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

(a) To show: That the equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is not exact. 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 .” Calculation: The differential equation is, ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ... ( 1 ) Compare ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 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 ) = 5 x 2 y + 6 x 3 y 2 + 4 x y 2 N ( x , y ) = 2 x 3 + 3 x 4 y + 3 x 2 y Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) = 5 x 2 + 12 x 3 y + 8 x y Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( 2 x 3 + 3 x 4 y + 3 x 2 y ) = 6 x 2 + 12 x 3 y + 6 x y Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the given differential equation is not exact. Hence, it is showed that the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is not exact. (b) To multiply: The equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 by x n y m and find the values of n and m that make the resulting equation exact. The product of x n y m and the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y = 0 _ where m = 1 and n = 2 _ . The resulting exact equation is ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 _ . The differential equation is, ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 Multiply both sides of the equation by x n y m , to make the equation is exact. x n y m [ ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ] = 0 x n y m ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + x n y m ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y = 0 [ ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y ] = 0 ... ( 2 ) Take the equation (2) to be exact, then ∂ M ∂ y = ∂ N ∂ x . Compare ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) 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 ) = 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 N ( x , y ) = 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y = ∂ ∂ y ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) = 5 ( m + 1 ) x n + 2 y m + 6 ( m + 2 ) x n + 3 y m + 2 − 1 + 4 ( m + 2 ) x n + 1 y m + 2 − 1 = 5 ( m + 1 ) x n + 2 y m + 6 ( m + 2 ) x n + 3 y m + 1 + 4 ( m + 2 ) x n + 1 y m + 1 Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x = ∂ ∂ x ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) = 2 ( n + 3 ) x n + 3 − 1 y m + 3 ( n + 4 ) x n + 4 − 1 y m + 1 + 3 ( n + 2 ) x n + 2 − 1 y m + 1 = 2 ( n + 3 ) x n + 2 y m + 3 ( n + 4 ) x n + 3 y m + 1 + 3 ( n + 2 ) x n + 1 y m + 1 Since ∂ M ∂ y = ∂ N ∂ x , the coefficients of x n + 2 y m , x n + 3 y m + 1 and x n + 1 y m + 1 must be equal to each other. That is, { 5 ( m + 1 ) = 2 ( n + 3 ) ⇒ 5 m + 5 = 2 n + 6 6 ( m + 2 ) = 3 ( n + 4 ) ⇒ 6 m + 12 = 3 n + 12 4 ( m + 2 ) = 3 ( n + 2 ) ⇒ 4 m + 8 = 3 n + 6 The above equations can be simplified as, 5 m − 2 n = 1 ... ( a ) 6 m − 3 n = 0 ... ( b ) 4 m − 3 n = − 2 ... ( c ) Multiply the equation (b) by − 1 and add it to (c). − 6 m + 3 n = 0 4 m − 3 n = − 2 _ − 2 m = − 2 m = 1 Substitute m = 1 in equation (a) 5 ( 1 ) − 2 n = 1 5 − 2 n = 1 − 2 n = 1 − 5 − 2 n = − 4 n = 2 Therefore, the product of x n y m and the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y = 0 _ where m = 1 and n = 2 _ . Substitute the values of m and n in x n y m . x n y m = x 2 y Multiply the equation (1) by integrating factor x 2 y . x 2 y ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + x 2 y ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 ... ( 3 ) Check whether the resulting equation (3) is exact or not. Compare ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 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 ) = 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 N ( x , y ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) = 10 x 4 y + 18 x 5 y 2 + 12 x 3 y 2 Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) = 10 x 4 y + 18 x 5 y 2 + 12 x 3 y 2 Since ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) , the new differential equation is exact. Thus, the resulting exact equation is ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 _ . (c) To solve: The original equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 by using the resulting exact equation ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 . The solution of the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is x 5 y 2 + x 6 y 3 + x 4 y 3 = C _ . 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: From part (b), the resulting exact equation is, ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 . Compare ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 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 ) = 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 N ( x , y ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 Integrate M ( x , y ) with respect to x and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + g ( y ) = 5 x 5 y 2 5 + 6 x 6 y 3 6 + 4 x 4 y 3 4 + g ( y ) = x 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) Differentiate F ( x , y ) = x 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) partially with respect to y and substitute 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 for N as shown below. ∂ F ∂ y = N ∂ F ∂ y ( x 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 + g ′ ( y ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 g ′ ( y ) = 0 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 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) and the following is obtained. F ( x , y ) = x 5 y 2 + x 6 y 3 + x 4 y 3 Obtain the solution of the differential equation ( 1 + 2 x y ) d x − ( x 2 y 2 ) d y = 0 implicitly as shown below. x 5 y 2 + x 6 y 3 + x 4 y 3 = C Hence, the solution of the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is x 5 y 2 + x 6 y 3 + x 4 y 3 = C _ .

