3 + 4 x y 3 ) d y = 0 implicitly as shown below. x 2 2 y 2 − x y − 2 x y 2 = C for y ≠ 0 Hence, the solution of the differential equation y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 is x 2 2 y 2 − x y − 2 x y 2 = C and y ≡ 0 _ .

The differential equation y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 . The solution of the differential equation y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 is x 2 2 y 2 − x y − 2 x y 2 = C and y ≡ 0 _ . Theorem used: Exact equation: “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 .” 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 .” Procedure used: “a) If M ( x , y ) d x + N ( x , y ) d y = 0 is exact, then ∂ F ∂ x = M . Integrate the last equation with respect to x to get F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) b) To determine g ( y ) , take the partial derivative with respect to y of both sides of equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) and substitute N for ∂ F ∂ y . We can now solve for g ′ ( y ) . c) Integrate g ′ ( y ) to obtain g ( y ) up to a numerical constant. Substituting g ( y ) into equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) gives F ( x , y ) . d) The solution to M d x + N d y = 0 is given implicitly by F ( x , y ) = C .” Calculation: The differential equation is, y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 ... ( 1 ) Compare y ( x − y − 2 ) d x + x ( y − x + 4 ) 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 ( x − y − 2 ) N ( x , y ) = x ( y − x + 4 ) Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( y ( x − y − 2 ) ) = ∂ ∂ y ( x y − y 2 − 2 y ) = x − 2 y − 2 Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( x ( y − x + 4 ) ) = ∂ ∂ x ( x y − x 2 + 4 x ) = y − 2 x + 4 Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the given differential equation is not exact, and it is also neither separable nor linear. So, use the method of special integrating factor to solve the equation. Calculate ( ∂ M ∂ y − ∂ N ∂ x ) N as follows. ( ∂ M ∂ y − ∂ N ∂ x ) N = x − 2 y − 2 − ( y − 2 x + 4 ) x ( y − x + 4 ) = 3 x − 3 y − 6 x y − x 2 + 4 x Note that, the function not only contains the variable x . So, calculate ( ∂ N ∂ x − ∂ M ∂ y ) M as follows. ( ∂ N ∂ x − ∂ M ∂ y ) M = y − 2 x + 4 − ( x − 2 y − 2 ) y ( x − y − 2 ) = 3 y − 3 x + 6 x y − y 2 − 2 y = − 3 ( − y + x − 2 ) y ( x − y − 2 ) = − 3 y Here, the function is continuous on ℝ − { 0 } and only depends on the variable y . Calculate μ ( y ) as follows. μ ( y ) = exp [ ∫ ( − 3 y ) d y ] = exp ( − 3 ln y ) = e ln y − 3 = y − 3 = 1 y 3 Multiply the equation (1) by μ ( y ) , to make the equation is exact. 1 y 3 [ y ( x − y − 2 ) d x + x ( y − x + 4 ) d y ] = 0 ( x − y − 2 ) y 2 d x + x ( y − x + 4 ) y 3 d y = 0 ( x y 2 − 1 y − 2 y 2 ) d x + ( x y 2 − x 2 y 3 + 4 x y 3 ) d y = 0 ... ( 2 ) Check whether the equation (2) is exact or not. M y ( x , y ) = ∂ ∂ y ( x y 2 − 1 y − 2 y 2 ) = − 2 x y 3 + 1 y 2 + 4 y 3 N x ( x , y ) = ∂ ∂ x ( x y 2 − x 2 y 3 + 4 x y 3 ) = 1 y 2 − 2 x y 3 + 4 y 3 Therefore, the differential equation ( x y 2 − 1 y − 2 y 2 ) d x + ( x y 2 − x 2 y 3 + 4 x y 3 ) d y = 0 is exact. Integrate M ( x , y ) with respect to x and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( x y 2 − 1 y − 2 y 2 ) d x + g ( y ) = x 2 2 y 2 − x y − 2 x y 2 + g ( y ) Differentiate F ( x , y ) = x 2 2 y 2 − x y − 2 x y 2 + g ( y ) partially with respect to y and substitute x y 2 − x 2 y 3 + 4 x y 3 for N as shown below. ∂ F ∂ y = N ∂ ∂ y ( x 2 2 y 2 − x y − 2 x y 2 + g ( y ) ) = x y 2 − x 2 y 3 + 4 x y 3 − 2 x 2 2 y 3 + x y 2 + 4 x y 3 + g ′ ( y ) = x y 2 − x 2 y 3 + 4 x y 3 − x 2 y 3 + x y 2 + 4 x y 3 + g ′ ( y ) = − x 2 y 3 + x y 2 + 4 x y 3 Thus, it is observed that g ′ ( y ) = 0 . Note that as the choice of constant of integration is not important obtain g ( y ) as follows. g ( y ) = ∫ ( 0 ) d y = 0 Substitute g ( y ) = 0 in F ( x , y ) = x 2 2 y 2 − x y − 2 x y 2 + g ( y ) and the following is obtained. F ( x , y ) = x 2 2 y 2 − x y − 2 x y 2 Obtain the solution of the differential equation ( x y 2 − 1 y − 2 y 2 ) d x + ( x y 2 − x 2 y 3 + 4 x y 3 ) d y = 0 implicitly as shown below. x 2 2 y 2 − x y − 2 x y 2 = C for y ≠ 0 Hence, the solution of the differential equation y ( x − y − 2 ) d x + x ( y − x + 4 ) d y = 0 is x 2 2 y 2 − x y − 2 x y 2 = C and y ≡ 0 _ .

Solution Summary: The author states that the differential equation y(x-y-2)dx+x (y-x +4 ) is an exact equation in R if the compatibility condition

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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original ssD 400x (100) 2 400 x (14)=100 34 10or (2)² = 100 × (0-85) = 100 x = 72.25 100x 40 72.25 =36.13 500 36.13 13.84 O. 7225 12x13.84≈ 166.08mAs 10) The dose of a radiograph was 100 mSv with a technique of: 10 mAs, 180 kV at 200 cm and tabletop. If the technique is changed to: 20 mAs, 153 kV at 100 cm using a 5:1 grid then find the new dose in rem to the Lungs. 11) A radiographic technique produces an exposure index of EI= 300 and TEI=600 at a source- to-image receptor distance (SID) of 200 cm, 100 kV, 5:1 grid using 10 mAs. If the technique is changed to 100 cm and 85 kV and 5 mAs with table top find the new exposure index? What is the value of DI? 10

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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. 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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