Problem 24 - Calculus

Question

Match solutions and differential equations. (Note: Each equation may have more than one solution.) (a) \(y^{\prime \prime}-y=0\) (b) \(x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=0\) (c) \(x^{2} y^{\prime \prime}-6 y=0\) (1) \(y=e^{x}\) (II) \(y=x^{3}\) (III) \(y=e^{-x}\) (IV) \(y=x^{-2}\)

Step-by-Step Solution

Answer

(a) matches with (1) and (III); (b) matches with (IV); (c) matches with (II).

Step 1: Identify Solution for Equation (a)

The differential equation \(y'' - y = 0\) suggests a solution involving exponentials, as this is a homogeneous linear equation with constant coefficients. Testing the given solutions, we find that substituting \(y = e^{x}\) gives \(y'' = e^{x}\) and hence \(y'' - y = e^{x} - e^{x} = 0\), confirming \(y = e^{x}\) is a solution. Similarly, substituting \(y = e^{-x}\) gives \(y'' = e^{-x}\) and hence \(y'' - y = e^{-x} - e^{-x} = 0\), confirming \(y = e^{-x}\) is another solution.

Step 2: Identify Solution for Equation (b)

The differential equation \(x^{2} y'' + 2x y' - 2y = 0\) suggests a solution involving powers of \(x\). Substituting \(y = x^{-2}\) yields \(y' = -2x^{-3}\) and \(y'' = 6x^{-4}\). Substituting into the equation gives: \(x^{2}(6x^{-4}) + 2x(-2x^{-3}) - 2x^{-2} = \)\[(6x^{-2} - 4x^{-2} - 2x^{-2}) = 0\]. This confirms that \(y = x^{-2}\) is a solution.

Step 3: Identify Solution for Equation (c)

The differential equation \(x^{2} y'' - 6y = 0\) typically involves polynomial solutions. Testing \(y = x^{3}\), we find that \(y' = 3x^{2}\) and \(y'' = 6x\). Substituting into the equation, we get: \[x^{2}(6x) - 6(x^{3}) = 6x^{3} - 6x^{3} = 0\]. Thus, \(y = x^{3}\) satisfies the equation.

Key Concepts

Solution Verification

Verifying a solution in the context of differential equations means checking whether a potential solution satisfies the given equation. It's a crucial step because it confirms that the solution fits the problem. Here’s how it works:

• Take the proposed solution and compute any necessary derivatives.
• Substitute these derivatives back into the original differential equation.
• Simplify the equation to verify it results in an identity, often zero.

This process ensures that the solution is valid and consistent with the differential equation, enhancing understanding and preventing errors.

Homogeneous Equations

Homogeneous differential equations have all terms involving the unknown function or its derivatives with consistent degree, meaning no standalone constants or differing operations. In linear homogeneous equations like \(y'' - y = 0\), you’ll often deal with solutions that are exponential functions. Recognizing homogeneity is key because it guides the types of assumed solutions:

• Look for functions like \(e^{x}\) or \(x^{n}\) as they often fit.
• Utilize characteristic equations for constant coefficient ones to find roots indicating solution forms.

These characteristics make solving homogeneous equations systematic, linking directly to the types of functions explored.

Exponential Functions

Exponential functions like \(e^{x}\) and \(e^{-x}\) play a pivotal role in solving linear homogeneous differential equations with constant coefficients. Their inherent property of retaining form under differentiation makes them ideal candidates:

• The derivative of \(e^{x}\) is itself, simplifying substitution.
• Exponential growth or decay in physical phenomena often modelled by such equations.

Knowing how to work with these functions enhances the solution verification process and allows for easily identifying valid solutions in many differential equations.

Polynomial Functions

Polynomial functions, represented as sums of powers of \(x\), like \(x^{3}\), are commonly examined when solving differential equations with variable coefficients. The characteristics that make them useful include:

• Ease of differentiation—each application of \(d/dx\) reduces the power by one, revealing patterns.
• Appearing naturally in problems involving processes with proportional growth, like area or volume changes.

Solving equations like \(x^{2} y'' - 6y = 0\) benefits from this understanding, as testing polynomial solutions often leads to successful resolution.