Question

Solve the initial-value problem for y as a function of x. $$ \left(x^{2}+36\right) \frac{d y}{d x}=1, y(6)=0 $$

Short answer

\(y = \frac{1}{6} \tan^{-1} \left( \frac{x}{6} \right) - \frac{\pi}{24}\)

Step-by-step solution

Convert Differential Equation

First, express the differential equation in the standard form by isolating \( \frac{dy}{dx} \). This gives us: \[ \frac{dy}{dx} = \frac{1}{x^2 + 36} \]

Integrate Both Sides

Integrate both sides of the equation with respect to \(x\). For the left side, integrate \(1 \cdot dx\) to obtain \(y\). For the right side, integrate \(\frac{1}{x^2 + 36}\): \[ y = \int \frac{1}{x^2 + 36} \, dx \]

Recognize and Apply Standard Integral

Recognize that the integral of \( \frac{1}{x^2 + a^2} \) is a standard form \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). Here, \(a = 6\). Thus, \[ y = \frac{1}{6} \tan^{-1} \left( \frac{x}{6} \right) + C \]

Apply Initial Condition

Use the initial condition \(y(6) = 0\) to determine \(C\). Substitute \(x=6\) and \(y=0\) into the equation:\[ 0 = \frac{1}{6} \tan^{-1} \left( \frac{6}{6} \right) + C \]\[ 0 = \frac{1}{6} \tan^{-1}(1) + C \]\[ \tan^{-1}(1) = \frac{\pi}{4} \]Thus,\[ 0 = \frac{1}{6} \cdot \frac{\pi}{4} + C \]\[ C = -\frac{\pi}{24} \]

Finalize the Function

Substitute \(C\) back into the general solution. The solution for \(y\) as a function of \(x\) is:\[ y = \frac{1}{6} \tan^{-1} \left( \frac{x}{6} \right) - \frac{\pi}{24} \]

Key concepts

Differential Equations

Differential equations are equations that relate a function with its derivatives. In simple terms, they describe how a particular quantity evolves over time or space. These equations play a crucial role in expressing various physical, biological, and economic phenomena. For example, the rate at which a population grows or how heat spreads across a material can be modeled using differential equations.

In the given problem, we are dealing with a first-order differential equation. It involves a single derivative, in this case, \( \frac{dy}{dx} \), which describes how the function \( y \) changes with respect to \( x \). The main goal here is to solve for \( y \) as a function of \( x \), which involves integrating the derivative back into a function form. This process of converting the derivative back into the original function is what solving a differential equation is all about. The initial condition \( y(6) = 0 \) adds a constraint that helps us find a unique solution among the infinite possibilities.

Integration Techniques

Integration is the reverse process of differentiation and helps find the original function from its derivative. There are several techniques for integration, and choosing the right one is crucial for solving differential equations effectively.

In our problem, we need to integrate \( \frac{1}{x^2 + 36} \). This seems tricky at first glance, but recognizing the form of the integral can simplify the task significantly. This integral matches the standard form \( \int \frac{1}{x^2 + a^2} \, dx \), which has a well-known solution. Integrating functions involving expressions like \( x^2 + a^2 \) often leads to inverse trigonometric functions, specifically the arctangent function. By applying this technique, we translate the problem into a tractable form, leading directly to our answer.

Initial Conditions

Initial conditions are essential for solving differential equations because they allow us to find specific solutions that fit a particular scenario. When you solve differential equations through integration, you often get a general solution with an arbitrary constant, denoted as \( C \). Without further information, this general solution could represent infinitely many possible functions.

In our exercise, the problem states an initial condition, \( y(6) = 0 \). This condition serves as a sort of anchor, letting us determine the exact value of \( C \). By plugging this point into our integrated function, we solve for \( C \) uniquely, thereby transforming the general solution into a particular one that satisfies the initial requirement. Initial conditions make the solution applicable to real-world problems where specific outcomes are desired.

Inverse Trigonometric Functions

Inverse trigonometric functions like \( \tan^{-1}(x) \), also known as arctan, are central to several integration problems. These functions serve as the inverse operations of regular trigonometric functions, meaning they tell us the angle whose trigonometric value is given.

In this exercise, recognizing that the integral of \( \frac{1}{x^2 + 36} \) results in an inverse trigonometric function is a key step. Specifically, the standard integral form \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C \) helps convert the integration into a simpler calculation.

• Here, \( a \) corresponds to the square root of 36, which is 6.
• Substituting \( a = 6 \) into the integral solution provides our result \( \frac{1}{6} \tan^{-1}(\frac{x}{6}) + C \).

These functions are important tools when solving problems that involve integrals of specific quadratic expressions, and they appear frequently in differential equations solutions.