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

(a) To show: The equation M ( x , y ) d x + N ( x , y ) d y = 0 has an integrating factor of the form μ ( x y ) if ∂ N / ∂ x − ∂ M / ∂ y x M − y N depends only on the product x y that is, ∂ N / ∂ x − ∂ M / ∂ y x M − y N = H ( x y ) ; Give the general formula for μ ( x y ) . The general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x y _ . The differential equation is, M ( x , y ) d x + N ( x , y ) d y = 0 ... ( 1 ) Assume that, the function ∂ N / ∂ x − ∂ M / ∂ y x M − y N depends only on the product x y that is, ∂ N / ∂ x − ∂ M / ∂ y x M − y N = H ( x y ) . Suppose μ ( x y ) be an integrating factor of the equation (1) such that ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ x ( ∂ F ∂ y ) . Multiply the equation (1) by μ ( x y ) . μ ( x y ) M d x + μ ( x y ) N d y = 0 ... ( 2 ) Here, ∂ F ∂ x = μ ( x y ) M and ∂ F ∂ y = μ ( x y ) N . Differentiate ∂ F ∂ x with respect to y as shown below. ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ y ( μ ( x y ) M ) = μ ( x y ) ∂ M ∂ y + M ∂ μ ( x y ) ∂ ( x y ) ⋅ ∂ ( x y ) ∂ y = μ ( x y ) ∂ M ∂ y + M μ ′ ( x y ) ⋅ x = μ ( x y ) ∂ M ∂ y + M x μ ′ ( x y ) Differentiate ∂ F ∂ y with respect to x as shown below. ∂ ∂ x ( ∂ F ∂ y ) = ∂ ∂ x ( μ ( x y ) N ) = μ ( x y ) ∂ N ∂ x + N ∂ μ ( x y ) ∂ ( x y ) ⋅ ∂ ( x y ) ∂ x = μ ( x y ) ∂ N ∂ x + N μ ′ ( x y ) ⋅ y = μ ( x y ) ∂ N ∂ x + N y μ ′ ( x y ) The given equation (2) to be exact, then ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ x ( ∂ F ∂ y ) . μ ( x y ) ∂ M ∂ y + M x μ ′ ( x y ) = μ ( x y ) ∂ N ∂ x + N y μ ′ ( x y ) μ ( x y ) ∂ M ∂ y − μ ( x y ) ∂ N ∂ x = N y μ ′ ( x y ) − M x μ ′ ( x y ) μ ( x y ) ( ∂ M ∂ y − ∂ N ∂ x ) = μ ′ ( x y ) ( N y − M x ) μ ′ ( x y ) μ ( x y ) = ( ∂ M ∂ y − ∂ N ∂ x ) ( N y − M x ) μ ′ ( x y ) μ ( x y ) = H ( x y ) ( by assumption ) Take z = x y , the above result becomes μ ' ( z ) μ ( z ) = H ( z ) Integrate on both sides of the equation. ∫ μ ' ( z ) μ ( z ) d z = ∫ H ( z ) d z ln μ ( z ) = ∫ H ( z ) d z μ ( z ) = e ∫ H ( z ) d z where z = x y Hence, it is showed that the equation M ( x , y ) d x + N ( x , y ) d y = 0 has an integrating factor of the form μ ( x y ) and the general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x y _ . (b) To find: An implicit solution to the IVP ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) d y = 0 , y ( 1 ) = 1 by using an answer to part (a). The solution of the differential equation ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) d y = 0 is ln x 3 y + 2 x y = 2 _ . 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 .” 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, ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) d y = 0 ... ( 1 ) Compare ( 3 y + 2 x y 2 ) d x + ( 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 ) = 3 y + 2 x y 2 N ( x , y ) = x + 2 x 2 y Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y = ∂ ∂ y ( 3 y + 2 x y 2 ) = 3 + 4 x y Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x = ∂ ∂ x ( x + 2 x 2 y ) = 1 + 4 x y From part (a), the general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x y . Calculate H ( x y ) as follows. H ( x y ) = ∂ N ∂ x − ∂ M ∂ y x M − y N = 1 + 4 x y − 3 − 4 x y x ( 3 y + 2 x y 2 ) − y ( x + 2 x 2 y ) = − 2 3 x y + 2 x 2 y 2 − x y − 2 x 2 y 2 = − 2 2 x y = − 1 x y Therefore, the function H ( x y ) = − 1 x y . Find the integrating factor μ ( x y ) as follows. μ ( x y ) = exp ( ∫ ( − 1 x y ) d ( x y ) ) μ ( z ) = exp ( − ∫ 1 z d z ) ( take z = x y ) μ ( z ) = e ln 1 z μ ( z ) = 1 z μ ( x y ) = 1 x y ( substitute back, z = x y ) Multiply the equation (1) by μ ( x y ) = 1 x y , to make the equation is exact 1 x y ( 3 y + 2 x y 2 ) d x + 1 x y ( x + 2 x 2 y ) d y = 0 ( 3 x + 2 y ) d x + ( 1 y + 2 x ) d y = 0 Integrate M ( x , y ) with respect to x and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( 3 x + 2 y ) d x + g ( y ) = ∫ 3 x d x + ∫ 2 y d x + g ( y ) = 3 ln x + 2 x y + g ( y ) Differentiate F ( x , y ) = 3 ln x + 2 x y + g ( y ) partially with respect to y and substitute 1 y + 2 x for N as shown below. ∂ F ∂ y = N ∂ F ∂ y ( 3 ln x + 2 x y + g ( y ) ) = 1 y + 2 x 2 x + g ′ ( y ) = 1 y + 2 x g ′ ( y ) = 1 y Thus, it is observed that g ′ ( y ) = 1 y . Note that as the choice of constant of integration is not important obtain g ( y ) as follows. g ( y ) = ∫ ( 1 y ) d y = ln y Substitute g ( y ) = ln y in F ( x , y ) = 3 ln x + 2 x y + g ( y ) and the following is obtained. F ( x , y ) = 3 ln x + 2 x y + ln y Obtain the solution of the differential equation ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) = 0 implicitly as shown below. 3 ln x + 2 x y + ln y = C Apply the initial condition y ( 1 ) = 1 in the above result to find the constant C . 3 ln ( 1 ) + 2 ( 1 ) + ln 1 = C C = 2 Substitute C = 2 in the solution 3 ln x + 2 x y + ln y = C and simplify further. 3 ln x + 2 x y + ln y = 2 ln x 3 + 2 x y + ln y = 2 ln x 3 y + 2 x y = 2 Hence, the solution of the differential equation ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) = 0 is ln x 3 y + 2 x y = 2 _ .

