Explanation Given: The system of differential equations is, x ′ = z − y , ( 1 ) y ′ = z , ( 2 ) z ′ = z − x .... ( 3 ) x ( 0 ) = 0 , y ( 0 ) = 0 , z ( 0 ) = 2 ( 4 ) Approach: 1) Use elimination method for solving algebraic systems. Step1: Add or subtract the given equations to get an equation in single variable. Step2: Solve the single variable equation. Step3: Substitute the value of variable obtained in Step2 in other equations to get the values of other unknowns. 2) Use method of undetermined coefficients to find particular solution of non homogeneous ordinary differential equation. In this method, find the form of particular solution and then solve according to that. Calculation: Differentiate both sides of equation ( 1 ) . x ″ = ( z − y ) ′ As, x ″ = ( x ′ ) ′ so, x ″ = z ′ − y ′ Substitute z − x for z ′ in above equation from equation ( 3 ) . x ″ = z − x − y ′ x ″ + x = z − y ′ Substitute 0 for z − y ' in above equation from equation ( 3 ) . x ″ + x = 0 ( 5 ) The auxiliary equation of Equation ( 5 ) is, r 2 + 1 = 0 ( 6 ) Roots of Equation ( 6 ) are r = ± i . So, the general solution of Equation ( 4 ) is, x ( t ) = A 1 cos ( t ) + B 1 sin ( t ) ( 7 ) From ( 1 ) x ′ = z − y Substitute y ' for z from equation ( 2 ) x ′ = y ′ − y Thus, y ′ − y = − A 1 sin ( t ) + B 1 cos ( t ) ( 8 ) The auxiliary equation of Equation ( 7 ) is, r − 1 = 0 ( 9 ) Root of Equation ( 9 ) is r = 1 . So, the complementary solution of Equation ( 8 ) is y h ( t ) = A 2 e t Suppose the particular solution of Equation ( 8 ) is y p ( t ) = k 1 cos ( t ) + k 2 sin ( t ) ( 10 ) Differentiate Equation ( 10 ) with respect to t . y p ′ ( t ) = − k 1 sin ( t ) + k 2 cos ( t ) ( 11 ) Substitute − k 1 sin ( t ) + k 2 cos ( t ) for y ' and k 1 cos ( t ) + k 2 sin ( t ) y in Equation ( 8 ) . − k 1 sin ( t ) + k 2 cos ( t ) − ( k 1 cos ( t ) + k 2 sin ( t ) ) = − A 1 sin ( t ) + B 1 cos ( t ) ( 12 ) Compare the coefficients of cos ( t ) and sin ( t ) in Equation ( 12 )

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