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Math Advanced Math Custom Kreyszig: Advanced Engineering Mathematics

The IVP x 3 ( y 1 ″ ) ′ + x y ′ − y = x 2 with initial condition y 1 ( 1 ) = 1 , y 1 ′ ( 1 ) = 3 and y 1 ″ ( 1 ) = 14 . The solution of the given IVP is y = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + x 2 _ . Formula used: The particular solution of the differential equation is y p ( x ) = y 1 ( x ) ∫ W 1 ( x ) W ( x ) r ( x ) d x + y 2 ( x ) ∫ W 2 ( x ) W ( x ) r ( x ) d x + y 3 ( x ) ∫ W 3 ( x ) W ( x ) r ( x ) d x . Where y 1 , y 2 , y 3 are basis of the homogenous differential equation. Calculation: The differential equation can be written as: x 3 ( y 1 ″ ) ′ + x y ′ − y = x 2 ( y 1 ″ ) ′ + x x 3 y ′ − 1 x 3 y = x 2 x 3 ( y 1 ″ ) ′ + 1 x 2 y ′ − 1 x 3 y = 1 x The characteristic equation of the differential equation is written as: m ( m − 1 ) ( m − 2 ) x m + m x m − x m = 0 m ( m − 1 ) ( m − 2 ) + m − 1 = 0 ( m − 1 ) ( m 2 − 2 m + 1 ) = 0 On further simplification, the following obtained. ( m − 1 ) ( m − 1 ) 2 = 0 m = 1 , 1 , 1 Since the auxiliary equation has three repeated roots, the general solution of the homogeneous is y h = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 . The value of W ( y 1 , y 2 , y 3 ) is calculated as: W ( y 1 , y 2 , y 3 ) = | y 1 y 2 y 3 y 1 ′ y 2 ′ y 3 ′ y 1 ″ y 2 ″ y 3 ″ | = | x x ln x x ( ln x ) 2 1 ln x + 1 ( ln x ) 2 + 2 ln x 0 1 x 2 + 2 ln x x | = { x [ ( ln x + 1 ) ( 2 + 2 ln x x ) − 1 x [ ( ln x ) 2 + 2 ln x ] ] − x ln x ( 2 + 2 ln x x ) + x ( ln x ) 2 [ 1 x + 0 ] } = { [ 2 x ( ln x + 1 ) ( 1 + ln x x ) − x x [ ( ln x ) 2 + 2 ln x ] ] − 2 x ln x ( 1 + ln x x ) + x ( ln x ) 2 [ 1 x ] } On further simplification, the following is obtained. W ( y 1 , y 2 , y 3 ) = [ 2 ( ln x + 1 ) 2 − [ ( ln x ) 2 + 2 ln x ] ] − 2 ln x ( 1 + ln x ) + ( ln x ) 2 = [ 2 ( ln x ) 2 + 2 + 4 ln x − ( ln x ) 2 − 2 ln x ] − 2 ln x − 2 ( ln x ) 2 + ( ln x ) 2 = 2 The value of W 1 is calculated as: W 1 = | 0 y 2 y 3 0 y 2 ′ y 3 ′ 1 y 2 ″ y 3 ″ | = | 0 x ln x x ( ln x ) 2 0 ln x + 1 ( ln x ) 2 + 2 ln x 1 1 x 2 + 2 ln x x | = x ( ln x ) ( ( ln x ) 2 + 2 ln x ) − [ x ( ln x ) 2 ( ln x + 1 ) ] = x ( ln x ) 2 The value of W 2 is calculated as: W 2 = | y 1 0 y 3 y 1 ′ 0 y 3 ′ y 1 ″ 1 y 3 ″ | = | x 0 x ( ln x ) 2 1 0 ( ln x ) 2 + 2 ln x 0 1 2 + 2 ln x x | = x [ 0 − ( ln x ) 2 + 2 ln x ] + x ( ln x ) 2 = − 2 x ln x The value of W 3 is calculated as: W 3 = | y 1 y 2 0 y 1 ′ y 2 ′ 0 y 1 ″ y 2 ″ 1 | = | x x ln x 0 1 ln x + 1 0 0 1 x 1 x | = x [ ( ln x + 1 ) ] − x ln x = x Now the value of y p is calculated as: y p ( x ) = x ∫ ( x ( ln x ) 2 ( 1 x ) 2 ) d x + x ln x ∫ ( − 2 x ( ln x ) ( 1 x ) 2 ) d x + x ( ln x ) 2 ∫ x ( 1 x ) 2 d x = x 2 ∫ ( ln x ) 2 d x − x ln x ∫ ( ln x ) d x + x 2 ( ln x ) 2 ∫ 1 2 d x = x 2 [ x ( ln x ) 2 − 2 x ( ln x − 1 ) ] − x ln x ( x ln x − x ) + x 2 2 ( ln x ) 2 = x 2 Therefore, the general solution of the differential equation is written as: y = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + y p = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + x 2 Therefore, the solution of the given IVP is y = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + x 2 _ .