(a) To show: The equation M ( x , y ) d x + N ( x , y ) d y = 0 has an integrating factor of the form μ ( x y ) if ∂ N / ∂ x − ∂ M / ∂ y x M − y N depends only on the product x y that is, ∂ N / ∂ x − ∂ M / ∂ y x M − y N = H ( x y ) ; Give the general formula for μ ( x y ) . The general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x y _ . The differential equation is, M ( x , y ) d x + N ( x , y ) d y = 0 ... ( 1 ) Assume that, the function ∂ N / ∂ x − ∂ M / ∂ y x M − y N depends only on the product x y that is, ∂ N / ∂ x − ∂ M / ∂ y x M − y N = H ( x y ) . Suppose μ ( x y ) be an integrating factor of the equation (1) such that ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ x ( ∂ F ∂ y ) . Multiply the equation (1) by μ ( x y ) . μ ( x y ) M d x + μ ( x y ) N d y = 0 ... ( 2 ) Here, ∂ F ∂ x = μ ( x y ) M and ∂ F ∂ y = μ ( x y ) N . Differentiate ∂ F ∂ x with respect to y as shown below. ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ y ( μ ( x y ) M ) = μ ( x y ) ∂ M ∂ y + M ∂ μ ( x y ) ∂ ( x y ) ⋅ ∂ ( x y ) ∂ y = μ ( x y ) ∂ M ∂ y + M μ ′ ( x y ) ⋅ x = μ ( x y ) ∂ M ∂ y + M x μ ′ ( x y ) Differentiate ∂ F ∂ y with respect to x as shown below. ∂ ∂ x ( ∂ F ∂ y ) = ∂ ∂ x ( μ ( x y ) N ) = μ ( x y ) ∂ N ∂ x + N ∂ μ ( x y ) ∂ ( x y ) ⋅ ∂ ( x y ) ∂ x = μ ( x y ) ∂ N ∂ x + N μ ′ ( x y ) ⋅ y = μ ( x y ) ∂ N ∂ x + N y μ ′ ( x y ) The given equation (2) to be exact, then ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ x ( ∂ F ∂ y ) . μ ( x y ) ∂ M ∂ y + M x μ ′ ( x y ) = μ ( x y ) ∂ N ∂ x + N y μ ′ ( x y ) μ ( x y ) ∂ M ∂ y − μ ( x y ) ∂ N ∂ x = N y μ ′ ( x y ) − M x μ ′ ( x y ) μ ( x y ) ( ∂ M ∂ y − ∂ N ∂ x ) = μ ′ ( x y ) ( N y − M x ) μ ′ ( x y ) μ ( x y ) = ( ∂ M ∂ y − ∂ N ∂ x ) ( N y − M x ) μ ′ ( x y ) μ ( x y ) = H ( x y ) ( by assumption ) Take z = x y , the above result becomes μ ' ( z ) μ ( z ) = H ( z ) Integrate on both sides of the equation. ∫ μ ' ( z ) μ ( z ) d z = ∫ H ( z ) d z ln μ ( z ) = ∫ H ( z ) d z μ ( z ) = e ∫ H ( z ) d z where z = x y Hence, it is showed that the equation M ( x , y ) d x + N ( x , y ) d y = 0 has an integrating factor of the form μ ( x y ) and the general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x y _ . (b) To find: An implicit solution to the IVP ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) d y = 0 , y ( 1 ) = 1 by using an answer to part (a). The solution of the differential equation ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) d y = 0 is ln x 3 y + 2 x y = 2 _ . 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 .” 