Problem 48 - Advanced Engineering Mathematics

Question

Radioactive Decay Series The following system of differential equations is encountered in the study of the decay of a special type of radioactive series of elements: $$ \begin{aligned} &\frac{d x}{d t}=-\lambda_{1} x \\ &\frac{d y}{d t}=\lambda_{1} x-\lambda_{2} y \end{aligned} $$ where \(\lambda_{1}\) and \(\lambda_{2}\) are constants. Discusshow to solve this system subject to \(x(0)=x_{0}, y(0)=y_{0}\). Carry out your ideas.

Step-by-Step Solution

Answer

The solutions are \( x(t) = x_0 e^{-\lambda_{1} t} \) and \( y(t) = \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} e^{-\lambda_{1} t} + \left(y_0 - \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1}\right) e^{-\lambda_{2} t} \).

Step 1: Solve the First Differential Equation

The first equation is \( \frac{d x}{d t} = -\lambda_{1} x \). This is a simple first-order linear differential equation. We can solve it by separation of variables: \( \frac{1}{x} dx = -\lambda_{1} dt \). Integrating both sides, we get \( \ln|x| = -\lambda_{1} t + C_1 \). Exponentiating both sides gives \( x = C e^{-\lambda_{1} t} \), where \( C \) is a constant. Using the initial condition \( x(0) = x_0 \), we find \( C = x_0 \), so the solution is \( x(t) = x_0 e^{-\lambda_{1} t} \).

Step 2: Substitute the Solution into the Second Equation

Now, substitute \( x(t) = x_0 e^{-\lambda_{1} t} \) into the second equation: \( \frac{d y}{d t} = \lambda_{1} x_0 e^{-\lambda_{1} t} - \lambda_{2} y \). This results in a first-order linear differential equation in terms of \( y \): \( \frac{d y}{d t} + \lambda_{2} y = \lambda_{1} x_0 e^{-\lambda_{1} t} \).

Step 3: Solve the Second Differential Equation Using Integrating Factor

To solve the differential equation \( \frac{d y}{d t} + \lambda_{2} y = \lambda_{1} x_0 e^{-\lambda_{1} t} \), we use the integrating factor method. The integrating factor is \( e^{\lambda_{2} t} \). Multiply the entire differential equation by this factor: \( e^{\lambda_{2} t} \frac{d y}{d t} + \lambda_{2} e^{\lambda_{2} t} y = \lambda_{1} x_0 e^{(\lambda_{2} - \lambda_{1}) t} \). Notice the left-hand side is the derivative of \( e^{\lambda_{2} t} y \), i.e., \( \frac{d}{dt}(e^{\lambda_{2} t} y) = \lambda_{1} x_0 e^{(\lambda_{2} - \lambda_{1}) t} \).

Step 4: Integrate and Solve for y(t)

Integrate both sides: \( e^{\lambda_{2} t} y = \int \lambda_{1} x_0 e^{(\lambda_{2} - \lambda_{1}) t} dt + C_2 \). Simplifying the integral \( \int \lambda_{1} x_0 e^{(\lambda_{2} - \lambda_{1}) t} dt \), we obtain \( \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} e^{(\lambda_{2} - \lambda_{1}) t} \). Thus, \( e^{\lambda_{2} t} y = \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} e^{(\lambda_{2} - \lambda_{1}) t} + C_2 \), and \( y = \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} e^{-\lambda_{1} t} + C_2 e^{-\lambda_{2} t} \).

Step 5: Determine the Constant C2 Using Initial Condition

Apply the initial condition \( y(0) = y_0 \): \( y_0 = \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} + C_2 \). Solving for \( C_2 \), we find \( C_2 = y_0 - \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} \). Thus, the solution for \( y(t) \) is \( y(t) = \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1} e^{-\lambda_{1} t} + \left(y_0 - \frac{\lambda_{1} x_0}{\lambda_2 - \lambda_1}\right) e^{-\lambda_{2} t} \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In our context, these functions describe how quantities such as radioactive substances change over time. For the exercise, the system of differential equations given describes how two substances, denoted by functions \(x(t)\) and \(y(t)\), decay. A differential equation like \(\frac{d x}{d t} = -\lambda_{1} x\) indicates that the rate of change of \(x\) with respect to time \(t\) is proportional to the value of \(x\) itself, multiplied by a decay constant \(-\lambda_{1}\). This means the quantity decreases exponentially over time, a typical behavior of radioactive substances.
Understanding the nature of differential equations is crucial because they model many real-world phenomena, like population growth, heat transfer, and, in our case, radioactive decay. Their solutions provide us with a function that predicts how a system behaves over time based on initial conditions. By setting initial values, such as \(x(0)=x_{0}\), we tailor the general solution to our specific scenario.
These equations can be classified generally into several types, such as ordinary or partial, linear or nonlinear. The equations in our exercise are linear and ordinary, as they involve ordinary derivatives of one variable with respect to \(t\) and have constant coefficients.

Separation of Variables

Separation of variables is a method used to solve simple differential equations. It's especially handy when a differential equation can be rearranged so that all terms involving one variable are on one side, and all terms with the other variable are on the opposite side.
In our radioactive decay problem, we use separation of variables to tackle the first equation, \(\frac{d x}{d t} = -\lambda_{1} x\). By separating, we reorganize it to \(\frac{1}{x} dx = -\lambda_{1} dt\). This rearrangement makes it possible to integrate both sides separately, resulting in a solution which relates \(x\) and \(t\).
The magic of separation of variables lies in its simplicity and effectiveness for certain types of differential equations. Once separated, integrating both sides with respect to their respective variables gives us new equations that are often easier to solve. In our case, integrating gives \(\ln|x| = -\lambda_{1} t + C_1\), and solving for \(x\) gives the exponential decay formula: \(x(t) = x_0 e^{-\lambda_{1} t}\). This demonstrates how separation helps us move from a complex-looking equation to a clear and useful solution.

Integrating Factor

The integrating factor method is a technique for solving linear first-order differential equations of the form \(\frac{dy}{dt} + P(t)y = Q(t)\). To make the equation easier to solve, we multiply every term by a cleverly chosen function, the integrating factor.
In our exercise, after solving the first differential equation for \(x(t)\), we're left to solve the second one: \(\frac{d y}{d t} + \lambda_{2} y = \lambda_{1} x_0 e^{-\lambda_{1} t}\). The integrating factor for this equation is \(e^{\lambda_{2} t}\), a function that simplifies solving by turning the left-hand side into the derivative of a product. This means the equation can be expressed as \(\frac{d}{dt}(e^{\lambda_{2} t} y) = \lambda_{1} x_0 e^{(\lambda_{2} - \lambda_{1}) t}\).
Integrating both sides with respect to \(t\) results in a solvable equation for \(y\). Specifically, it leads to finding \(y(t)\), showing how the substance \(y\) changes over time. This method reduces the complexity of the problem by leveraging the characteristics of exponential functions to reveal solutions more transparently.