Problem 2 - Applied Calculus

Question

Decide whether or not each of the following is a solution to the differential equation \(x y^{\prime}-2 y=0\). (a) \(y=x^{2}\) (b) \(y=x^{3}\)

Step-by-Step Solution

Answer

(a) is a solution; (b) is not a solution.

Step 1: Understand the Differential Equation

The differential equation given is \(x y' - 2y = 0\). To determine if a function is a solution, it should satisfy this equation when \(y\) and \(y'\) are substituted into it.

Step 2: Differentiate Function (a)

Consider \(y = x^2\). Find the derivative: \(y' = \frac{d}{dx}(x^2) = 2x\).

Step 3: Substitute and Check Function (a)

Substitute \(y = x^2\) and \(y' = 2x\) into the differential equation: \(x(2x) - 2(x^2) = 0\) which simplifies to \(2x^2 - 2x^2 = 0\). This holds true, so \(y = x^2\) is a solution.

Step 4: Differentiate Function (b)

Consider \(y = x^3\). Find the derivative: \(y' = \frac{d}{dx}(x^3) = 3x^2\).

Step 5: Substitute and Check Function (b)

Substitute \(y = x^3\) and \(y' = 3x^2\) into the differential equation: \(x(3x^2) - 2(x^3) = 0\) which simplifies to \(3x^3 - 2x^3 = x^3\). This does not equal zero, so \(y = x^3\) is not a solution.

Key Concepts

Solution Verification

When faced with a differential equation, one of the key challenges is verifying whether a proposed function is indeed a solution. The first step towards verification is substitution. To see if a function like \( y = x^2 \) is a solution to the differential equation \( xy' - 2y = 0 \), you must substitute both the function and its derivative back into the equation. If the left-hand side simplifies to zero, then the function is a solution.Verification Steps:

• Identify the function: Start with the given function, such as \( y = x^2 \).
• Differentiate to find \( y' \) — this derivative is necessary for substitution.
• Substitute: Place both \( y \) and \( y' \) into the differential equation.
• Check the equality: If both sides are equal (resulting in an identity), then the function is a solution.

Verification simplifies the problem to a test of equality, ensuring that the function satisfies the differential equation throughout its domain.

Derivative Calculation

Calculating derivatives is crucial for differential equations, as it helps find the rate of change of a function. In problems like these, you start with your function and calculate its derivative to verify solutions of differential equations.Basics of Derivative Calculation:

• Power rule: For functions of the form \( y = x^n \), the derivative \( y' = nx^{n-1} \).
• Example calculation: For \( y = x^2 \), the derivative is calculated as \( \frac{d}{dx}(x^2) = 2x \). This is straightforward using the power rule.
• Consistency: Ensure that each step in finding the derivative is checked and aligns with standard norms, reducing calculation errors.

Derivative calculation provides the functional component needed in substitution, allowing for direct comparison and verification within differential equations.

Substitution Method

The substitution method is essential when solving or verifying solutions to differential equations. It involves replacing the function and its derivative into the equation and simplifying.Using Substitution:

• Preparation: Have your function (\( y \)) and its derivative (\( y' \)) ready after differentiation.
• Substitute: Replace \( y' \) and \( y \) in the differential equation, such as substituting \( y = x^2 \) and \( y' = 2x \) into \( xy' - 2y = 0 \).
• Simplify: Carry out algebraic simplifications to ensure both sides of the equation match. For example, \( x(2x) - 2(x^2) = 0 \) simplifies to \( 2x^2 - 2x^2 = 0 \), verifying the function.

Substitution confirms if the differential equation holds true for a given function, thereby validating or excluding it as a solution.