Problem 73 - Calculus - AP Edition

Question

Show that (a) \(y=x e^{-x}\) satisfies the equation \(x y^{\prime}=(1-x) y\) (b) \(y=x e^{-x^{2} / 2}\) satisfies the equation \(x y^{\prime}=\left(1-x^{2}\right) y\)

Step-by-Step Solution

Answer

Both functions satisfy their respective differential equations through differentiation and substitution, confirming the given solutions.

Step 1: Differentiate the given function for Part (a)

For the function given in part (a), \( y = x e^{-x} \), we need to differentiate it with respect to \( x \). Use the product rule for differentiation: if \( u = x \) and \( v = e^{-x} \), then \( y = uv \). The derivative \( y' \) is given by:\[y' = u'v + uv'\]First, find the derivatives \( u' = 1 \) and \( v' = -e^{-x} \). Now substitute back:\[y' = 1 imes e^{-x} + x imes (-e^{-x}) = e^{-x} - xe^{-x}\]

Step 2: Substitute and simplify for Part (a)

Next, substitute \( y = x e^{-x} \) and its derivative \( y' = e^{-x} - xe^{-x} \) into the equation \( x y' = (1-x) y \):- Left side: \( x y' = x(e^{-x} - xe^{-x}) = xe^{-x} - x^2e^{-x} \).- Right side: \( (1-x) y = (1-x)(x e^{-x}) = x e^{-x} - x^2e^{-x} \).Both sides are equal, showing that \( y = x e^{-x} \) satisfies the equation \( x y^{ ext{'}} = (1-x) y \).

Step 3: Differentiate the given function for Part (b)

For the function given in part (b), \( y = x e^{-x^2/2} \), we differentiate it using the product rule:- Let \( u = x \) and \( v = e^{-x^2/2} \), so \( y = uv \).- Thus, \( y' = u'v + uv' \).- Calculate \( u' = 1 \) and \( v' = e^{-x^2/2} \times (-x) = -x e^{-x^2/2} \).So, we have:\[y' = e^{-x^2/2} + (-x^2 e^{-x^2/2}) = e^{-x^2/2} - x^2 e^{-x^2/2}\]

Step 4: Substitute and simplify for Part (b)

Substitute \( y = x e^{-x^2/2} \) and its derivative \( y' = e^{-x^2/2} - x^2 e^{-x^2/2} \) into the equation \( x y' = (1-x^2) y \):- The left side is \( x y' = x(e^{-x^2/2} - x^2 e^{-x^2/2}) = x e^{-x^2/2} - x^3 e^{-x^2/2} \).- The right side is \( (1-x^2) y = (1-x^2)(x e^{-x^2/2}) = x e^{-x^2/2} - x^3 e^{-x^2/2} \).Both sides are equal, confirming \( y = x e^{-x^2/2} \) satisfies the equation \( x y^{ ext{'}} = (1-x^2) y \).

Key Concepts

Product Rule

The product rule is an essential tool in calculus, especially for differentiating functions that are products of two or more expressions. If you have two functions, say \( u(x) \) and \( v(x) \), their product \( y(x) = u(x)v(x) \) can be differentiated using the product rule. The formula is given by:

• \( y' = u'v + uv' \)

This means you take the derivative of the first function \( u \) and multiply it by the second function \( v \), then add the product of the first function \( u \) and the derivative of the second function \( v \).

For example, in the solution to part (a) of the original exercise, we applied the product rule to \( y = x e^{-x} \) by letting \( u = x \) and \( v = e^{-x} \). Differentiating gives \( u' = 1 \) and \( v' = -e^{-x} \), leading to the derivative \( y' = e^{-x} - x e^{-x} \).

The product rule's usefulness lies in its ability to simplify the differentiation of product expressions, helping solve complex equations found in calculus problems.

Differentiation

Differentiation is the process in calculus used to find the rate at which a function changes. It essentially measures how the function's output value changes as its input changes.

To differentiate a function, you apply rules such as the product rule, chain rule, or quotient rule, depending on the type of function you have. This process lets you find the derivative, which is the function that represents the rate of change.

In the original exercise, differentiation was our primary tool. We differentiated \( y = x e^{-x} \) for part (a) using the product rule. This provided us with the life derivative \( y' \) needed to substitute into the given equation. Similarly, differentiation of \( y = x e^{-x^2/2} \) for part (b) involved applying the differentiation process to obtain \( y' = e^{-x^2/2} - x^2 e^{-x^2/2} \).

By understanding differentiation, you can unravel the behavior of functions, especially when confirming if a function satisfies a given differential equation.

Mathematical Proof

Mathematical proofs are a structured way of demonstrating the truth of a statement or theorem. In calculus, proving that a function satisfies a differential equation involves verifying that both sides of the equation are equivalent.

For example, in part (a) of the exercise, the proof involved showing that both sides of the equation \( x y' = (1-x) y \) are equivalent when \( y = x e^{-x} \). By substituting and simplifying both the left and right hand sides, we can see they become identical: \( x e^{-x} - x^2e^{-x} \). This confirms that the original function truly satisfies the differential equation.

Mathematical proofs help ensure the correctness of mathematical statements, grounding abstract ideas in logical argumentation. In calculus, particularly, they validate solutions and enhance our understanding of how mathematical principles work.