Problem 9 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the particular solution corresponding to the initial conditions given. \(\frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}=2 x, \quad x(0)=-1, \quad x^{\prime}(0)=0\)

Step-by-Step Solution

Answer

The particular solution of the differential equation \( \frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}=2 x \) with the initial conditions \( x(0)=-1 \), \( x^{\prime}(0)=0 \) is \( x(t) = -2 + (1/2) e^{-t} + 1/2.\)

Step 1: Solve the Homogeneous Equation

First, solve the homogeneous equation by considering the equation without the constant.\[ \frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}=0 \]The characteristic equation of above differential equation is \( r^2 + r = 0 \). Solving for \( r \), we get two roots, \( r = 0, -1 \). The general solution can then be written as \( x(t) = c_1 e^{0 \cdot t} + c_2 e^{-1 \cdot t} = c_1 + c_2 e^{-t}\).

Step 2: Find a Particular Solution

For the non-homogeneous part of the equation, we try a solution of the form \( x = A \), where \( A \) is a constant. Substituting into the equation gives \( A = 1/2 \). Thus the particular solution is \( x(t) = 1/2 \).

Step 3: Combine The Homogeneous and Particular Solution

The final solution to the entire non-homogeneous differential equation is obtained by combining the general solution to the homogeneous equation and the particular solution. In this case, it is \( x(t) = c_1 + c_2 e^{-t} + 1/2\).

Step 4: Apply the Initial Conditions

To find the values of \( c_1 \) and \( c_2 \), we apply initial conditions \( x(0) = -1 \) and \( x^{\prime}(0) = 0 \). From \( x(0) = -1 \), we get \( c_1 + c_2 = -3/2 \). And from \( x^{\prime}(0) = 0 \), by differentiating \( x(t) \) and substituting \( t = 0 \), we get \( c_2 = 1/2\). Consequently, substituting \( c_2 = 1/2 \) into the first equation yields \( c_1 = -2 \).

Step 5: Combine the Coefficients and The Function

Finally, substituting \( c_1 = -2 \) and \( c_2 = 1/2 \) into the general solution of non-homogeneous differential equation gives the particular solution \( x(t) = -2 + (1/2) e^{-t} + 1/2.\)

Key Concepts

Differential Equation Initial Conditions

Understanding initial conditions is crucial when solving differential equations. They are the values that the unknown function and its derivatives have at the start of the problem, acting as constraints that allow us to find a particular solution. A differential equation may have an infinite number of solutions, but initial conditions enable us to pinpoint the specific solution that fits the problem's requirements.

In the provided exercise, the initial conditions are given as \(x(0) = -1\) and \(x'(0) = 0\). These conditions are applied after finding the general solution to the differential equation, which includes arbitrary constants. By substituting these initial values into the general solution and its derivative, we determine the exact values of the constants, resulting in the unique solution that satisfies both the differential equation and the initial conditions. This particular solution is essential in various applications such as physics and engineering, where initial states often determine the subsequent behavior of a system.

Characteristic Equation of Differential Equations

The characteristic equation is a key concept when solving linear homogeneous differential equations. It transforms the problem of finding the solution of a differential equation into the simpler problem of solving an algebraic equation.

In our step-by-step solution, the characteristic equation is derived from the homogeneous part \( \frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}=0 \) as \( r^2 + r = 0 \). This equation comes from assuming a solution of the form \( e^{rt} \) for the differential equation, where \( r \) is a constant that we need to find. Solving the characteristic equation gives us the roots \( r = 0, -1 \), which guide us to the multiple solutions to the homogeneous equation, typically in terms of exponential functions with base \(e\) raised to the power of these roots multiplied by the variable \(t\). These solutions are the building blocks for constructing the general solution to the entire differential equation, including the particular solution.

Homogeneous and Non-Homogeneous Solutions

Differential equations can often be categorized as either homogeneous or non-homogeneous. A homogeneous differential equation is one where all the terms are a function of the unknown function and its derivatives, and there are no extra constant or forcing terms.

In contrast, a non-homogeneous differential equation includes additional terms that are not functions of the unknown variable or its derivatives. These terms typically represent external forces or inputs in a system. To solve a non-homogeneous differential equation, we find the general solution to the homogeneous part and then add a particular solution that accounts for the non-homogeneous part.

In our problem, after finding the homogeneous solution, we searched for a particular solution of the form \(x = A\) to satisfy the non-homogenous term, resulting in \( A = 1/2 \). Combining the homogeneous solution with this particular solution gives the complete solution to the non-homogeneous differential equation. This combined solution is then tailored with the initial conditions to yield the specific solution that not only satisfies the differential equation's constraints but also fulfills the specific requirements imposed at the outset of the problem.