Problem 7 - 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. $$ \frac{d y}{d x}=y+x^{2}, y(x)=C e^{x}-x^{2}-2 x-2 $$

Step-by-Step Solution

Answer

The given function satisfies the differential equation.

Step 1: Compute the derivative of y

We need to find \( \frac{dy}{dx} \) of the given function \( y(x) = Ce^{x} - x^{2} - 2x - 2 \). Differentiate each term separately. For \( Ce^{x} \), the derivative is \( Ce^{x} \). For \( -x^{2} \), the derivative is \( -2x \). For \( -2x \), the derivative is \( -2 \), and for \( -2 \), the derivative is \( 0 \). Thus, \( \frac{dy}{dx} = Ce^{x} - 2x - 2 \).

Step 2: Substitute y into the differential equation

The differential equation given is \( \frac{dy}{dx} = y + x^2 \). Substitute \( y(x) = Ce^{x} - x^{2} - 2x - 2 \) into the right side of the equation: \( y + x^2 = (Ce^{x} - x^{2} - 2x - 2) + x^2 \). Simplify this expression to get \( Ce^{x} - 2x - 2 \).

Step 3: Verify equality

Now compare the expressions from Step 1 and Step 2. We have \( \frac{dy}{dx} = Ce^{x} - 2x - 2 \) and \( y + x^2 = Ce^{x} - 2x - 2 \). Since both expressions are equal, \( \frac{dy}{dx} = y + x^2 \) holds.

Key Concepts

Understanding Derivatives

The concept of derivatives is crucial in calculus and differential equations. A derivative represents the rate of change of a function concerning one of its variables. In simpler terms, it tells you how fast or slow the function is changing at any given point. This is why you often see derivatives used in physics to describe motion or in economics to describe how much supply or demand changes in response to other variables.
For example, for a function like \( y(x) = Ce^{x} - x^{2} - 2x - 2 \), computing the derivative means finding \( \frac{dy}{dx} \). You break down the function into separate terms and differentiate each. Here's a recap of how this is done with our example:

• The derivative of \( Ce^{x} \) is \( Ce^{x} \) since the derivative of \( e^x \) is itself \( e^x \).
• The derivative of \( -x^{2} \) is \(-2x\) because of the power rule \( nx^{n-1} \).
• The derivative of \( -2x \) is \(-2\) as linear terms decrease to their coefficients.
• A constant like \(-2\) just becomes zero.

Altogether, you get \( \frac{dy}{dx} = Ce^{x} - 2x - 2 \). This derivative plays a key role in understanding how the function behaves.

Function Verification Explained

Function verification involves proving that a given function is a solution to a specific equation, like a differential equation. It's like a mathematical check to ensure everything lines up. In this case, you're checking if our function \( y(x) = Ce^{x} - x^{2} - 2x - 2 \) correctly satisfies the differential equation \( \frac{dy}{dx} = y + x^2 \).
The process begins by substituting the function into the equation. Replacing \( y(x) \) with the actual expression gives you:
\[ y + x^2 = (Ce^{x} - x^2 - 2x - 2) + x^2 \]
Simplifying this expression confirms that both sides equal \( Ce^{x} - 2x - 2 \), just like the derivative we found. If everything matches perfectly, then you've verified that the function is indeed a solution. This process is essential because it confirms the functionality and correctness of the solution in relation to the given differential equation.

Significance of the Constant of Integration

The constant of integration, often represented as \( C \), shows up in solutions to differential equations. When you solve these equations, you integrate to find a general solution. This process introduces an unknown constant because integration effectively "reverses" differentiation, where part of the original information might be lost.
The constant \( C \) reflects the fact that there are infinitely many functions that differentiate to give the same rate of change. For example, \( y(x) = Ce^{x} - x^{2} - 2x - 2 \) can take different shapes depending on the value of \( C \).
In practical terms, you determine \( C \) using initial conditions or extra bits of information. Knowing specifics about the function at a particular point allows you to find \( C \) and tailor the general solution to one that's precise and unique. Understanding this constant's role helps in appreciating the broader scope of solutions possible for differential equations.