Question

Verify that the following functions are solutions to the given differential equation.\(y=\frac{1}{1-x}\) solves \(y^{\prime}=y^{2}\)

Short answer

Yes, the function satisfies the equation \( y' = y^2 \).

Step-by-step solution

Differentiate the Given Function

Let us first differentiate the function \( y = \frac{1}{1-x} \) with respect to \( x \). Using the chain rule, we have:\[ \frac{dy}{dx} = \frac{d}{dx}(1 - x)^{-1} = (-1)(1-x)^{-2}(-1) = \frac{1}{(1-x)^2}. \]

Calculate \( y^2 \)

Now, we need to calculate \( y^2 \). Substitute \( y = \frac{1}{1-x} \) into \( y^2 \), we have:\[ y^2 = \left( \frac{1}{1-x} \right)^2 = \frac{1}{(1-x)^2}. \]

Verify the Differential Equation

We check if the function \( y \) satisfies the differential equation \( y' = y^2 \). From Step 1, \( \frac{dy}{dx} = \frac{1}{(1-x)^2} \). From Step 2, \( y^2 = \frac{1}{(1-x)^2} \).Since \( \frac{dy}{dx} = y^2 \), the given function \( y = \frac{1}{1-x} \) is a solution to the differential equation \( y' = y^2 \).

Key concepts

Chain Rule

The concept of the chain rule in calculus refers to a method for differentiating composite functions. When you have a function within another function, you use the chain rule to differentiate it easily. Here, we took the function \( y = \frac{1}{1-x} \) and needed to find its derivative.
To apply the chain rule:

• Identify the outer and inner functions. For \( y = \frac{1}{1-x} \), the outer function is \( u^{-1} \), where \( u = 1-x \).
• Differentiate the outer function while keeping the inner function intact. This gives us \( -1(1-x)^{-2} \).
• Multiply by the derivative of the inner function \( (1-x) \), which is \( -1 \).

With these steps, we ultimately reach the derivative \( \frac{dy}{dx} = \frac{1}{(1-x)^2} \). This showcases how the chain rule simplifies differentiation of nested functions.

Differentiation

Differentiation is a fundamental operation in calculus, allowing us to find the rate at which a quantity changes. In this problem, we differentiated the function \( y = \frac{1}{1-x} \) to satisfy the differential equation.
Differentiation involves:

• Applying rules such as the chain rule, product rule, or quotient rule, depending on the function's structure.
• Finding the derivative provides insight into the behavior of \( y \) as \( x \) changes.

Here, we focused on obtaining \( \frac{dy}{dx} \), representing the rate of change of \( y \) with respect to \( x \). Using calculus tools ensures accurate modeling of dynamic behaviors in mathematical expressions, such as this differential equation.

Verifying Solutions

Verifying solutions to differential equations is crucial for confirming that a proposed function actually satisfies the given equation. For the function \( y = \frac{1}{1-x} \) and the differential equation \( y' = y^2 \), our task was to ensure they align.
Verification process:

• First, differentiate the function, obtaining \( y' \).
• Calculate \( y^2 \) by squaring the original function.
• Compare \( y' \) and \( y^2 \). If both results match, as in \( \frac{dy}{dx} = \frac{1}{(1-x)^2} = y^2 \), then the function is a valid solution to the equation.

This step is essential because it confirms the theoretically derived solution works practically – ensuring correctness and reliability in mathematical models.