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

The differential equation d y d x = ( x − y + 5 ) 2 . The solution of the differential equation d y d x = ( x − y + 5 ) 2 is y = 2 C e 2 x + 1 + x + 4 _ . Definition used: “The equation is of the form d y d x = G ( a x + b y ) . Let z = a x + b y . Then d z d x = a + b ( d y d x ) , and the transformed equation in the variables z and x is separable.” Calculation: The differential equation is, d y d x = ( x − y + 5 ) 2 The above equation is of the form d y d x = G ( x + y ) . So take z = x − y + 5 . Then d y d x = 1 − d z d x , the transformed equation is 1 − d z d x = ( z ) 2 d z d x = 1 − z 2 Separate the variables and rewrite the differential equation as follows. 1 1 − z 2 d z = d x Integrate on both sides of the equation. ∫ 1 1 − z 2 d z = ∫ d x ∫ 1 1 − z 2 d z = x + C 1 − ∫ 1 ( z − 1 ) ( z + 1 ) d z = x + C 1 ... ( 1 ) Consider the term 1 ( z − 1 ) ( z + 1 ) . Split the term by using the method of partial fractions. 1 ( z − 1 ) ( z + 1 ) = A ( z − 1 ) + B ( z + 1 ) ... ( 2 ) Multiply ( z − 1 ) ( z + 1 ) on both sides of the equation. ( z − 1 ) ( z + 1 ) ( z − 1 ) ( z + 1 ) = A ( z − 1 ) ( z + 1 ) ( z − 1 ) + B ( z − 1 ) ( z + 1 ) ( z + 1 ) 1 = A ( z + 1 ) + B ( z − 1 ) Put z = 1 results A = 1 2 and put z = − 1 results B = − 1 2 . Substitute A = 1 2 and B = − 1 2 in the equation (2). 1 ( z − 1 ) ( z + 1 ) = 1 2 ( z − 1 ) − 1 2 ( z + 1 ) Thus, the integral (1) becomes, − ∫ ( 1 2 ( z − 1 ) − 1 2 ( z + 1 ) ) d z = x + C 1 − 1 2 ∫ 1 z − 1 d z + 1 2 ∫ 1 z + 1 d z = x + C 1 − 1 2 ln | z − 1 | + 1 2 ln | z + 1 | = x + C 1 1 2 ln | z + 1 z − 1 | = x + C 1 Simplify further as follows. ln | z + 1 z − 1 | = 2 x + 2 C 1 | z + 1 z − 1 | = e 2 x + 2 C 1 z + 1 z − 1 = ± e 2 C 1 e 2 x z + 1 z − 1 = C 2 e 2 x ( Put ± e 2 C 1 = C 2 ) Substitute back, z = x − y + 5 . x − y + 5 + 1 x − y + 5 − 1 = C 2 e 2 x x − y + 6 x − y + 4 = C 2 e 2 x ( x − y + 4 ) + 2 x − y + 4 = C 2 e 2 x 1 + 2 x − y + 4 = C 2 e 2 x Simplify further as follows. 2 x − y + 4 = C 2 e 2 x − 1 x − y + 4 = 2 C 2 e 2 x − 1 − y = 2 C 2 e 2 x − 1 − x − 4 y = 2 − C 2 e 2 x + 1 + x + 4 y = 2 C e 2 x + 1 + x + 4 ( Put − C 2 = C , C is an arbitrary constant ) Thus, the solution of the differential equation d y d x = ( x − y + 5 ) 2 is y = 2 C e 2 x + 1 + x + 4 _ .

The differential equation d y d x = ( x − y + 5 ) 2 . The solution of the differential equation d y d x = ( x − y + 5 ) 2 is y = 2 C e 2 x + 1 + x + 4 _ . Definition used: “The equation is of the form d y d x = G ( a x + b y ) . Let z = a x + b y . Then d z d x = a + b ( d y d x ) , and the transformed equation in the variables z and x is separable.” Calculation: The differential equation is, d y d x = ( x − y + 5 ) 2 The above equation is of the form d y d x = G ( x + y ) . So take z = x − y + 5 . Then d y d x = 1 − d z d x , the transformed equation is 1 − d z d x = ( z ) 2 d z d x = 1 − z 2 Separate the variables and rewrite the differential equation as follows. 1 1 − z 2 d z = d x Integrate on both sides of the equation. ∫ 1 1 − z 2 d z = ∫ d x ∫ 1 1 − z 2 d z = x + C 1 − ∫ 1 ( z − 1 ) ( z + 1 ) d z = x + C 1 ... ( 1 ) Consider the term 1 ( z − 1 ) ( z + 1 ) . Split the term by using the method of partial fractions. 1 ( z − 1 ) ( z + 1 ) = A ( z − 1 ) + B ( z + 1 ) ... ( 2 ) Multiply ( z − 1 ) ( z + 1 ) on both sides of the equation. ( z − 1 ) ( z + 1 ) ( z − 1 ) ( z + 1 ) = A ( z − 1 ) ( z + 1 ) ( z − 1 ) + B ( z − 1 ) ( z + 1 ) ( z + 1 ) 1 = A ( z + 1 ) + B ( z − 1 ) Put z = 1 results A = 1 2 and put z = − 1 results B = − 1 2 . Substitute A = 1 2 and B = − 1 2 in the equation (2). 1 ( z − 1 ) ( z + 1 ) = 1 2 ( z − 1 ) − 1 2 ( z + 1 ) Thus, the integral (1) becomes, − ∫ ( 1 2 ( z − 1 ) − 1 2 ( z + 1 ) ) d z = x + C 1 − 1 2 ∫ 1 z − 1 d z + 1 2 ∫ 1 z + 1 d z = x + C 1 − 1 2 ln | z − 1 | + 1 2 ln | z + 1 | = x + C 1 1 2 ln | z + 1 z − 1 | = x + C 1 Simplify further as follows. ln | z + 1 z − 1 | = 2 x + 2 C 1 | z + 1 z − 1 | = e 2 x + 2 C 1 z + 1 z − 1 = ± e 2 C 1 e 2 x z + 1 z − 1 = C 2 e 2 x ( Put ± e 2 C 1 = C 2 ) Substitute back, z = x − y + 5 . x − y + 5 + 1 x − y + 5 − 1 = C 2 e 2 x x − y + 6 x − y + 4 = C 2 e 2 x ( x − y + 4 ) + 2 x − y + 4 = C 2 e 2 x 1 + 2 x − y + 4 = C 2 e 2 x Simplify further as follows. 2 x − y + 4 = C 2 e 2 x − 1 x − y + 4 = 2 C 2 e 2 x − 1 − y = 2 C 2 e 2 x − 1 − x − 4 y = 2 − C 2 e 2 x + 1 + x + 4 y = 2 C e 2 x + 1 + x + 4 ( Put − C 2 = C , C is an arbitrary constant ) Thus, the solution of the differential equation d y d x = ( x − y + 5 ) 2 is y = 2 C e 2 x + 1 + x + 4 _ .

Solution Summary: The author explains how to solve the differential equation dx=(x-y+5)2.

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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