Problem 44 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(41-44\) for \(x\) as a function of \(t .\) $$ (t+1) \frac{d x}{d t}=x^{2}+1 \quad(t>-1), \quad x(0)=\pi / 4 $$

Step-by-Step Solution

Answer

The solution is \( x = \tan(\ln|t+1| + 0.665) \).

Step 1: Separate Variables

To solve the differential equation \((t+1) \frac{dx}{dt} = x^2 + 1\), first separate variables. Rewrite it as \( \frac{dx}{x^2 + 1} = \frac{dt}{t+1} \). This separates the variables \(x\) and \(t\) to different sides of the equation.

Step 2: Integrate Both Sides

Integrate both sides to solve for \(x\) in terms of \(t\). Integrating the left side, \( \int \frac{dx}{x^2 + 1} = \tan^{-1}(x) + C_1 \). Integrating the right side, \( \int \frac{dt}{t+1} = \ln|t+1| + C_2 \). Combine constants: \( C = C_2 - C_1 \).

Step 3: Set Up General Solution

Combine the integrations to set up the general solution: \( \tan^{-1}(x) = \ln|t+1| + C \). Isolate \(x\) by applying the tangent function: \( x = \tan(\ln|t+1| + C) \).

Step 4: Apply Initial Condition

Apply the initial condition \(x(0) = \pi/4\) to find the constant \(C\). Substituting, \( \frac{\pi}{4} = \tan(\ln|0+1| + C) \). Since \(\ln(1) = 0\), we have \( \frac{\pi}{4} = \tan(C)\). Thus, \( C = \tan^{-1}(\frac{\pi}{4}) \).

Step 5: Solve for the Constant

Evaluate \( C = \tan^{-1}(\frac{\pi}{4}) \). Recalling the angle whose tangent is \(\frac{\pi}{4}\), we approximate it numerically, as \(\tan^{-1}(\frac{\pi}{4}) \approx 0.665\).

Step 6: Write the Particular Solution

Substitute the constant \(C\) back into the general solution: \( x = \tan(\ln|t+1| + 0.665) \). This is the particular solution for the initial value problem \((t+1) \frac{dx}{dt} = x^2 + 1\), with the initial condition \(x(0) = \frac{\pi}{4}\).

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In simple terms, it expresses how the rate of change of a quantity depends on other variables. These equations are fundamental in describing various physical systems and phenomena, such as growth models, motions, or decay processes.
For example, in our exercise, the differential equation is given by \[(t+1) \frac{dx}{dt} = x^2 + 1\] Here, the derivative \(\frac{dx}{dt}\) represents how the variable \(x\) changes with respect to \(t\).
The equation suggests that this rate of change is related to both the value of \(x\) itself and the function \(t + 1\). Differential equations can often be complex to solve, but understanding them is crucial for translating a wide range of real-world problems into solvable mathematical forms.

Separation of Variables

Separation of variables is a straightforward and effective method for solving some types of differential equations. The main idea is to rearrange the equation so that functions of each variable appear on opposite sides of the equation. This allows for easier integration.
In our example, the equation \[(t+1) \frac{dx}{dt} = x^2 + 1\] can be rearranged to separate \(x\) and \(t\) as follows:

• \(\frac{dx}{x^2 + 1} = \frac{dt}{t+1}\)

This separation is the key to tackling the problem using integration. By isolating each variable, we can independently integrate both sides. This enables us to transform a challenging problem into more manageable pieces.

Integration

Integration is the mathematical process of finding the antiderivative of a function. It is the inverse operation of differentiation and is crucial in solving equations once variables are separated.
Once we've separated our equation, we integrate both sides:

• Left side: \(\int \frac{dx}{x^2 + 1} = \tan^{-1}(x) + C_1\)
• Right side: \(\int \frac{dt}{t+1} = \ln|t+1| + C_2\)

This step heavily relies on knowing standard integration formulas. The functions \(\tan^{-1}(x)\) and \(\ln|t+1|\) arise from specific integral results. The constants \(C_1\) and \(C_2\) are known as integration constants and are important for determining the particular solution to an initial value problem.

Initial Condition

The initial condition is an additional piece of information that specifies the value of the function at a certain point. This is crucial because differential equations typically have an infinite number of possible solutions. The initial condition helps us pinpoint a unique solution.
In our case, the initial condition is given as \(x(0) = \frac{\pi}{4}\). We incorporate this into our solution by substituting back after integrating:

• \(\frac{\pi}{4} = \tan(\ln|0+1| + C)\)
• Since \(\ln(1) = 0\), we simplify to find \(C = \tan^{-1}(\frac{\pi}{4})\)

The initial condition allows us to solve for the constant \(C\), thus refining the general solution into a specific one that precisely fits the problem's requirements. Through this step, the solution becomes applicable to real-world scenarios where specific starting values are known.