Problem 6 - Calculus Single Variable

Question

Verify that the given function \(y\) satisfies the given differential equation. In each expression for \(y(x)\) the letter \(C\) denotes a constant. $$ \frac{d y}{d x}=x+x y, y(x)=C e^{x^{2} / 2}-1 $$

Step-by-Step Solution

Answer

The function \( y(x) = Ce^{x^2/2} - 1 \) satisfies the differential equation.

Step 1: Identify the given differential equation

The differential equation provided is \( \frac{dy}{dx} = x + xy \). Our task is to verify if the given function, \( y(x) = Ce^{x^2/2} - 1 \), is a solution to this differential equation.

Step 2: Differentiate the given function

Start by differentiating the function \( y(x) = Ce^{x^2/2} - 1 \) with respect to \( x \). Use the chain rule: \[ \frac{d}{dx}(e^{x^2/2}) = e^{x^2/2} \cdot \frac{d}{dx}(x^2/2) = e^{x^2/2} \cdot x \].Therefore, \( \frac{dy}{dx} = C \cdot e^{x^2/2} \cdot x \).

Step 3: Substitute and verify in the differential equation

Replace \( y \) with \( Ce^{x^2/2} - 1 \) and \( \frac{dy}{dx} \) with \( Cxe^{x^2/2} \) in the differential equation:Left-hand side: \( \frac{dy}{dx} = Cxe^{x^2/2} \).Right-hand side: \( x + x(Ce^{x^2/2} - 1) = x + x \cdot Ce^{x^2/2} - x = Cxe^{x^2/2} \).Both sides are equal, confirming \( y(x) \) is indeed a solution.

Key Concepts

Verifying Solutions in Differential Equations

Verifying a function as a solution for a differential equation is like fact-checking in math. This process ensures that a proposed function truly satisfies the equation, confirming its validity. To verify a solution:

• Start by inserting the given function into the differential equation.
• Differentiate the function as required by the equation.
• Compare both sides of the equation once the function is substituted.

Let's see this in action with our original example. We are given the differential equation \( \frac{dy}{dx} = x + xy \) and the function \( y(x) = Ce^{x^2/2} - 1 \). By differentiating and substituting, we ensure both sides match. In this example, both sides become \( Cxe^{x^2/2} \) after substitution, confirming the function is indeed a solution. This validation process is vital when working with differential equations as it provides certainty that the function operates as intended under the constraints provided by the equation.

Applying the Chain Rule

The chain rule is a powerful tool in calculus for finding the derivative of composite functions. It is especially handy when dealing with exponential functions like the one in our example. The principle of the chain rule is simple:

• The derivative of a composed function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
• Identify the inner function \( g(x) \) and the outer function \( f(u) \), where \( u = g(x) \).
• Differentiate both functions separately and multiply them together.

In the example, the function \( e^{x^2/2} \) is a composition of the base \( e^x \) and the function \( g(x) = x^2/2 \). We differentiate \( e^{x^2/2} \) using the chain rule, identifying \( g(x) \) as \( x^2/2 \), which differentiates to \( x \). Thus, the derivative is \( e^{x^2/2} \cdot x \), an essential step in verifying the solution to our differential equation.

Understanding First-Order Differential Equations

First-order differential equations are equations that involve the first derivative of a function. These equations often take the form \( \frac{dy}{dx} = f(x, y) \), where the rate of change of \( y \) with respect to \( x \) is given by some function involving both parameters. Their applications span across fields such as:

• Physics: Describing motion and energy.
• Biology: Modeling population dynamics.
• Economics: Predicting market behaviors.

The objective in solving these equations is to find a function \( y(x) \) that satisfies the equation over a particular range. In our example, the equation \( \frac{dy}{dx} = x + xy \) requires finding \( y(x) \) such that both sides match after substitution. This process involves understanding both algebraic manipulation and calculus principles to resolve the equation and verify solutions.