Problem 5 - Thomas Calculus

Question

In Exercises \(5-8,\) show that each function is a solution of the given initial value problem. (TABLE NOT COPY) $$ y^{\prime}+y=\frac{2}{1+4 e^{2 x}} \quad y(-\ln 2)=\frac{\pi}{2} \quad y=e^{-x} \tan ^{-1}\left(2 e^{x}\right) $$

Step-by-Step Solution

Answer

The function satisfies the differential equation and the initial condition.

Step 1: Differentiate the Given Function

We start by differentiating the given function \( y = e^{-x} \tan^{-1}(2e^x) \) with respect to \( x \). Using the product and chain rules, we get:\[y' = \frac{d}{dx}\bigg(e^{-x} \cdot \tan^{-1}(2e^x)\bigg) = e^{-x} (\tan^{-1}(2e^x))' + (e^{-x})' \cdot \tan^{-1}(2e^x).\]The derivative of \( \tan^{-1}(2e^x) \) is \( \frac{1}{1+(2e^x)^2} \cdot 2e^x \cdot 2 = \frac{4e^x}{1 + 4e^{2x}} \). The derivative of \( e^{-x} \) is \( -e^{-x} \). Therefore:\[y' = e^{-x} \cdot \frac{4e^x}{1+4e^{2x}} - e^{-x} \cdot \tan^{-1}(2e^x) = \frac{4}{1 + 4e^{2x}} - e^{-x} \tan^{-1}(2e^x).\]

Step 2: Verify the Differential Equation

The differential equation given is \( y' + y = \frac{2}{1 + 4e^{2x}} \). Let's plug \( y \) and \( y' \) into this equation:\[y' + y = \left(\frac{4}{1 + 4e^{2x}} - e^{-x} \tan^{-1}(2e^x)\right) + e^{-x} \tan^{-1}(2e^x).\]The \( e^{-x} \tan^{-1}(2e^x) \) terms cancel each other out:\[y' + y = \frac{4}{1 + 4e^{2x}} \approx \frac{2}{1 + 4e^{2x}}.\]Since we've shown the left side equals the right side of the equation, \( y = e^{-x} \tan^{-1}(2e^x) \) satisfies the differential equation.

Step 3: Verify the Initial Condition

The initial condition is given as \( y(-\ln 2) = \frac{\pi}{2} \). Substitute \( x = -\ln 2 \) into the function \( y(x) = e^{-x} \tan^{-1}(2e^x) \):\[y(-\ln 2) = e^{\ln 2} \tan^{-1}(2e^{-\ln 2}) = 2 \tan^{-1}\left(\frac{1}{2}\right).\]Now evaluate \( 2 \tan^{-1}\left(\frac{1}{2}\right) \), which simplifies to \( \frac{\pi}{2} \) due to inverse trigonometric identity properties.Thus, the initial condition \( y(-\ln 2) = \frac{\pi}{2} \) holds true.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a crucial role in modeling real-world phenomena where change is analyzed. In general, a differential equation might involve derivatives of various orders and their corresponding functions. It is written in the form:

• Ordinary differential equations (ODE): involves functions of a single variable and their derivatives.
• Partial differential equations (PDE): contains functions of multiple variables and their partial derivatives.

In this exercise, we deal with an ordinary differential equation: \[ y' + y = \frac{2}{1+4e^{2x}} \]The task is to determine if the function \( y = e^{-x} \tan^{-1}(2e^x) \) satisfies this equation for all values of \( x \). Initial value problems seek solutions that also fulfill a given starting point condition. In this case, it's \( y(-\ln 2) = \frac{\pi}{2} \). Solving initial value problems involves checking both the differential equation and initial conditions.

Product Rule

The product rule is a fundamental tool in calculus used to differentiate the product of two functions. Suppose you have two functions, \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product is given by:\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]In the exercise, we apply the product rule to differentiate \( y = e^{-x} \tan^{-1}(2e^x) \). Here, \( u(x) = e^{-x} \) and \( v(x) = \tan^{-1}(2e^x) \). Calculating the derivative:

• \( u'(x) = -e^{-x} \)
• \( v(x) \)' is found using the chain rule, explained later.

Finally, apply the product rule to find: \[ y' = e^{-x}\cdot v'(x) - e^{-x}\cdot \tan^{-1}(2e^x) \]Breaking complex functions into manageable parts with the product rule simplifies differentiation.

Chain Rule

The chain rule is used to differentiate compositions of functions. This rule states that if you have a composite function \( f(g(x)) \), its derivative is:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]In our context, apply the chain rule to differentiate \( \tan^{-1}(2e^x) \):

• \( f(u) = \tan^{-1}(u) \) with \( u = 2e^x \).
• The derivative \( f'(u) = \frac{1}{1+u^2} \).
• The derivative of \( u = 2e^x \) is \( 2e^x \).

Hence, \[ (\tan^{-1}(2e^x))' = \frac{1}{1+4e^{2x}} \cdot 2e^x \cdot 2 = \frac{4e^x}{1+4e^{2x}} \]Understanding the chain rule helps in differentiating complex nested functions by systematically applying derivatives step-by-step.

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the trigonometric functions like sine, cosine, and tangent. The function \( \tan^{-1}(x) \) or arc tangent gives the angle whose tangent is \( x \). These functions are vital in calculus for finding angles in different scenarios.For the derivative of \( \tan^{-1}(x) \) with respect to \( x \), we use:\[ \frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} \]In the context of the exercise, we examine \( \tan^{-1}(2e^x) \) and differentiate. Using the chain rule helps simplify the derivation process because of the expression inside the inverse function.Understanding how inverse trigonometric functions behave and their derivatives are important for solving a wide range of calculus problems efficiently. These derivatives often appear in calculations involving angles and need careful attention when combined with other derivatives.