(a) To show: That the equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is not exact. 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 .” Calculation: The differential equation is, ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ... ( 1 ) Compare ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 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 ) = 5 x 2 y + 6 x 3 y 2 + 4 x y 2 N ( x , y ) = 2 x 3 + 3 x 4 y + 3 x 2 y Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) = 5 x 2 + 12 x 3 y + 8 x y Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( 2 x 3 + 3 x 4 y + 3 x 2 y ) = 6 x 2 + 12 x 3 y + 6 x y Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the given differential equation is not exact. Hence, it is showed that the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is not exact. (b) To multiply: The equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 by x n y m and find the values of n and m that make the resulting equation exact. The product of x n y m and the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y = 0 _ where m = 1 and n = 2 _ . The resulting exact equation is ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 _ . The differential equation is, ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 Multiply both sides of the equation by x n y m , to make the equation is exact. x n y m [ ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ] = 0 x n y m ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + x n y m ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y = 0 [ ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y ] = 0 ... ( 2 ) Take the equation (2) to be exact, then ∂ M ∂ y = ∂ N ∂ x . Compare ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) 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 ) = 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 N ( x , y ) = 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y = ∂ ∂ y ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) = 5 ( m + 1 ) x n + 2 y m + 6 ( m + 2 ) x n + 3 y m + 2 − 1 + 4 ( m + 2 ) x n + 1 y m + 2 − 1 = 5 ( m + 1 ) x n + 2 y m + 6 ( m + 2 ) x n + 3 y m + 1 + 4 ( m + 2 ) x n + 1 y m + 1 Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x = ∂ ∂ x ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) = 2 ( n + 3 ) x n + 3 − 1 y m + 3 ( n + 4 ) x n + 4 − 1 y m + 1 + 3 ( n + 2 ) x n + 2 − 1 y m + 1 = 2 ( n + 3 ) x n + 2 y m + 3 ( n + 4 ) x n + 3 y m + 1 + 3 ( n + 2 ) x n + 1 y m + 1 Since ∂ M ∂ y = ∂ N ∂ x , the coefficients of x n + 2 y m , x n + 3 y m + 1 and x n + 1 y m + 1 must be equal to each other. That is, { 5 ( m + 1 ) = 2 ( n + 3 ) ⇒ 5 m + 5 = 2 n + 6 6 ( m + 2 ) = 3 ( n + 4 ) ⇒ 6 m + 12 = 3 n + 12 4 ( m + 2 ) = 3 ( n + 2 ) ⇒ 4 m + 8 = 3 n + 6 The above equations can be simplified as, 5 m − 2 n = 1 ... ( a ) 6 m − 3 n = 0 ... ( b ) 4 m − 3 n = − 2 ... ( c ) Multiply the equation (b) by − 1 and add it to (c). − 6 m + 3 n = 0 4 m − 3 n = − 2 _ − 2 m = − 2 m = 1 Substitute m = 1 in equation (a) 5 ( 1 ) − 2 n = 1 5 − 2 n = 1 − 2 n = 1 − 5 − 2 n = − 4 n = 2 Therefore, the product of x n y m and the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is ( 5 x n + 2 y m + 1 + 6 x n + 3 y m + 2 + 4 x n + 1 y m + 2 ) d x + ( 2 x n + 3 y m + 3 x n + 4 y m + 1 + 3 x n + 2 y m + 1 ) d y = 0 _ where m = 1 and n = 2 _ . Substitute the values of m and n in x n y m . x n y m = x 2 y Multiply the equation (1) by integrating factor x 2 y . x 2 y ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + x 2 y ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 ... ( 3 ) Check whether the resulting equation (3) is exact or not. Compare ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 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 ) = 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 N ( x , y ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) = 10 x 4 y + 18 x 5 y 2 + 12 x 3 y 2 Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) = 10 x 4 y + 18 x 5 y 2 + 12 x 3 y 2 Since ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) , the new differential equation is exact. Thus, the resulting exact equation is ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 _ . (c) To solve: The original equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 by using the resulting exact equation ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 . The solution of the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is x 5 y 2 + x 6 y 3 + x 4 y 3 = C _ . 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: From part (b), the resulting exact equation is, ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 ) d y = 0 . Compare ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + ( 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 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 ) = 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 N ( x , y ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 Integrate M ( x , y ) with respect to x and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( 5 x 4 y 2 + 6 x 5 y 3 + 4 x 3 y 3 ) d x + g ( y ) = 5 x 5 y 2 5 + 6 x 6 y 3 6 + 4 x 4 y 3 4 + g ( y ) = x 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) Differentiate F ( x , y ) = x 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) partially with respect to y and substitute 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 for N as shown below. ∂ F ∂ y = N ∂ F ∂ y ( x 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 + g ′ ( y ) = 2 x 5 y + 3 x 6 y 2 + 3 x 4 y 2 g ′ ( y ) = 0 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 5 y 2 + x 6 y 3 + x 4 y 3 + g ( y ) and the following is obtained. F ( x , y ) = x 5 y 2 + x 6 y 3 + x 4 y 3 Obtain the solution of the differential equation ( 1 + 2 x y ) d x − ( x 2 y 2 ) d y = 0 implicitly as shown below. x 5 y 2 + x 6 y 3 + x 4 y 3 = C Hence, the solution of the differential equation ( 5 x 2 y + 6 x 3 y 2 + 4 x y 2 ) d x + ( 2 x 3 + 3 x 4 y + 3 x 2 y ) d y = 0 is x 5 y 2 + x 6 y 3 + x 4 y 3 = C _ .

Solution Summary: The author explains that the differential equation (x,y)dx+Nlmtnrdy=0 is not exact, 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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Fundamentals of Differential Equations (9th Edition)

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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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