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B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order EXERCISE # 10.1 Question # 1:( ) √( ) ( Solution: )( ) ) √ Given equation is ( ( ) () √ The characteristics equation of ( ) will be √ ( ) √( ) ( )( ) √ ( ) √ √ √ √ Therefore, the complementary function is ( ) ( ) ( ) ( ) Therefore, the complementary function is ( ( √ √ ( ) )) is required solution. Question # 2:( ) ( √ ( ) √ ( )) Solution: is required solution. Given equation is ( ) () The characteristics equation of ( ) will be Question # 3:( ) Solution: Given equation is ( ) 1 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) The characteristics equation of ( ) will be Chapter # 10: Differential Equations of Higher Order Therefore, the complementary function is Since is a root of characteristic equation. So we use synthetic division in order to find the other roots. is required solution. -1 1 -4 1 6 0 -1 5 -6 1 -5 6 0 Question # 5:( ) Solution: Now, the residual equation will be Given equation is ( ( ) ( ( )( ) () The characteristics equation of ( ) will be ) ) Therefore, the complementary function is Since is a root of characteristic equation. So we use synthetic division in order to find the other roots. 1 0 1 2 -6 2 -4 12 -8 4 -8 8 0 is required solution. Question # 4:( ) Now the residual equation will be Solution: Given equation is ( ) () ( ) The characteristics equation of ( ) will be Therefore, the complementary function is ( ( ) )( ( ) ) ( ) is required solution. 2 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) Question # 6:( Chapter # 10: Differential Equations of Higher Order ) Solution: ( ) Given equation is ( )( ( ) ( ) () ( ) ) ( ) The characteristics equation of ( ) will be ( ) √( ) Since is a root of characteristic equation. So we use synthetic division in order to find the other roots. ( )( ) √ 2 1 -6 3 10 0 2 -8 -10 1 -4 -5 0 √ √ √ Now, the residual equation will be Therefore, the complementary solution is ( )( * ) ( * Therefore, the complementary solution is ( √ √ ) ( ) ( √ √ )+ )+ is required solution. ) Question # 8:( Solution: is required solution. Question # 7:( Given equation is ) Solution: ( ) () The characteristics equation of ( ) will be Given equation is ( ) () The characteristics equation of ( ) will be Since are the roots of characteristic equation. So we use synthetic division in order 3 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) to find the other roots. The synthetic division is as follow:- 1 -1 4 0 4 0 4 -4 4 0 -4 -4 -3 0 -3 4 1 4 -3 1 -1 0 -1 1 0 Chapter # 10: Differential Equations of Higher Order 1 1 1 0 1 0 1 2 1 3 1 4 )( -6 1 -5 5 0 5 -5 0 Now, the residual equation will be √ Now, the residual equation will be ( -2 3 1 4 5 √ ) Therefore, the complementary solution is Therefore, the complementary solution is ( ( ( ) ) is required solution. is required solution. Question # 9:( ) Question # 10 ( ) Solution: Solution: Given equation is Given equation is ( ) () The characteristics equation of ( ) will be Since are the roots of characteristic equation. So we use synthetic division in order to find the other roots. The synthetic division is as follow:- ) ( ) ( ) () The characteristics equation of ( ) will be Since are the roots of characteristic equation. So we use synthetic division in order to find the other roots. The synthetic division is as follow:- 4 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) 1 0 1 0 1 -1 2 -5 -1 -6 2 -4 6 6 12 -8 4 4 -12 -8 8 0 -8 8 0 Chapter # 10: Differential Equations of Higher Order 1 0 1 0 1 -1 -2 -4 -1 -5 -2 -7 -7 5 -2 14 12 22 2 24 -24 0 24 -24 0 Now, the residual equation will be Now, the residual equation will be ( ( ) ( )( ) ) ( ) ( ( )( ) ) Therefore, the complementary solution is Therefore, the complementary solution is ( ) Or Or ( ) ( ) is required solution. is required solution. Question # 11 Question # 12 ( ( ) Solution: ) Solution: Given equation is ( Given equation is ( ) () ) The characteristics equation of ( ) will be () The characteristics equation of ( ) will be Since are the roots of characteristic equation. So we use synthetic division in order to find the other roots. The synthetic division is as follow:- ( ) ( ) ( ) ( ) ( ( ( )( )( ) ( )( ) ) ) 