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by Vtu Rangers
Type: NoteInstitute: Visvesvaraya Technological University VTU Specialization: Computer Science EngineeringOffline Downloads: 57Views: 396Uploaded: 1 month agoAdd to Favourite

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Vtu Rangers
Vtu Rangers
Module – I Engineering Mathematics - IV 2017 Numerical Solution of Ordinary Differential Equations of First order and first degree Introduction Many ordinary differential equations can be solved by analytical methods discussed earlier giving closed form solutions i.e. expressing y in terms of a finite number of elementary functions of x. However, a majority of differential equations appearing in physical problems cannot be solved analytically. Thus it becomes imperative to discuss their solution by numerical methods. Numerical methods for Initial value problem: Consider the first order and first degree differential equations the initial condition problem. ( that is ) and ( )with called initial value We discuss the following numerical methods for solving an initial value problem. 1. Taylor’s series method 2. Modified Euler’s method 3. Runge - Kutta method of order IV 4. Milne’s Predictor - Corrector Method 5. Adams – Bashforth Predictor - Corrector Method Type -1 Taylor’s series method Consider ( the ) first condition order ( ) and first degree equations . Taylor’s series expansion of ( ) in powers of ( Dr. A.H.Srinivasa, MIT, Mysore differential ) is Page 1
Module – I Engineering Mathematics - IV 2017 ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) ( ) Where at the point( ) Worked Examples 1. Using Taylor’s Series method, find the value of y at x = 0.1, and x = 0.2 for the initial value problem ( ) Solution: Taylor’s Series expansion of ( ) about a point ( ( ) ) Here, compare ( ) ( ( ) )  ( ) ( ) ( ) w.r.t ( ) ( w.r.t ) Differentiate ( ( )( )] ( ) ( ) ( ) ( ) ( ) and ( ) ( ) [ ( ) ] ( ) ( )( ) [ ( ) ] we get, ( ) ( ( ) ( ) [( )( ) ( ) ] [ ( ) ] we get, ) ( ) ) we get, ( ) w.r.t ( ( ) ( ) ( ) Differentiate ) , then ( ) ( ) Differentiate ( is given by ( ) Substitute the values of ( ) ( ) ( ) ( ( ( ) ) ( ) [( )( ) ( ) ) [ ( ) ] ( ) in equation (*) ( ) Dr. A.H.Srinivasa, MIT, Mysore Page 2
Module – I Engineering Mathematics - IV 2017 ( ) This is called Taylors series expansion up to fourth degree term. Put x = 0.1. x = 0.2 2. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Using Taylor’s Series method, find the value of y at x = 0.1, and x = 0.2 for the initial value problem ( ) Solution: Taylor’s Series expansion of ( ) about a point ( ) ( ) Here, compare ( ) ( ) ( )  ( ) ( ) ( ) ( ) ( ) w.r.t w.r.t ( Differentiate )( ) w.r.t ( )( ) ) ( ( ) ( ) and ( ) ( ) )( ) ( ) ( ) [ ( ) ] we get, ( ) Differentiate ( ) , then ( ) ( ) Differentiate ( is given by ( ) ( ) ( )( ) ( )( ) [ ( ) ] we get, ( ) ( )( ( ) ) ( ) [ ( ) ( ) ( ) ] ( ) we get, ( ) ( )( ) Dr. A.H.Srinivasa, MIT, Mysore ( ) ( ) [ ( ) ( ) ( ) ( ) ] Page 3
Module – I Engineering Mathematics - IV 2017 Substitute the values of ( ) ( ) ( ) ( ) ( ) ( ) ( ) in equation (*) ( ) ( ) ( ) This is called Taylors series expansion up to fourth degree term. Put x = 0.1 and x = 0.2 3. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Using Taylor’s Series method, find the value of y at x = 0.1, and x = 0.2, for the initial value problem ( ) Solution: Taylor’s Series expansion of ( ) about a point ( ) ( Here, compare ( ) ) ( ) (  ( ) ( ) ( ) ( ) Differentiate Differentiate w.r.t we get, ( ) ( ) w.r.t we get, w.r.t ( ( ) ] ( ) ) ( , then ( ) ( ) ) ( ) ( ) and ( ) ( ) ( ) [ ( ) Differentiate ) is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( [ ( ) ] [ ( ) ] ( ) we get, Dr. A.H.Srinivasa, MIT, Mysore Page 4

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