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

The differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 . The solution of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ . Definition used: Linear coefficients: “The equations of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) , where a i 's, b i 's, and c i 's are constants. For a 1 b 2 ≠ a 2 b 1 and b 1 ≠ a 2 , let x = u + h and y = v + k , where h and k satisfy a 1 h + b 1 k + c 1 = 0 , a 2 h + b 2 k + c 2 = 0. Then the transformed equation in the variables u and v is homogeneous. That is, d v d u = − a 1 u + b 1 v a 2 u + b 2 v = − a 1 + b 1 ( v u ) a 2 + b 2 ( v u ) .” Calculation: The differential equation is, ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 ... ( 1 ) The equation is of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) . Here, a 1 = 2 , b 1 = − 1 , c 1 = 4 , a 2 = 1 , b 2 = − 2 and c 2 = − 2 . Check the condition a 1 b 2 ≠ a 2 b 1 . a 1 b 2 = ( 2 ) ( − 2 ) = − 4 a 2 b 1 = ( 1 ) ( − 1 ) = − 1 Thus, the condition a 1 b 2 ≠ a 2 b 1 is satisfied. Let x = u + h and y = v + k , where h and k satisfy the system. 2 h − k + 4 = 0 ... ( 2 ) h − 2 k − 2 = 0 ... ( 3 ) Multiply the equation (3) by − 2 and add it to the equation (2). 2 h − k + 4 = 0 − 2 h + 4 k + 4 = 0 _ 3 k + 8 = 0 k = − 8 3 Substitute k = − 8 3 in equation (3). h − 2 k − 2 = 0 h − 2 ( − 8 3 ) − 2 = 0 h + 16 3 − 2 = 0 h + 10 3 = 0 h = − 10 3 Thus, the translation of the axes becomes x = u − 10 3 and y = v − 8 3 . Then d x = d u and d y = d v . The equation can be translated as, ( 2 ( u − 10 3 ) − ( v − 8 3 ) + 4 ) d u + ( ( u − 10 3 ) − 2 ( v − 8 3 ) − 2 ) d v = 0 ( 2 ( 3 u − 10 3 ) − ( 3 v − 8 3 ) + 4 ) d u + ( ( 3 u − 10 3 ) − 2 ( 3 v − 8 3 ) − 2 ) d v = 0 ( 6 u − 20 − 3 v + 8 + 12 3 ) d u + ( 3 u − 10 − 6 v + 16 − 6 3 ) d v = 0 ( 6 u − 3 v 3 ) d u + ( 3 u − 6 v 3 ) d v = 0 Simplify further as follows. ( 2 u − v ) d u + ( u − 2 v ) d v = 0 d v d u = − 2 u − v u − 2 v d v d u = − 2 + v u 1 − 2 ( v u ) This is homogeneous equation of the form d v d u = G ( v u ) . So take z = v u . Then d v d u = z + u d z d u , the transformed equation is, z + u d z d u = − 2 + z 1 − 2 z u d z d u = − 2 + z 1 − 2 z − z u d z d u = − 2 + z − z + 2 z 2 1 − 2 z Separate the variables and rewrite the differential equation as follows. u d z d u = − 2 + 2 z 2 1 − 2 z u d z d u = 2 ( z 2 − 1 ) 1 − 2 z 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u Integrate on both sides of the equation. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u ... ( 4 ) Consider the term 1 − 2 z 2 ( z + 1 ) ( z − 1 ) Split the term by using the method of partial fractions. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( A ( z + 1 ) + B ( z − 1 ) ) ... ( 5 ) Multiply 2 ( z + 1 ) ( z − 1 ) on both sides of the equation. 1 − 2 z = A ( z − 1 ) + B ( z + 1 ) Put z = 1 results B = − 1 2 and put z = − 1 results A = − 3 2 . Substitute A = − 3 2 and B = − 1 2 in the equation (5). 