Problem 21 - Calculus for Biology and Medicine

Question

Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 3 & -2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=1 \text { and } x_{2}(0)=1 \text { . } $$

Step-by-Step Solution

Answer

The solution is \( x_1(t) = 2e^{3t} - e^t \) and \( x_2(t) = e^t \).

Step 1: Write the System of Differential Equations

The given system can be rewritten as a set of two equations: \( \frac{d x_1}{d t} = 3x_1 - 2x_2 \) and \( \frac{d x_2}{d t} = x_2 \).

Step 2: Solve the Second Equation

The equation \( \frac{d x_2}{d t} = x_2 \) is separable. Integrate both sides to solve: \( \int \frac{1}{x_2} \, dx_2 = \int \, dt \). This results in \( \ln |x_2| = t + C_2 \), thus \( x_2(t) = e^{t+C_2} = C_2e^t \).

Step 3: Solve the First Equation Using Variation of Parameters

Substitute \( x_2(t) = C_2e^t \) into the first equation: \( \frac{d x_1}{d t} = 3x_1 - 2C_2e^t \). To solve this non-homogeneous equation, first solve the homogeneous part \( T'(t) = 3T(t) \), giving \( T(t) = C_1e^{3t} \). Assume a particular solution \( x_1(t) = C_1e^{3t} + u(t)e^{3t} \) and determine \( u'(t) \) by substituting back and comparing terms.

Step 4: Integrate to Find u(t) and Resolve Constants

To find \( u(t) \), solve the equation obtained by balancing terms: \( u'(t)e^{3t} = 2C_2e^t \). This gives \( u'(t) = 2C_2e^{-2t} \), and integrating \( u(t) = -C_2e^{-2t} + C_3 \). So, the solution for \( x_1(t) \) is \( x_1(t) = C_1e^{3t} - C_2e^{t} + C_3e^{3t} \).

Step 5: Apply Initial Conditions

Use the initial conditions \( x_1(0) = 1 \) and \( x_2(0) = 1 \) to solve for the constants \( C_1 \), \( C_2 \), and \( C_3 \). Substitute \( x_2(0) = 1 \) into \( C_2e^0 = 1 \), hence \( C_2 = 1 \). Use \( x_1(0) = 1 \) in \( C_1 + 0(C_3) - C_2 \) to find \( C_1 = 2 \), and note \( C_3 \) is not needed due to the form of solutions.

Step 6: State the General Solution

With the constants resolved as \( C_1 = 2 \) and \( C_2 = 1 \), continue with \( x_1(t) = 2e^{3t} - e^{t} \) and \( x_2(t) = e^t \). This forms a complete solution satisfying the given differential equations and the initial conditions.

Key Concepts

System of Differential Equations

A system of differential equations consists of multiple differential equations that often involve two or more functions with respect to one or more variables. In this context, the given problem involves two such equations:

• \( \frac{d x_1}{d t} = 3x_1 - 2x_2 \)
• \( \frac{d x_2}{d t} = x_2 \)

The system encapsulates interactions between variables \( x_1(t) \) and \( x_2(t) \), where the change of each function with respect to time depends on the function itself and possibly other functions in the system. By solving this system, we can discover how \( x_1 \) and \( x_2 \) evolve over time given their initial conditions. To approach the solution, it helps to split the system into simpler equations or find suitable methods to manage their relationships.

Variation of Parameters

Variation of parameters is a method for finding particular solutions to differential equations, especially useful for non-homogeneous linear differential equations. Unlike the method of undetermined coefficients, variation of parameters doesn't limit the type of non-homogeneous terms we can handle. In step 3 of solving the initial-value problem, after finding the homogeneous solution \( T(t) = C_1e^{3t} \), a particular solution is assumed in the form:\[ x_1(t) = C_1e^{3t} + u(t)e^{3t} \]Here, \( u(t) \) is an unknown function adjusted to fit the particular solution into the non-homogeneous part. Through substitution and differentiation, \( u'(t) \) is derived, integrated, and consequently used to form the overall solution. This method is a powerful and flexible tool, allowing for the accommodation of diverse functions within the system.

Separable Equations

Separable equations are differential equations that can be rearranged so that all terms involving one variable are on one side, with terms involving the other variable on the other side. They are among the simplest differential equations to solve and are characterized by the structure:\[ \frac{dy}{dx} = g(y) h(x) \]This can be rearranged as:\[ g(y) \, dy = h(x) \, dx \]Such separation allows integration on both sides. In our particular problem, the equation \( \frac{d x_2}{d t} = x_2 \) is separable. Solving involves:

• Rewriting: \( \int \frac{1}{x_2} \, dx_2 = \int \, dt \)
• Integration leads to logarithmic terms: \( \ln |x_2| = t + C_2 \)
• Exponentiation yields the solution: \( x_2(t) = C_2e^t \)

With the constant \( C_2 \) determined by initial conditions, separable equations provide a straightforward path to solutions of certain differential equations.

Homogeneous and Non-Homogeneous Solutions

Differential equations may be classified as homogeneous or non-homogeneous, based on their structure. A homogeneous differential equation contains terms that only include the dependent variable and its derivatives, zero on the right side. Conversely, a non-homogeneous equation includes additional terms.For example, the first equation in our system \( \frac{d x_1}{d t} = 3x_1 - 2x_2 \) is non-homogeneous due to the \(-2x_2\) term. Solving such equations typically involves:

• Finding the homogeneous solution, \( T(t) \), which solves the associated homogeneous equation: \( \frac{d T}{d t} = 3T \), resulting in \( T(t) = C_1e^{3t} \).
• Determining a particular solution that satisfies the entire equation using methods like variation of parameters.

Combining these solutions gives a complete answer to the initial-value problem, as seen with \( x_1(t) = 2e^{3t} - e^{t} \) and \( x_2(t) = e^t \). Understanding these solutions shapes much of differential equation analysis.