Program Explanation Given information: The linear system of the differential equation x ′ + 2 y ′ = 4 x + 5 y , 2 x ′ − y ′ = 3 x and the initial condition is x ( 0 ) = 1 , y ( 0 ) = − 1 . Calculation: Assume that the tank 1contain x ( t ) pound of salt in 100gal of brine and tank 2 contain y ( t ) pounds of salt in 200 gal of brine the two tanks at time t The concentration of the sat on the two tanks are x 100 pounds per gallon and y 200 pounds per gallon. Now find the rate of change of the amount of salt in first tank can be calculated as, x ′ = − 30 ⋅ x 100 + 10 ⋅ y 200 x ′ = − 3 x 10 + y 20 20 x ′ = − 6 x + y Therefore, the differential equation for the first tank is as follows, 20 x ′ = − 6 x + y ...............(1) Now find the rate of change of the amount of salt in second tank can be calculated as, y ′ = − 30 ⋅ x 100 − 10 ⋅ y 200 − 20 ⋅ y 200 y ′ = − 3 x 10 − 3 y 20 20 y ′ = 6 x − 3 y Therefore, thelinear system of the differential equation 20 x ′ = − 6 x + y , 20 y ′ = 6 x − 3 y . The value of x can be found by the second equation of the differential equation. 20 y ′ = 6 x − 3 y − 6 x = − 20 y ′ − 3 y x = 20 6 y ′ + 1 2 y x = 10 3 y ′ + 1 2 y Therefore, the equation for x ( t ) is as follows, x = 10 3 y ′ + 1 2 y ..............(2) Now differentiate the above equation to obtain the value of x ′ . x ′ = 20 6 y ″ + 1 2 y ′ Now substitute the value of x , x ′ in equation (1) as, 20 ( 20 6 y ″ + 1 2 y ′ ) = − 69 ( 20 6 x ′ + 1 2 y ) + y ( 200 3 y ″ + 10 y ′ ) = ( − 20 y ′ − 3 y ) + y 200 3 y ″ + 30 y ′ + 2 y = 0 100 y ″ + 45 y ′ + 3 y = 0 Now solve the characteristic equation to obtain the roots of the characteristic equation as, 100 r 2 + 45 r + 3 = 0 a = 100 , b = 45 , c = 3 r = − 9 40 ± 33 40 Therefore, the roots of the equations are r 1 = − 9 40 + 33 40 , r 2 = − 9 40 − 33 40 . The solution of the differential equation y ( t ) can be written as, y ( t ) = c 1 e ( − 9 40 + 33 40 ) t + c 2 e ( − 9 40 − 33 40 ) t ..............(3) Differentiate y ( t ) with respect to t both sides. y ′ ( t ) = c 1 ( − 9 40 + 33 40 ) e ( − 9 40 + 33 40 ) t + c 2 ( − 9 40 − 33 40 ) e ( − 9 40 − 33 40 ) t Now substitute y ( t ) and y ′ ( t ) in equation (2). x = [ 10 3 ( c 1 ( − 9 40 + 33 40 ) e ( − 9 40 + 33 40 ) t + c 2 ( − 9 40 − 33 40 ) e ( − 9 40 − 33 40 ) t ) + 1 2 ( c 1 e ( − 9 40 + 33 40 ) t + c 2 e ( − 9 40 − 33 40 ) t ) ] x = ( c 1 ( − 1 4 + 11 4 3 ) e ( − 9 40 + 33 40 ) t + c 2 ( − 1 4 − 11 4 3 ) e ( − 9 40 − 33 40 ) t ) Therefore, the amount of the salt in tank 1 is as follows, x ( t ) = ( c 1 ( − 1 4 + 11 4 3 ) e ( − 9 40 + 33 40 ) t + c 2 ( − 1 4 − 11 4 3 ) e ( − 9 40 − 33 40 ) t )

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Chapter 4.2, Problem 30P

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Program Description: Purpose of problem is to obtain the amount x(t) and y(t) of salt in the two tanks at time t .

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Chapter 4.2, Problem 29P

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Program Description: Purpose of problem is to obtain the amount x ( t ) and y ( t ) of salt in the two tanks at time t .

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Program Description: Purpose of problem is to obtain the amount x(t) and y(t) of salt in the two tanks at time t .

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Chapter 4 Solutions