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Verify that the given function \(y\) satisfies the given differential equation. In each expression for \(y(x)\) the letter \(C\) denotes a constant. $$ \frac{d y}{d x}=\frac{2 x-y}{x+y}, y=-x+\sqrt{3 x^{2}+2 C} $$

Short Answer

Expert verified
The given function satisfies the differential equation as both sides of the equation are equivalent upon substitution and differentiation.

Step by step solution

01

Differentiate the given function

Start with the function given: \( y = -x + \sqrt{3x^2 + 2C} \). We need to differentiate it with respect to \( x \). Use the chain rule to differentiate \( \sqrt{3x^2 + 2C} \). The derivative is: \( \frac{dy}{dx} = -1 + \frac{1}{2\sqrt{3x^2 + 2C}} \cdot 6x \). Simplifying this, we get \( \frac{dy}{dx} = -1 + \frac{3x}{\sqrt{3x^2 + 2C}} \).
02

Express terms in the differential equation form

Now, substitute \( y = -x + \sqrt{3x^2 + 2C} \) back into the right side of the given differential equation \( \frac{2x - y}{x + y} \). This results in: \( \frac{2x - (-x + \sqrt{3x^2 + 2C})}{x + (-x + \sqrt{3x^2 + 2C})} \). Simplify the terms in both the numerator and the denominator.
03

Simplification

Simplify the expression from Step 2: The numerator becomes \( 3x - \sqrt{3x^2 + 2C} \), and the denominator simplifies to \( \sqrt{3x^2 + 2C} \). Hence, the expression becomes \( \frac{3x - \sqrt{3x^2 + 2C}}{\sqrt{3x^2 + 2C}} \).
04

Compare derivatives

The derivative we previously found in Step 1 is \( -1 + \frac{3x}{\sqrt{3x^2 + 2C}} \). The expression obtained in Step 3 from the differential equation is \( \frac{3x - \sqrt{3x^2 + 2C}}{\sqrt{3x^2 + 2C}} = -1 + \frac{3x}{\sqrt{3x^2 + 2C}} \). Both expressions are identical, indicating the solution satisfies the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. It allows us to find the derivative of a function that is composed of other functions. In this exercise, the function \( y \) is given by \( y = -x + \sqrt{3x^2 + 2C} \).To differentiate \( \sqrt{3x^2 + 2C} \), we use the Chain Rule. The outer function is the square root \( \sqrt{x} \), and the inner function is \( 3x^2 + 2C \).
  • Differentiate the outer function: The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).
  • Differentiate the inner function: The derivative of \( 3x^2 \) is \( 6x \), and the constant \( 2C \) has a derivative of zero.
Combine these using the Chain Rule formula, giving \( \frac{d}{dx}(\sqrt{3x^2 + 2C}) = \frac{1}{2\sqrt{3x^2 + 2C}} \cdot 6x \). This illustrates how the Chain Rule links the derivatives of the inner and outer functions to provide the final derivative.
Verification of Solutions
Verification of solutions involves checking if a proposed function satisfies a given differential equation. In our example, we need to show that the function \( y = -x + \sqrt{3x^2 + 2C} \) fulfills the differential equation \( \frac{d y}{d x} = \frac{2 x-y}{x+y} \).To verify, we:1. **Differentiate**: Begin by finding the derivative \( \frac{dy}{dx} \) of the proposed solution.2. **Substitute**: Replace \( y \) in the differential equation with the given function.3. **Simplify and Compare**: Simplify both the derivative and the expression from substitution, and check if they are identical.If both the expressions from differentiation and substitution match, the function is a solution to the differential equation.
Differentiation
Differentiation is the process of finding the derivative of a function, which is essential in solving differential equations. The derivative represents the rate at which a function is changing at any given point. In our exercise, we derived \( \frac{dy}{dx} \) for \( y = -x + \sqrt{3x^2 + 2C} \).Differentiation involves applying rules such as:- **The Power Rule**: Used when differentiating terms like \( x^n \).- **The Sum Rule**: Used for the sum of multiple functions.- **The Chain Rule**: Essential for composite functions, as previously discussed.By applying these rules step by step, we arrive at the derivative: \( \frac{dy}{dx} = -1 + \frac{3x}{\sqrt{3x^2 + 2C}} \). This reflects how each part of the function contributes to the change in \( y \) with respect to \( x \).
Simplification
Simplification is a crucial step in verifying solutions to differential equations. After differentiation and substitution, simplifying expressions allows us to compare terms more easily. For the differential equation and proposed solution given, several simplifications occur:- In Step 2, after substituting \( y = -x + \sqrt{3x^2 + 2C} \) into the differential equation, both the numerator and denominator are separately simplified.- The expression \( 2x - (-x + \sqrt{3x^2 + 2C}) \) simplifies to \( 3x - \sqrt{3x^2 + 2C} \).- Similarly, \( x + (-x + \sqrt{3x^2 + 2C}) \) simplifies to \( \sqrt{3x^2 + 2C} \).These simplifications lead to the form \( \frac{3x - \sqrt{3x^2 + 2C}}{\sqrt{3x^2 + 2C}} \), showing how simplification clarifies mathematical steps.
Calculus
Calculus forms the backbone of differential equations, enabling us to model and solve problems involving rates of change. Key concepts in calculus include differentiation and integration, which help analyze dynamic systems. - **Differentiation**: Central to solving differential equations, it finds how a function changes, as highlighted in the differentiation section above. - **Integration**: Although not directly used here, integration is the reverse process of differentiation, often needed to solve differential equations of that form solutions. Understanding these principles allows us to confirm that a function meets a given differential equation, as in this exercise, and explore diverse mathematical phenomena efficiently. Calculus evolves as a powerful tool for solving complex real-world problems.

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Most popular questions from this chapter

Suppose that \(\alpha\) and \(\beta\) are positive constants. The differential equation $$ P^{\prime}(t)=\alpha \cdot e^{-\beta t} \cdot P(t) $$ for a positive function \(P\) is known as the Gompertz growth equation. (It is named for Benjamin Gompertz \((1779-1865)\), a self-educated scholar of wide- ranging interests.) Find an explicit formula for \(P(t) .\) Use your explicit solution to show that there is a number \(P_{\infty}\) (known as the carrying capacity) such that $$ \lim _{t \rightarrow \infty} P(t)=P_{\infty} $$ Show that the Gompertz growth equation may be written in the form $$ P^{\prime}(t)=k \cdot P(t) \cdot \ln \left(\frac{P_{\infty}}{P(t)}\right) $$

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