Explanation Approach: Consider the linear homogeneous n th order differential equation, a n y ( n ) ( x ) + a n − 1 y ( n − 1 ) ( x ) + a n − 2 y ( n − 2 ) ( x ) + … … + a 0 y ( x ) = 0 . The general solution for this differential equation is given by the steps as follows: Step 1 Assume y = e r x to be the solution of the n th order linear homogeneous differential equation. Step 2 Calculate the subsequent derivatives of y and substitute in the differential equation. Step 3 An Auxiliary equation in terms of r is formed. Here r is the root of the equation. Step 4 If the roots r 1 , … … , r n of the auxiliary equation are real and distinct, then the general solution to the differential equation is given as, y ( x ) = C 1 e r 1 x + C 2 e r 2 x + … … + C n e r n x . Here C 1 , C 2 , … … , C n are arbitrary constants. Step-5 If the roots of the auxiliary equation are complex numbers such as r = ( α ± i β ) , then the general solution to the differential equation is given as, y ( x ) = C 1 e α x ( C 2 cos β x ± C 3 sin β x ) . Here C 1 , C 2 , … … , C n are arbitrary constants. Step-6 If the roots of the auxiliary equation are repeated such as r 1 is a repeated root of multiplicity m , then the general solution to the differential equation is given as, y ( x ) = C 1 e r 1 x + C 2 x e r 1 x + C 3 x 2 e r 1 x + … … + C m x m − 1 e r 1 x . Here C 1 , C 2 , … … , C n are arbitrary constants. Calculation: Consider the system of equations, d 2 x d t 2 − x + 5 y = 0 2 x + d 2 y d t 2 + 2 y = 0 Define the operator D = d d t D 2 = d 2 d t 2 Thus, ( D 2 − 1 ) [ x ] + 5 [ y ] = 0 ... ( 1 ) 2 [ x ] + ( D 2 + 2 ) [ y ] = 0 ( 2 ) Multiply equation ( 1 ) by D 2 + 2 and equation ( 2 ) by − 5 and then add both the equations. This gives, { ( D 2 − 1 ) ( D 2 + 2 ) } [ x ] = 0 { D 4 + 2 D 2 − D 2 − 2 − 10 } [ x ] = 0 { D 4 + D 2 − 12 } [ x ] = 0 Consider the auxiliary solution x = e r t therefore the auxiliary solutions of the equation,

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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Chapter 6.2, Problem 22E

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To find:

The general solution of the system of equations

d2xdt2−x+5y=02x+d2ydt2+2y=0

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Chapter 6.2, Problem 21E

Chapter 6.2, Problem 23E

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Fundamentals of Differential Equations [With CDROM] - 7th Edition

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The general solution of the system of equations d 2 x d t 2 − x + 5 y = 0 2 x + d 2 y d t 2 + 2 y = 0

Solution Summary: The author explains that the general solution of the system of equations is rlld2x

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To find:

The general solution of the system of equations

d2xdt2−x+5y=02x+d2ydt2+2y=0

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Chapter 6 Solutions

The general solution of the system of equations d 2 x d t 2 − x + 5 y = 0 2 x + d 2 y d t 2 + 2 y = 0

Solution Summary: The author explains that the general solution of the system of equations is rlld2x

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To find: The general solution of the system of equations d2xdt2−x+5y=02x+d2ydt2+2y=0

Want to see the full answer?

Chapter 6 Solutions

To find:

The general solution of the system of equations

d2xdt2−x+5y=02x+d2ydt2+2y=0