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( − 3 2 ( z + 1 ) − 1 2 ( z − 1 ) ) 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) Thus, the integral becomes, ∫ 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) d z = ∫ 1 u d u − 1 4 ∫ 3 ( z + 1 ) d z − 1 4 ∫ 1 ( z − 1 ) d z = ∫ 1 u d u − 3 4 ln | z + 1 | − 1 4 ln | z − 1 | = ln | u | + C − 1 4 ln | ( z + 1 ) 3 ( z − 1 ) | = ln | u | + C Simplify further as follows. ln | ( z + 1 ) 3 ( z − 1 ) | = − 4 ln | u | − 4 C 1 ln | ( z + 1 ) 3 ( z − 1 ) | = ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = e ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = C 2 u 4 Substitute back, z = v u . ( v u + 1 ) 3 ( v u − 1 ) = C 2 u 4 ( v + u u ) 3 ( v − u u ) = C 2 u 4 ( v + u ) 3 ( v − u ) = C 2 Substitute back, u = x + 10 3 and v = y + 8 3 . ( ( y + 8 3 ) + ( x + 10 3 ) ) 3 ( ( y + 8 3 ) − ( x + 10 3 ) ) = C 2 ( ( 3 y + 8 3 ) + ( 3 x + 10 3 ) ) 3 ( ( 3 y + 8 3 ) − ( 3 x + 10 3 ) ) = C 2 ( 3 x + 3 y + 18 3 ) 3 ( − 3 x + 3 y − 2 3 ) = C 2 1 81 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C 2 Simplify further as follows. ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = 81 C 2 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C ( C is an arbitrary constant ) Thus, the solutions of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ .

The differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 . The solution of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ . Definition used: Linear coefficients: “The equations of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) , where a i 's, b i 's, and c i 's are constants. For a 1 b 2 ≠ a 2 b 1 and b 1 ≠ a 2 , let x = u + h and y = v + k , where h and k satisfy a 1 h + b 1 k + c 1 = 0 , a 2 h + b 2 k + c 2 = 0. Then the transformed equation in the variables u and v is homogeneous. That is, d v d u = − a 1 u + b 1 v a 2 u + b 2 v = − a 1 + b 1 ( v u ) a 2 + b 2 ( v u ) .” Calculation: The differential equation is, ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 ... ( 1 ) The equation is of the form ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 d y = 0 ) . Here, a 1 = 2 , b 1 = − 1 , c 1 = 4 , a 2 = 1 , b 2 = − 2 and c 2 = − 2 . Check the condition a 1 b 2 ≠ a 2 b 1 . a 1 b 2 = ( 2 ) ( − 2 ) = − 4 a 2 b 1 = ( 1 ) ( − 1 ) = − 1 Thus, the condition a 1 b 2 ≠ a 2 b 1 is satisfied. Let x = u + h and y = v + k , where h and k satisfy the system. 2 h − k + 4 = 0 ... ( 2 ) h − 2 k − 2 = 0 ... ( 3 ) Multiply the equation (3) by − 2 and add it to the equation (2). 