Solving Mixture and Differential Equations: Step-by-Step
Mixture Problems and Differential Equations
This document presents solutions to two types of mathematical problems: mixture problems and differential equations. The solutions are presented step-by-step, with explanations.
2.3.35 Mixture Problems
A tank contains 500L of water with 0.2 kg of salt. Initially, there are 5 kg of salt in the tank. The inflow and outflow rates are 5 L/min. We will determine the concentration at 10 minutes and after a leak develops, at 20 minutes.
a) Concentration at 10 minutes
Let A(t) denote the mass of salt in the tank at t minutes after the process begins. Then:
- Rate of input = 5 L/min × 0.2 kg/L = 1 kg/min
- Rate of exit = 5 L/min × A(t)/ 500 kg/L = A(t)/ 100 kg/min
Therefore, dA /dt = 1 – A/ 100 = (100 – A)/ 100.
Separating this differential equation yields dA/(100 − A) = dt/100. Integrating, we obtain:
− ln |100 − A| = (t /100) + C1 ⇒ |100 − A| = e ^(−t/100−C1) = e ^(−C1) e ^(−t/100) ⇒ 100 − A = Ce^(−t/100) (C = ±e^−C1) ⇒ A = 100 – Ce^−t/100.
The initial condition, A(0) = 5 (initially, there were 5 kg of salt in the tank) implies that 5 = A(0) = 100 − C ⇒ C = 95.
Substituting this value of C into the solution, we have A(t) = 100 − 95e^−t/100.
Thus, the mass of salt in the tank after 10 min is A(10) = 100 − 95e^(−10/100) ≈ 14.04 kg, which gives the concentration 14.04 kg/500 L ≈ 0.0281 kg/L.
b) Concentration after the leak
After the leak develops, the system satisfies a new differential equation. While the rate of input remains the same, 1 kg/min, the rate of exit is now different. Since, every minute, 5 liters of the solution is coming in and 5 + 1 = 6 liters are going out, the volume of the solution in the tank decreases by 6 − 5 = 1 liter per minute. Thus, for t ≥ 10, the volume of the solution in the tank is 500 − 1 · (t − 10) = 510 − t liters. This gives the concentration of salt in the tank A(t)/( 510 – t) kg/L and rate of exit = 6 L/min × A(t)/( 510 – t) kg/L = 6A(t)/ (510 – t) kg/min
Hence, the differential equation, for t > 10, becomes dA /dt = 1 – (6A/ (510 – t)) ⇒ dA/ dt + (6A /(510 – t) = 1 with the initial condition A(10) = 14.04 (the value found in (a) ). This equation is a linear equation. We have µ(t) = exp (∫ 6 /(510 – t) dt)= exp (−6 ln |510 − t|) = (510 − t) ^ −6 ⇒ d /dt ((510 − t) ^(−6) A) = ∫ (510 − t) ^−6 = (510 – t)^−6 ⇒ (510 − t) ^−6 A =∫ (510 − t) ^−6 dt = (1/5) (510 − t) ^−5 + C ⇒ A = 1/5 (510 − t) + C(510 − t) ^6
Using the initial condition, A(10) = 14.04, we compute C. 14.04 = A(10) = 1/5 (510 − 10) + C(510 − 10) ^6 ⇒ C = − 85.96/((500) ^6)
Therefore, A(t) = (1/5) (510 − t) – (85.96/ (500) ^6) (510 − t) ^6 = (1/5) (510 − t) − 85.96 (510 – t/ 500) ^6 and, according to (2.11), the concentration of salt is given by A(t)/(510 – t) = (1/5) –( 85.96 /510 − t ) *((510 – t)/ 500) ^6 .
20 minutes after the leak develops, that is, when t = 30, the concentration will be (1/5) –( 85.96/ (510 – 30) )*((510 – 30)/ 500) ^6 ≈ 0.0598 kg/L.
2.4.29 Differential Equation Example
Consider the differential equation (y^2 + 2xy)dx – x^2 dy=0
a) Showing the equation is not exact
We have M(x, y) = y ^2 + 2xy and N(x, y) = −x ^2. Therefore My(x, y)=2y + 2x and Nx(x, y) = −2x. Thus My(x, y) ≠ Nx(x, y), so the differential equation is not exact.
b) Making the equation exact
If we multiply (y^2 + 2xy)dx − x^2 dy = 0 by y^−2, we obtain (1 +( 2x /y) dx – (x^2/y^2) dy = 0. In this equation we have M(x, y)=1+2xy^−1 and N(x, y) = −x^2*y^−2. Therefore, (∂M(x, y))/ ∂y = −2x/ y^2 = (∂N(x, y))/ ∂x . So the new differential equation is exact.
c) Solving the exact equation
Following the method for solving exact equations we integrate M(x, y) in part (b) with respect to x to obtain F(x, y) = ∫(1+2* x/ y) dx = x + (x^2/ y) + g(y). To determine g(y), take the partial derivative of both sides of the above equation with respect to y to obtain ∂F/∂y = (−x2 /y2) + g´ (y). Substituting N(x, y) (given in part (b)) for ∂F/∂y, we can now solve for g´ (y) to obtain N(x, y) = (−x2/ y2) = (−x2/ y2) + g´ (y) ⇒ g´ (y)=0. The integral of g´(y) will yield a constant and the choice of the constant of integration is not important so we can take g(y) = 0. Hence we have F(x, y) = x + (x^2/y) and the solution to the equation is given implicitly by x +( x^2 /y )= C. Solving the above equation for y, we obtain y = x^2 /C – x.
d) Lost solution
By dividing both sides by y^2 we lost the solution y ≡ 0.