Explanation Procedure used: Solution of system of differential equations by elimination method. 1 Consider the system of differential equations. 2 Express the equations in differential operator form. 3 Find the DE by elimination. 4 Obtain the complementary function. 5 Further, Obtain the particular solution. 6 Ultimately write the general solution in form y ( x ) = y c + y p . Calculation: Consider the system of differential equations d x d t = y − 1 , d y d t = − 3 x + 2 y . Indicate the set of differential equation in differential operator form. D x − y = 1 … ( 1 ) 3 x + ( D − 2 ) y = 0 … ( 2 ) Solve the equations ( 1 ) and ( 2 ) as ( 1 ) × ( D − 2 ) + ( 2 ) . Therefore, ( D − 2 ) D x − ( D − 2 ) y + 3 x + ( D − 2 ) y = 0 ( D 2 − 2 D + 3 ) x = 2 The roots of auxiliary equation associated to homogeneous differential equation ( D 2 − 2 D + 3 ) x = 2 . Thus, m 2 − 2 m + 3 = 0 m = − ( b ) ± b 2 − 4 a c 2 a = 2 ± 2 2 i 2 = 1 + 2 i Thus, the complementary solution is x c = e t ( c 1 cos 2 t + c 2 sin 2 t ) . Use the method of undermined coefficients to find the particular solution. Let x p = A be the particular solution. Substitute x p = A into ( D 2 − 2 D + 3 ) x = 2 . Thus, x p = A 3 A = 2 A = 2 3 Hence, the particular solution is x ( t ) = e t ( c 1 cos 2 t + c 2 sin 2 t ) + 2 3 . From D x − y = 1

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.12, Problem 22E

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To solve: The initial value problem dxdt=y−1,dydt=−3x+2y,x(0)=0,y(0)=0.

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Chapter 3.12, Problem 21E

Chapter 3.12, Problem 23E

Chapter 3 Solutions

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The initial value problem d x d t = y − 1 , d y d t = − 3 x + 2 y , x ( 0 ) = 0 , y ( 0 ) = 0 .

Solution Summary: The author explains the solution of the system of differential equations by elimination method.

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To solve: The initial value problem dxdt=y−1,dydt=−3x+2y,x(0)=0,y(0)=0.

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