2 h − k + 4 = 0 − 2 h + 4 k + 4 = 0 _ 3 k + 8 = 0 k = − 8 3 Substitute k = − 8 3 in equation (3). h − 2 k − 2 = 0 h − 2 ( − 8 3 ) − 2 = 0 h + 16 3 − 2 = 0 h + 10 3 = 0 h = − 10 3 Thus, the translation of the axes becomes x = u − 10 3 and y = v − 8 3 . Then d x = d u and d y = d v . The equation can be translated as, ( 2 ( u − 10 3 ) − ( v − 8 3 ) + 4 ) d u + ( ( u − 10 3 ) − 2 ( v − 8 3 ) − 2 ) d v = 0 ( 2 ( 3 u − 10 3 ) − ( 3 v − 8 3 ) + 4 ) d u + ( ( 3 u − 10 3 ) − 2 ( 3 v − 8 3 ) − 2 ) d v = 0 ( 6 u − 20 − 3 v + 8 + 12 3 ) d u + ( 3 u − 10 − 6 v + 16 − 6 3 ) d v = 0 ( 6 u − 3 v 3 ) d u + ( 3 u − 6 v 3 ) d v = 0 Simplify further as follows. ( 2 u − v ) d u + ( u − 2 v ) d v = 0 d v d u = − 2 u − v u − 2 v d v d u = − 2 + v u 1 − 2 ( v u ) This is homogeneous equation of the form d v d u = G ( v u ) . So take z = v u . Then d v d u = z + u d z d u , the transformed equation is, z + u d z d u = − 2 + z 1 − 2 z u d z d u = − 2 + z 1 − 2 z − z u d z d u = − 2 + z − z + 2 z 2 1 − 2 z Separate the variables and rewrite the differential equation as follows. u d z d u = − 2 + 2 z 2 1 − 2 z u d z d u = 2 ( z 2 − 1 ) 1 − 2 z 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u Integrate on both sides of the equation. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) d z = 1 u d u ... ( 4 ) Consider the term 1 − 2 z 2 ( z + 1 ) ( z − 1 ) Split the term by using the method of partial fractions. 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( A ( z + 1 ) + B ( z − 1 ) ) ... ( 5 ) Multiply 2 ( z + 1 ) ( z − 1 ) on both sides of the equation. 1 − 2 z = A ( z − 1 ) + B ( z + 1 ) Put z = 1 results B = − 1 2 and put z = − 1 results A = − 3 2 . Substitute A = − 3 2 and B = − 1 2 in the equation (5). 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 2 ( − 3 2 ( z + 1 ) − 1 2 ( z − 1 ) ) 1 − 2 z 2 ( z + 1 ) ( z − 1 ) = 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) Thus, the integral becomes, ∫ 1 4 ( − 3 ( z + 1 ) − 1 ( z − 1 ) ) d z = ∫ 1 u d u − 1 4 ∫ 3 ( z + 1 ) d z − 1 4 ∫ 1 ( z − 1 ) d z = ∫ 1 u d u − 3 4 ln | z + 1 | − 1 4 ln | z − 1 | = ln | u | + C − 1 4 ln | ( z + 1 ) 3 ( z − 1 ) | = ln | u | + C Simplify further as follows. ln | ( z + 1 ) 3 ( z − 1 ) | = − 4 ln | u | − 4 C 1 ln | ( z + 1 ) 3 ( z − 1 ) | = ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = e ln 1 u 4 − 4 C 1 ( z + 1 ) 3 ( z − 1 ) = C 2 u 4 Substitute back, z = v u . ( v u + 1 ) 3 ( v u − 1 ) = C 2 u 4 ( v + u u ) 3 ( v − u u ) = C 2 u 4 ( v + u ) 3 ( v − u ) = C 2 Substitute back, u = x + 10 3 and v = y + 8 3 . ( ( y + 8 3 ) + ( x + 10 3 ) ) 3 ( ( y + 8 3 ) − ( x + 10 3 ) ) = C 2 ( ( 3 y + 8 3 ) + ( 3 x + 10 3 ) ) 3 ( ( 3 y + 8 3 ) − ( 3 x + 10 3 ) ) = C 2 ( 3 x + 3 y + 18 3 ) 3 ( − 3 x + 3 y − 2 3 ) = C 2 1 81 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C 2 Simplify further as follows. ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = 81 C 2 ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C ( C is an arbitrary constant ) Thus, the solutions of the differential equation ( 2 x − y + 4 ) d x + ( x − 2 y − 2 ) d y = 0 is ( 3 x + 3 y + 18 ) 3 ( − 3 x + 3 y − 2 ) = C _ .

Solution Summary: The author explains the solution of the differential equation (2x-y+4)dx+ (x-2y-2)dy=0.

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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