Suppose tank A consist of 500 liter brine with 500gm salt dissolved in it, whereas in tank B containing 500 liter of water. Water is allowed to enter tank A at a rate of 30 liter/minute, and that mixed solution flow from tank A to tank B at a rate of 40 liter/minute. At the same time, 10 liter/minute were pumped back from tank B to tank A, while the mixed solution also is drained from tank B at a rate of 30 liter/minute as shown in figure. Let’s represent the amount of salt in tank A and tank B respectively, so the rate of change of the amount of salt in each tank can be formulated as following system:



Find the particular solution for



dx/dt = - (40/500)x + (10/500)y dy/dt = 40/500x - (10/500 + 30/500)y

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To find the particular solution for the given system of differential equations, we can use the method of integrating factors.

First, we write the system in matrix form as:

d/dt [x; y] = [-(40/500) (10/500); (40/500) -(10/500 + 30/500)] [x; y]

Now, let's find the integrating factor for this system. The integrating factor, denoted as "IF," is calculated as:

IF = e^( ∫ (a_12 - a_21) dt)

where a_12 and a_21 are the elements of the coefficient matrix.

In this case, a_12 = (10/500) and a_21 = (40/500), so the integrating factor becomes:

IF = e^( ∫ [(10/500) - (40/500)] dt)
= e^( ∫ (-30/500) dt)
= e^((-30/500)t)
= e^(-0.06t)

Next, we multiply the whole system by the integrating factor:

e^(-0.06t) * d/dt [x; y] = e^(-0.06t) * [-(40/500) (10/500); (40/500) -(10/500 + 30/500)] [x; y]

Simplifying further, we have:

d/dt [e^(-0.06t) * x; e^(-0.06t) * y] = [-(40/500) * e^(-0.06t) * x + (10/500) * e^(-0.06t) * y; (40/500) * e^(-0.06t) * x - (10/500 + 30/500) * e^(-0.06t) * y]

Now, we can integrate both sides with respect to t:

∫ d/dt [e^(-0.06t) * x; e^(-0.06t) * y] dt = ∫ [-(40/500) * e^(-0.06t) * x + (10/500) * e^(-0.06t) * y; (40/500) * e^(-0.06t) * x - (10/500 + 30/500) * e^(-0.06t) * y] dt

This simplifies to:

[e^(-0.06t) * x; e^(-0.06t) * y] = ∫ [-(40/500) * e^(-0.06t) * x + (10/500) * e^(-0.06t) * y; (40/500) * e^(-0.06t) * x - (10/500 + 30/500) * e^(-0.06t) * y] dt

Integrating each component separately, we get:

e^(-0.06t) * x = ∫ (-(40/500) * e^(-0.06t) * x + (10/500) * e^(-0.06t) * y) dt
e^(-0.06t) * y = ∫ ((40/500) * e^(-0.06t) * x - (10/500 + 30/500) * e^(-0.06t) * y) dt

Now, we can evaluate these integrals. The right-hand side of the first equation can be integrated with respect to t, and the same for the second equation.

After evaluating the integrals, we will have two equations in terms of x, y, and t. We can simplify these equations further to get the particular solution for x and y.