Problem 24 - Calculus

Question

Find a function \(y(x)\) that solves the differential equation. \(d y / d x=1 / \sqrt{1-2 x}\)

Step-by-Step Solution

Answer

The solution is \( y(x) = -(1 - 2x)^{1/2} + C \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = \frac{1}{\sqrt{1 - 2x}} \). This is a first-order separable differential equation since we can separate the variables.

Step 2: Separate the Variables

Rearrange the equation to get all terms involving \( y \) on one side and all terms involving \( x \) on the other side to prepare for integration. We rearrange it as follows: \[ dy = \frac{1}{\sqrt{1 - 2x}} \, dx \]

Step 3: Integrate Both Sides

Integrate both sides of the equation:\[ \int dy = \int \frac{1}{\sqrt{1 - 2x}} \, dx \] The left-hand side integrates to \( y \). For the right-hand side, use the substitution \( u = 1 - 2x \), which gives \( du = -2 \, dx \), or \( dx = \frac{-1}{2} du \). So the integral becomes: \[ \int \frac{1}{\sqrt{u}} \cdot \frac{-1}{2} \, du = -\frac{1}{2} \int u^{-1/2} \, du \]

Step 4: Evaluate the Integrals

The integral \( \int u^{-1/2} \, du \) is evaluated as \( 2u^{1/2} \). Substituting back \( u = 1 - 2x \), we get: \[ -\frac{1}{2} \cdot 2(1 - 2x)^{1/2} = -(1 - 2x)^{1/2} \]Hence, \[ y = -(1 - 2x)^{1/2} + C \] where \( C \) is the constant of integration.

Step 5: Express the General Solution

The function \( y(x) \) that solves the differential equation is given by:\[ y(x) = -(1 - 2x)^{1/2} + C \] where \( C \) is an arbitrary constant that can be determined if an initial condition is provided.

Key Concepts

Integration Techniques

Integrating functions is a fundamental process in solving differential equations. In our exercise, we encountered the need to integrate both sides to find the solution to the differential equation. This technique involves two main steps:

• Identifying the function we need to integrate.
• Choosing the right integration method for the problem.

In this exercise, the left-hand side was straightforward as it integrated directly to give us the function \(y\). However, the right-hand side required a more nuanced approach using substitution. Depending on the integration problem, different techniques might be employed, such as:

• Direct integration, which can be applied when the antiderivative is immediately recognizable.
• Substitution method, often used when a simple transformation can simplify the integrand.
• Integration by parts, useful for products of functions.
• Partial fraction decomposition, applied when dealing with rational functions.

Each technique has its own strategic use, often depending on the form and complexity of the function to be integrated. Understanding these methods is crucial for proceeding with solutions in differential calculus.

Substitution Method

The substitution method is a powerful technique in integration, especially useful when the integrand contains a composition of functions. It essentially transforms a difficult integral into a simpler one by changing variables. In the exercise, the substitution \( u = 1 - 2x \) was used.
This substitution served several purposes:

• It simplified the expression in the integrand, making it easier to integrate \( \int u^{-1/2} \, du \).
• It helped in dealing with the square root in the denominator, which would be cumbersome to integrate directly.

To execute a substitution, we also need to adjust the differential \( dx \) to \( du \), achieved by differentiating the substitution formula. It's crucial to reverse the substitution at the end to express the solution in terms of the original variable \( x \). This ensures continuity and proper evaluation of the integral in its original context.

First-Order Differential Equations

First-order differential equations, which involve only the first derivative of the function, are among the simplest types to solve. They provide a foundational understanding of differential equations in mathematics. Our given differential equation \( \frac{dy}{dx} = \frac{1}{\sqrt{1 - 2x}} \) can be solved using separation of variables.
This technique involves:

• Rearranging the equation to isolate differentials of \( y \) and \( x \) on opposite sides.
• Integrating both sides separately to find the general solution.
• Applying an initial condition, if provided, to find the specific solution.

Separable differential equations are particularly straightforward for they allow separation of the dependent variable and its derivative. It's useful to remember that the final solution will typically include a constant of integration \( C \), which represents the family of solutions that satisfy the differential equation. This constant is crucial when matching the solution to initial conditions or specific boundary values.