Question

Solve the initial-value problems. $$ \frac{d r}{d \theta}+r \tan \theta=\cos ^{2} \theta, \quad r\left(\frac{\pi}{4}\right)=1 $$

Short answer

The solution to the initial-value problem is: \[ r(\theta) = \frac{1}{\sec\theta} \left(\ln|\sec\theta + \tan\theta| - \int \sec\theta \sin^2\theta\, d\theta + \frac{1}{\sqrt{2}} - \ln(1 + \sqrt{2})\right) \]

Step-by-step solution

Identify the integrating factor

First, let's rewrite the given differential equation to better recognize its standard form: \[ \frac{dr}{d\theta} + (\tan\theta) r = \cos^2\theta \] Now, we can see that it is a first-order linear ordinary differential equation (ODE) with the dependent variable \(r(\theta)\) and independent variable \(\theta\). The integrating factor \(\mu(\theta)\) is given by the exponential of the integral of the function multiplying \(r\): \[ \mu(\theta) = e^{\int \tan \theta\, d\theta} \]

Calculate the integrating factor

Now we need to find the integral of \(\tan\theta\): \[ \int \tan \theta\, d\theta = \ln|\sec\theta| \] Thus, the integrating factor \(\mu(\theta)\) is: \[ \mu(\theta) = e^{\ln|\sec\theta|} = \sec\theta \]

Multiply the differential equation by the integrating factor and integrate

By multiplying both sides of the equation by the integrating factor \(\mu(\theta) = \sec\theta\), we obtain: \[ \sec\theta \frac{dr}{d\theta} + r \sec\theta \tan\theta = \sec\theta \cos^2\theta \] Now, recognize the left side as the derivative of a product: \[ \frac{d}{d\theta}(r\sec\theta) = \sec\theta(1 - \sin^2\theta) \] Integrating both sides with respect to \(\theta\): \[ \int \frac{d}{d\theta}(r\sec\theta)\, d\theta = \int \sec\theta(1 - \sin^2\theta)\, d\theta \] We can perform the integration on both sides: \[ r\sec\theta = \int \sec\theta\, d\theta - \int \sec\theta \sin^2\theta\, d\theta + C \]

Solve for r(θ)

As we know, \(\int \sec\theta\, d\theta = \ln|\sec\theta + \tan\theta|\), so we have: \[ r\sec\theta = \ln|\sec\theta + \tan\theta| - \int \sec\theta \sin^2\theta\, d\theta + C \] Solving for \(r(\theta)\): \[ r(\theta) = \frac{1}{\sec\theta} \left(\ln|\sec\theta + \tan\theta| - \int \sec\theta \sin^2\theta\, d\theta + C\right) \]

Use initial condition to find the constant

Now we need to apply the initial condition \(r(\frac{\pi}{4}) = 1\): \[ 1 = \frac{1}{\sec(\pi / 4)} \left(\ln|\sec(\pi / 4) + \tan(\pi / 4)| - \int \sec(\pi / 4) \sin^2(\pi / 4)\, d(\pi / 4) + C\right) \] This simplifies to: \[ 1 = \sqrt{2} \left(\ln(1 + \sqrt{2}) + C\right) \] Therefore, the constant \(C\) can be determined as: \[ C = \frac{1}{\sqrt{2}} - \ln(1 + \sqrt{2}) \]

Write the final solution

Now we can substitute the constant \(C\) back into \(r(\theta)\) to obtain the final solution: \[ r(\theta) = \frac{1}{\sec\theta} \left(\ln|\sec\theta + \tan\theta| - \int \sec\theta \sin^2\theta\, d\theta + \frac{1}{\sqrt{2}} - \ln(1 + \sqrt{2})\right) \] This is the solution to the initial-value problem for the given differential equation and initial condition.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are a fundamental category of equations that describe the relationships between functions and their derivatives. These equations dictate how a quantity changes over time or in response to different variables, making them vital in physics, engineering, biology, economics, and various other scientific disciplines.

The differential equation in our exercise, \( \frac{dr}{d\theta} + r \tan \theta = \cos^2 \theta \), exemplifies an ODE as it includes a derivative with respect to a single variable, indicating how the function \(r\) changes with the angle \(\theta\). These types of equations can describe a broad range of phenomena, from the growth of populations to the motion of celestial bodies. Mastering the methods to solve ODEs, like the integrating factor used in this exercise, is an essential skill for scientists and mathematicians.

Integrating Factor Method

The integrating factor method is a systematic approach to solving linear first-order ordinary differential equations of the form \( \frac{dy}{dx} + p(x)y = q(x) \). The technique is particularly powerful because it transforms a non-exact differential equation into an exact one, which can be solved more straightforwardly.

To apply this method, one must correctly identify an integrating factor, which is typically a function of \( \mu(x) = e^{\int p(x) \text{d}x} \). When we multiply every term of the ODE by this integrating factor, the left side of the equation becomes the derivative of a product of the integrating factor and the dependent variable, \( y \). This property allows us to integrate the left side directly, and thus move closer to acquiring the solution for \( y(x) \). The process can be easier to comprehend through practice and examples, such as the initial-value problem provided here.

First-Order Linear ODE

A first-order linear ODE has the general form \( \frac{dx}{dt} + p(t)x = q(t) \), where \( x \) is the dependent variable, \( t \) is the independent variable, and \( p(t) \) and \( q(t) \) are known functions. This equation is first-order because it involves only the first derivative of \( x \) and no higher derivatives. It is linear because the dependent variable \( x \) and its derivative \( \frac{dx}{dt} \) appear linearly (without being raised to any power or being inside of other functions).

Our particular problem involves the function \( r(\theta) \) and its derivative with respect to \( \theta \). The most distinctive characteristic of linear first-order ODEs is that they can always be solved analytically using the integrating factor method. This robust framework provides a clear pathway to finding the explicit form of the solution, especially when an initial condition is provided, determining a unique solution to the problem.