5 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) √ Chapter # 10: Differential Equations of Higher Order Therefore, the complementary solution is √ Therefore, ( ) ( ) is required solution. Question # 14 ( Therefore, the complementary solution is ( ) ( ) ) Solution: Given equation is is required solution. Question # 13 ( ) ( ) () The characteristics equation of ( ) will be Solution: Given equation is ( ( () ) ) ( The characteristics equation of ( ) will be ) ( Since are the roots of characteristic equation. So we use synthetic division in order to find the other roots. The synthetic division is as follow:1 0 1 0 1 1 2 -1 1 0 2 2 -3 0 -3 4 1 1 -3 -2 2 0 2 -2 0 ( )( √ ) ) √ Now, the residual equation will be Therefore, the complementary solution is ( ) ( )( ) 6 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) ( ) ( ( Chapter # 10: Differential Equations of Higher Order Therefore, the complementary solution is ) ( ) ) ( √ √ ) ( √ √ ) Therefore, Therefore, ( ( ) ( ( ) ) is required solution. ) Solve each of the following initial problem. is required solution. Question # 16 ( Question # 15 ( ) ( ) ( ) Solution: ) Given equation is Solution: ( Given equation is ( ) () ) () The characteristics equation of ( ) will be The characteristics equation of ( ) will be Since are the roots of characteristic equation. So we use synthetic division in order to find the other roots. The synthetic division is as follow:- -2 -2 1 0 1 0 1 6 -2 4 -2 2 15 -8 7 -4 3 20 -14 6 -6 0 12 -12 0 ( ( ) )( ( ) ) Therefore, the complementary solution is Therefore ( ) Now, the residual equation will be Differentiating w.r.t we have ( ) Applying ( ) √ √ √ on ( ) we have, ( ) Applying ( ) on ( ) we have, 7 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order ( )( ) ( ) Using ( ) Therefore, the complementary solution is ( ) we have Therefore ( ) Differentiating w.r.t we have ( ) Applying ( ) on ( ) we have, Now ( ) Applying ( ) on ( ) we have, Thus equation ( ) becomes Thus equation ( ) becomes ( ) ( ) is required solution. is required solution. Question # 17 Question # 18 ( ) ( ) ( ) ( ) Solution: Solution: Given equation is Given equation is ( ) () The characteristics equation of ( ) will be ( ) ( ) ( ) () The characteristics equation of ( ) will be √ ( ) ( ) 8 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) √ Chapter # 10: Differential Equations of Higher Order Question # 19 ( ) ( ) ( ( ) ( ) Solution: ) Given equation is ( Therefore, the complementary solution is ( ) () The characteristics equation of ( ) will be ) Therefore ( ) Differentiating w.r.t ( ) Since is a root of characteristic equation. So we use synthetic division in order to find the other roots. we have ( ) ( 1 ( ) ) Applying ( ) on ( ) we have, ( 1 -6 11 -6 0 1 -5 6 1 -5 6 0 ) Now, the residual equation will be Applying ( ) on ( ) we have, ( ( ) ( ) )( ( ) ) Therefore, the complementary solution is Thus equation ( ) becomes ( ) Therefore ( ) `is required solution. Differentiating w.r.t we have ( ) 9 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) Applying ( ) Thus equation ( ) becomes on ( ) we have, is required solution. ( ) Applying ( ) Chapter # 10: Differential Equations of Higher Order on ( ) we have, Question # 20 ( ( ) ( ) Now, differentiating ( ) we have ( ) Applying ( ) ) on ( ) we have, ( ) ( ) ( ) Solution: Given equation is ( ) () The characteristics equation of ( ) will be ( ) By ( ) ( ) we have ( ) By ( ) ( ) we have ( ) * + Therefore, the complementary solution is Therefore ( ) ( ) Differentiating w.r.t ( we have ) ( ) Again differentiating w.r.t we have ( ) Again differentiating w.r.t we have ( ) Applying ( ) on ( ) we have, 10 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678 B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order ( ) Applying ( ) Applying ( ) on ( ) we have, Written by: on ( ) we have, Muhammad umer asghar ( ) Applying ( ) on ( ) we have, Sese (math) Govt. elementary school malwal Special Thanks: ( ) (i) ( ) ( ) Prof. Muhammad Ali(www.houseofphy.co m) Dr. Attique Ur Rehman Sb(www.mathcity.org) Thus equation ( ) becomes (ii) is required solution. Contact #: The end 03074896454 11 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678