Explanation Given: The given equations are, x ′ = 5 x − 3 y − 2 ; x ( 0 ) = 2 y ′ = 4 x − 3 y − 1 ; y ( 0 ) = 0 Procedure used: Consider a linear system of two first-order differential equations of the form a 1 x ′ ( t ) + a 2 x ( t ) + a 3 y ′ ( t ) + a 4 y ( t ) = f 1 ( t ) a 5 x ′ ( t ) + a 6 x ( t ) + a 7 y ′ ( t ) + a 8 y ( t ) = f 2 ( t ) Here, a 1 , a 2 , a 3 , ....... , a 8 are constants and x ( t ) and y ( t ) are pair of function to be determined. In operator notation the above two equations become ( a 1 D + a 2 ) [ x ] + ( a 3 D + a 4 ) [ y ] = f 1 ( a 5 D + a 6 ) [ x ] + ( a 7 D + a 8 ) [ y ] = f 2 . Calculation: Consider the given linear system of equations. x ′ = 5 x − 3 y − 2 ; x ( 0 ) = 2 y ′ = 4 x − 3 y − 1 ; y ( 0 ) = 0 Write the above linear system of equations in operator form that is, ( D − 5 ) [ x ] + 3 y + 2 = 0 ... ( 1 ) 4 x − ( D + 3 ) [ y ] = 0 ( 2 ) Multiply the equation ( 1 ) with 4 and equation ( 2 ) with ( D − 5 ) ( 4 D − 20 ) [ x ] + 12 y = − 8 ( 4 D − 20 ) [ x ] + ( D 2 − 2 D − 15 ) [ y ] = − 8 ( − ) ( + ) ( + ) ( D 2 − 2 D − 3 ) [ y ] = − 3 … ( 3 ) The auxiliary equation of the above equation is D 2 − 2 D − 3 = 0 Solve the above equation. D = ± 4 + 12 2 D = 2 ± 4 2 = 1 ± 2 = 3 , − 1 The general solution of the equation is given by, y = c 1 e 3 t + c 2 e − t Here, c 1 and c 2 are arbitrary constants. Assume the particular solution of above equation as, y p = λ Differentiate above equation with respect to t

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 5.2, Problem 22E

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The solution for the initial value problem satisfies [x(t)+y(t)] as t→∞.

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Chapter 5.2, Problem 21E

Chapter 5.2, Problem 23E

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

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To verify: The solution for the initial value problem satisfies [ x ( t ) + y ( t ) ] as t → ∞ .

Solution Summary: The author explains that the solution for the initial value problem satisfies (t)+yl (t)to (infty) .

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To verify:

The solution for the initial value problem satisfies [x(t)+y(t)] as t→∞.

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Chapter 5 Solutions

To verify: The solution for the initial value problem satisfies [ x ( t ) + y ( t ) ] as t → ∞ .

Solution Summary: The author explains that the solution for the initial value problem satisfies (t)+yl (t)to (infty) .

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To verify: The solution for the initial value problem satisfies [x(t)+y(t)] as t→∞.

Want to see the full answer?

Chapter 5 Solutions

To verify:

The solution for the initial value problem satisfies [x(t)+y(t)] as t→∞.