Question

Solve the initial-value problem for y as a function of x. $$ \left(64-x^{2}\right) \frac{d y}{d x}=1, y(0)=3 $$

Short answer

\(y = \frac{1}{8} \arcsin\left(\frac{x}{8}\right) + 3\).

Step-by-step solution

Separate Variables

First, rewrite the given differential equation to separate the variables. The original equation is \((64-x^2) \frac{dy}{dx} = 1\).Divide both sides by \(64-x^2\) to isolate \(\frac{dy}{dx}\) on one side: \[ \frac{dy}{dx} = \frac{1}{64-x^2}. \]

Integrate Both Sides

Integrate both sides with respect to \(x\). The left side is \(\int dy = y\). For the right side, integrate \(\int \frac{1}{64-x^2} dx\). This integral can be solved using the inverse trigonometric identities:\[ y = \frac{1}{8} \arcsin\left(\frac{x}{8}\right) + C. \]

Apply the Initial Condition

Use the initial condition \(y(0) = 3\) to solve for \(C\). Substitute \(x = 0\) and \(y = 3\) into the integrated equation:\[ 3 = \frac{1}{8} \arcsin\left(\frac{0}{8}\right) + C. \]Since \(\arcsin(0) = 0\), the equation simplifies to \(3 = C\).

Write the Solution

Substitute \(C = 3\) back into the equation:\[ y = \frac{1}{8} \arcsin\left(\frac{x}{8}\right) + 3. \]This equation defines \(y\) as a function of \(x\) with the given initial condition.

Key concepts

Initial-Value Problems

Initial-value problems are a common type of differential equation where the solution not only satisfies the differential equation but also meets certain specified conditions. These conditions are known as initial conditions. An example could be finding a function that not only satisfies the differential equation \((64-x^2) \frac{dy}{dx} = 1\) but also holds for a specific value, such as \(y(0) = 3\).
Initial conditions like this help to determine the constant of integration, ensuring the solution fits specific criteria. These problems are essential in fields such as physics and engineering, where boundary conditions define behavior at certain points.
An important step is applying the initial condition after integrating the equation. This helps to solve for any unknown constants, thus providing a specific solution that adheres to the given conditions.

Separation of Variables

Separation of variables is a method used to solve differential equations by isolating each variable on different sides of the equation. In the exercise given, the equation \((64-x^2) \frac{dy}{dx} = 1\) was rearranged so that all terms involving \(y\) were on one side and all terms involving \(x\) were on the other.
You achieve this by calculating

• \(\frac{dy}{dx} \rightarrow \frac{dy}{(64-x^2)} = \frac{1}{64-x^2}\)

By integrating each side independently, you can then find the solution to the equation.
While seemingly simple, separation of variables is powerful in that it allows complex differential equations to be broken down piece by piece. It is particularly effective when the function and its derivative can be expressed in terms of separate variables.

Inverse Trigonometric Functions

Inverse trigonometric functions are the functions that undo trigonometric functions such as sine, cosine, and tangent. In solving the integral \(\int \frac{1}{64-x^2} dx\), we used the identity of the inverse sine function, \(\arcsin(x)\).
The relationship \(\frac{1}{\sqrt{1-u^2}} = \arcsin(u)\) was key to finding the integral solution. Here, \(u\) was substituted with \(\frac{x}{8}\), making the integral become \(\frac{1}{8}\arcsin(\frac{x}{8})\).
Inverse trigonometric functions are extremely useful in integration, especially when dealing with equations containing terms like \(1/(a^2 - x^2)\). Being familiar with these functions allows you to tackle many calculus problems more efficiently.

Integration Techniques

Integration techniques are vital tools that are used to solve integrals, which is a fundamental part of calculus. In the exercise, integrating both sides of the equation was crucial in finding the solution.
The integral \(\int dy = y\) is a straightforward example. However, \(\int \frac{1}{64-x^2} dx\) required more complex methods. When dealing with such expressions, some potential techniques involve:

• Inverse trigonometric functions
• Substitution method

Making smart use of these techniques can simplify the process and lead to a correct answer. Different problems will call for different methods, so knowing a variety of techniques allows for flexibility in problem-solving. This is especially useful when standard methods might not work directly.