Explanation Given: Fresh water flows into tank A at a rate of 6 km/hr and fluid is drained out of tank B at the same rate. From tank A, 8 km/hr fluid is pumped to tank B. The volume of liquid in each tank remains constant at 24 L . Initial masses are given as x ( 0 ) = 0.1 kg, y ( 0 ) = 0 .3 kg . Approach: Use method of undetermined coefficients to find particular solution of non homogeneous ordinary differential equation. In this method, guess the form of particular solution. The form takes the terms of the consecutive derivatives. The guess has to be modified if it contains the terms from complementary solution. Calculation: Let x ( t ) be the mass of salt in tank A. Let y ( t ) be the mass of salt in tank B. Equate the rate of change of salt in each tank with the net rate at which salt is transferred to that tank. So, d x d t = 6 ( 0.2 ) + 2 ( y 24 ) − 8 ( x 24 ) = 1.2 + y 12 − x 3 … ( 1 ) d y d t = 8 ( x 24 ) − 2 ( y 24 ) − 6 ( y 24 ) = x 3 − y 3 … ( 2 ) This can be written as x = 3 y ′ + y … ( 3 ) Differentiate Equation ( 3 ) with respect to t. x ′ = 3 y ″ + y ′ Substitute 3 y ′ + y for x and 3 y ″ + y ′ for d x d t in Equation ( 1 ) . 3 y ″ + y ′ = 1.2 + y 12 − ( 3 y ′ + y ) 3 3 y ″ + 2 y ′ + y 4 = 1.2 … ( 4 ) The auxiliary equation of Equation ( 4 ) is, 3 r 2 + 2 r + 1 4 = 0 12 r 2 + 8 r + 1 = 0 ( 2 r + 1 ) ( 6 r + 1 ) = 0 … ( 5 ) Roots of Equation ( 5 ) are r = − 1 2 , − 1 6 . So, the complementary solution of Equation ( 4 ) is, y h ( t ) = A 1 e − t 2 + B 1 e − t 6 Let the particular solution of Equation ( 4 ) be y p ( t ) = λ

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ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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The mass of salt in each tank at time t for the given initial conditions.

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