The IVP x 3 ( y 1 ″ ) ′ + x y ′ − y = x 2 with initial condition y 1 ( 1 ) = 1 , y 1 ′ ( 1 ) = 3 and y 1 ″ ( 1 ) = 14 . The solution of the given IVP is y = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + x 2 _ . Formula used: The particular solution of the differential equation is y p ( x ) = y 1 ( x ) ∫ W 1 ( x ) W ( x ) r ( x ) d x + y 2 ( x ) ∫ W 2 ( x ) W ( x ) r ( x ) d x + y 3 ( x ) ∫ W 3 ( x ) W ( x ) r ( x ) d x . Where y 1 , y 2 , y 3 are basis of the homogenous differential equation. Calculation: The differential equation can be written as: x 3 ( y 1 ″ ) ′ + x y ′ − y = x 2 ( y 1 ″ ) ′ + x x 3 y ′ − 1 x 3 y = x 2 x 3 ( y 1 ″ ) ′ + 1 x 2 y ′ − 1 x 3 y = 1 x The characteristic equation of the differential equation is written as: m ( m − 1 ) ( m − 2 ) x m + m x m − x m = 0 m ( m − 1 ) ( m − 2 ) + m − 1 = 0 ( m − 1 ) ( m 2 − 2 m + 1 ) = 0 On further simplification, the following obtained. ( m − 1 ) ( m − 1 ) 2 = 0 m = 1 , 1 , 1 Since the auxiliary equation has three repeated roots, the general solution of the homogeneous is y h = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 . The value of W ( y 1 , y 2 , y 3 ) is calculated as: W ( y 1 , y 2 , y 3 ) = | y 1 y 2 y 3 y 1 ′ y 2 ′ y 3 ′ y 1 ″ y 2 ″ y 3 ″ | = | x x ln x x ( ln x ) 2 1 ln x + 1 ( ln x ) 2 + 2 ln x 0 1 x 2 + 2 ln x x | = { x [ ( ln x + 1 ) ( 2 + 2 ln x x ) − 1 x [ ( ln x ) 2 + 2 ln x ] ] − x ln x ( 2 + 2 ln x x ) + x ( ln x ) 2 [ 1 x + 0 ] } = { [ 2 x ( ln x + 1 ) ( 1 + ln x x ) − x x [ ( ln x ) 2 + 2 ln x ] ] − 2 x ln x ( 1 + ln x x ) + x ( ln x ) 2 [ 1 x ] } On further simplification, the following is obtained. W ( y 1 , y 2 , y 3 ) = [ 2 ( ln x + 1 ) 2 − [ ( ln x ) 2 + 2 ln x ] ] − 2 ln x ( 1 + ln x ) + ( ln x ) 2 = [ 2 ( ln x ) 2 + 2 + 4 ln x − ( ln x ) 2 − 2 ln x ] − 2 ln x − 2 ( ln x ) 2 + ( ln x ) 2 = 2 The value of W 1 is calculated as: W 1 = | 0 y 2 y 3 0 y 2 ′ y 3 ′ 1 y 2 ″ y 3 ″ | = | 0 x ln x x ( ln x ) 2 0 ln x + 1 ( ln x ) 2 + 2 ln x 1 1 x 2 + 2 ln x x | = x ( ln x ) ( ( ln x ) 2 + 2 ln x ) − [ x ( ln x ) 2 ( ln x + 1 ) ] = x ( ln x ) 2 The value of W 2 is calculated as: W 2 = | y 1 0 y 3 y 1 ′ 0 y 3 ′ y 1 ″ 1 y 3 ″ | = | x 0 x ( ln x ) 2 1 0 ( ln x ) 2 + 2 ln x 0 1 2 + 2 ln x x | = x [ 0 − ( ln x ) 2 + 2 ln x ] + x ( ln x ) 2 = − 2 x ln x The value of W 3 is calculated as: W 3 = | y 1 y 2 0 y 1 ′ y 2 ′ 0 y 1 ″ y 2 ″ 1 | = | x x ln x 0 1 ln x + 1 0 0 1 x 1 x | = x [ ( ln x + 1 ) ] − x ln x = x Now the value of y p is calculated as: y p ( x ) = x ∫ ( x ( ln x ) 2 ( 1 x ) 2 ) d x + x ln x ∫ ( − 2 x ( ln x ) ( 1 x ) 2 ) d x + x ( ln x ) 2 ∫ x ( 1 x ) 2 d x = x 2 ∫ ( ln x ) 2 d x − x ln x ∫ ( ln x ) d x + x 2 ( ln x ) 2 ∫ 1 2 d x = x 2 [ x ( ln x ) 2 − 2 x ( ln x − 1 ) ] − x ln x ( x ln x − x ) + x 2 2 ( ln x ) 2 = x 2 Therefore, the general solution of the differential equation is written as: y = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + y p = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + x 2 Therefore, the solution of the given IVP is y = ( c 1 x + c 2 x ln x ) + c 3 x ( ln x ) 2 + x 2 _ .