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, ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) d y = 0 ... ( 1 ) Compare ( 3 y + 2 x y 2 ) d x + ( 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 ) = 3 y + 2 x y 2 N ( x , y ) = x + 2 x 2 y Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y = ∂ ∂ y ( 3 y + 2 x y 2 ) = 3 + 4 x y Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x = ∂ ∂ x ( x + 2 x 2 y ) = 1 + 4 x y From part (a), the general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x y . Calculate H ( x y ) as follows. H ( x y ) = ∂ N ∂ x − ∂ M ∂ y x M − y N = 1 + 4 x y − 3 − 4 x y x ( 3 y + 2 x y 2 ) − y ( x + 2 x 2 y ) = − 2 3 x y + 2 x 2 y 2 − x y − 2 x 2 y 2 = − 2 2 x y = − 1 x y Therefore, the function H ( x y ) = − 1 x y . Find the integrating factor μ ( x y ) as follows. μ ( x y ) = exp ( ∫ ( − 1 x y ) d ( x y ) ) μ ( z ) = exp ( − ∫ 1 z d z ) ( take z = x y ) μ ( z ) = e ln 1 z μ ( z ) = 1 z μ ( x y ) = 1 x y ( substitute back, z = x y ) Multiply the equation (1) by μ ( x y ) = 1 x y , to make the equation is exact 1 x y ( 3 y + 2 x y 2 ) d x + 1 x y ( x + 2 x 2 y ) d y = 0 ( 3 x + 2 y ) d x + ( 1 y + 2 x ) d y = 0 Integrate M ( x , y ) with respect to x and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( 3 x + 2 y ) d x + g ( y ) = ∫ 3 x d x + ∫ 2 y d x + g ( y ) = 3 ln x + 2 x y + g ( y ) Differentiate F ( x , y ) = 3 ln x + 2 x y + g ( y ) partially with respect to y and substitute 1 y + 2 x for N as shown below. ∂ F ∂ y = N ∂ F ∂ y ( 3 ln x + 2 x y + g ( y ) ) = 1 y + 2 x 2 x + g ′ ( y ) = 1 y + 2 x g ′ ( y ) = 1 y Thus, it is observed that g ′ ( y ) = 1 y . Note that as the choice of constant of integration is not important obtain g ( y ) as follows. g ( y ) = ∫ ( 1 y ) d y = ln y Substitute g ( y ) = ln y in F ( x , y ) = 3 ln x + 2 x y + g ( y ) and the following is obtained. F ( x , y ) = 3 ln x + 2 x y + ln y Obtain the solution of the differential equation ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) = 0 implicitly as shown below. 3 ln x + 2 x y + ln y = C Apply the initial condition y ( 1 ) = 1 in the above result to find the constant C . 3 ln ( 1 ) + 2 ( 1 ) + ln 1 = C C = 2 Substitute C = 2 in the solution 3 ln x + 2 x y + ln y = C and simplify further. 3 ln x + 2 x y + ln y = 2 ln x 3 + 2 x y + ln y = 2 ln x 3 y + 2 x y = 2 Hence, the solution of the differential equation ( 3 y + 2 x y 2 ) d x + ( x + 2 x 2 y ) = 0 is ln x 3 y + 2 x y = 2 _ .

Solution Summary: The author explains the formula for the equation M(x,y)dx+N

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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)5. The operator T on a vector space V over Q has as its sequence of invariant factors the sequence (x-1)3(x-2)5(x-3)2, (x-1)(x-2)(x-3)2, (x-3). Find the Jordan normal form of T.

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Fundamentals of Differential Equations (9th Edition)

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