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

Whether the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 is exact and solve it. The differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 is exact _ and the solution is y = − 2 t e t + 2 _ . 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 .” 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, ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 Compare ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 with the equation M ( t , y ) d t + N ( t , y ) d y = 0 and the following is obtained. M ( t , y ) = e t y + t e t y N ( t , y ) = t e t + 2 Differentiate M ( t , y ) with respect to y as shown below. M y ( t , y ) = ∂ ∂ y ( e t y + t e t y ) = e t + t e t Differentiate N ( t , y ) with respect to t as shown below. N t ( x , y ) = ∂ ∂ t ( t e t + 2 ) = t e t + e t Note that as ∂ M ∂ y ( t , y ) = ∂ N ∂ t ( t , y ) , the given differential equation is exact. Therefore, the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 is exact _ . Integrate M ( t , y ) with respect to x and obtain F ( t , y ) as shown below. F ( t , y ) = ∫ ( e t y + t e t y ) d t + g ( y ) = y t e t + g ( y ) Differentiate F ( t , y ) = y t e t + g ( y ) partially with respect to y and substitute t e t + 2 for N as shown below. ∂ F ∂ y = N ∂ ∂ y ( y t e t + g ( y ) ) = t e t + 2 t e t + g ′ ( y ) = t e t + 2 Thus, it is observed that g ′ ( y ) = 2 . Note that as the choice of integration is not important obtain g ( y ) as follows. g ( y ) = ∫ 2 d y = 2 y Substitute g ( y ) = 2 y in F ( t , y ) = y t e t + g ( y ) and the following is obtained. F ( t , y ) = y t e t + 2 y Obtain the solution of the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 implicitly as shown below. y t e t + 2 y = C y ( t e t + 2 ) = C y = C t e t + 2 Use the initial condition y ( 0 ) = − 1 in the solution of the differential equation as shown. y = C t e t + 2 − 1 = C ( 0 ) e 0 + 2 − 1 = C 2 C = − 2 Hence, the solution of the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 is y = − 2 t e t + 2 _ .

Whether the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 is exact and solve it. The differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 is exact _ and the solution is y = − 2 t e t + 2 _ . 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 .” 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, ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 Compare ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 with the equation M ( t , y ) d t + N ( t , y ) d y = 0 and the following is obtained. M ( t , y ) = e t y + t e t y N ( t , y ) = t e t + 2 Differentiate M ( t , y ) with respect to y as shown below. M y ( t , y ) = ∂ ∂ y ( e t y + t e t y ) = e t + t e t Differentiate N ( t , y ) with respect to t as shown below. N t ( x , y ) = ∂ ∂ t ( t e t + 2 ) = t e t + e t Note that as ∂ M ∂ y ( t , y ) = ∂ N ∂ t ( t , y ) , the given differential equation is exact. Therefore, the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 is exact _ . Integrate M ( t , y ) with respect to x and obtain F ( t , y ) as shown below. F ( t , y ) = ∫ ( e t y + t e t y ) d t + g ( y ) = y t e t + g ( y ) Differentiate F ( t , y ) = y t e t + g ( y ) partially with respect to y and substitute t e t + 2 for N as shown below. ∂ F ∂ y = N ∂ ∂ y ( y t e t + g ( y ) ) = t e t + 2 t e t + g ′ ( y ) = t e t + 2 Thus, it is observed that g ′ ( y ) = 2 . Note that as the choice of integration is not important obtain g ( y ) as follows. g ( y ) = ∫ 2 d y = 2 y Substitute g ( y ) = 2 y in F ( t , y ) = y t e t + g ( y ) and the following is obtained. F ( t , y ) = y t e t + 2 y Obtain the solution of the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 implicitly as shown below. y t e t + 2 y = C y ( t e t + 2 ) = C y = C t e t + 2 Use the initial condition y ( 0 ) = − 1 in the solution of the differential equation as shown. y = C t e t + 2 − 1 = C ( 0 ) e 0 + 2 − 1 = C 2 C = − 2 Hence, the solution of the differential equation ( e t y + t e t y ) d t + ( t e t + 2 ) d y = 0 , y ( 0 ) = − 1 is y = − 2 t e t + 2 _ .

Solution Summary: The author explains how to solve the differential equation (ety), if the compatibility condition partial x=M holds.

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)

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(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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