Custom Kreyszig: Advanced Engineering Mathematics

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

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Custom Kreyszig: Advanced Engineering Mathematics

Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P Ch. 3.1 - Prob. 11P

Ch. 3.1 - Prob. 12P Ch. 3.1 - Prob. 13P Ch. 3.1 - Prob. 14P Ch. 3.1 - Prob. 15P Ch. 3.1 - Prob. 16P Ch. 3.2 - Prob. 1P Ch. 3.2 - Prob. 2P Ch. 3.2 - Solve the given ODE. Show the details of your...Ch. 3.2 - Solve the given ODE. Show the details of your...Ch. 3.2 - Solve the given ODE. Show the details of your...Ch. 3.2 - Solve the given ODE. Show the details of your...Ch. 3.2 - Solve the IVP by a CAS, giving a general solution...Ch. 3.2 - Prob. 8P Ch. 3.2 - Solve the IVP by a CAS, giving a general solution...Ch. 3.2 - Solve the IVP by a CAS, giving a general solution...Ch. 3.2 - Solve the IVP by a CAS, giving a general solution...Ch. 3.2 - CAS EXPERIMENT. Reduction of Order. Starting with...Ch. 3.3 - Solve the following ODEs, showing the details of...Ch. 3.3 - Solve the following ODEs, showing the details of...Ch. 3.3 - Solve the following ODEs, showing the details of...Ch. 3.3 - Solve the following ODEs, showing the details of...Ch. 3.3 - Prob. 5P Ch. 3.3 - Prob. 6P Ch. 3.3 - Solve the following ODEs, showing the details of...Ch. 3.3 - Solve the given IVP, showing the details of your...Ch. 3.3 - Solve the given IVP, showing the details of your...Ch. 3.3 - Prob. 10P Ch. 3.3 - Solve the given IVP, showing the details of your...Ch. 3.3 - Solve the given IVP, showing the details of your...Ch. 3.3 - Solve the given IVP, showing the details of your...Ch. 3 - Prob. 1RQ Ch. 3 - List some other basic theorems that extend from...Ch. 3 - If you know a general solution of a homogeneous...Ch. 3 - What form does an initial value problem for an...Ch. 3 - What is the Wronskian? What is it used for? Ch. 3 - Prob. 6RQ Ch. 3 - Solve the given ODE. Show the details of your...Ch. 3 - Solve the given ODE. Show the details of your...Ch. 3 - Prob. 9RQ Ch. 3 - Solve the given ODE. Show the details of your...Ch. 3 - Solve the given ODE. Show the details of your...Ch. 3 - Prob. 12RQ Ch. 3 - Solve the given ODE. Show the details of your...Ch. 3 - Prob. 14RQ Ch. 3 - Prob. 15RQ Ch. 3 - Solve the IVP. Show the details of your work. Ch. 3 - Solve the IVP. Show the details of your work. y‴ +...Ch. 3 - Solve the IVP. Show the details of your work. Ch. 3 - Solve the IVP. Show the details of your work. Ch. 3 - Solve the IVP. Show the details of your work.

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