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Magazine Chapter 4: Problem 24
Which of the following equation(s) is/are linear? (a) \(\frac{d y}{d x}+\frac{y}{x}=\ln x\) (b) \(y\left(\frac{d y}{d x}\right)+4 x=0\) (c) \(d x+d y=0\) (d) \(\frac{d^{2} y}{d x^{2}}=\cos x\)
Short Answer
Expert verified
Equations (a), (c), and (d) are linear; equation (b) is not linear.
Step by step solution
01
Understanding Linear Equations
A linear differential equation is of the first order if it can be expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). Higher-order linear differential equations have the form \( a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x) \), where the functions \( a_i(x) \) are continuous and \( g(x) \) is any function of \( x \).
02
Analyzing Equation (a)
Equation (a) is \( \frac{d y}{d x} + \frac{y}{x} = \ln x \). This is a first-order linear differential equation. It is in the standard form of \( \frac{dy}{dx} + P(x)y = Q(x) \) with \( P(x) = \frac{1}{x} \) and \( Q(x) = \ln x \). Thus, equation (a) is a linear equation.
03
Analyzing Equation (b)
Equation (b) is \( y\left(\frac{d y}{d x}\right) + 4x = 0 \). This can be rewritten as \( y \frac{dy}{dx} = -4x \). This equation cannot be rearranged into the form \( \frac{dy}{dx} + P(x)y = Q(x) \) due to the presence of \( y \frac{dy}{dx} \). Therefore, this equation is not linear.
04
Analyzing Equation (c)
Equation (c) is \( dx + dy = 0 \). This can be rewritten as \( \frac{dy}{dx} = -1 \). It is in the form of \( \frac{dy}{dx} = constant \) which is the simplest form of a first-order linear differential equation. Thus, equation (c) is a linear equation.
05
Analyzing Equation (d)
Equation (d) is \( \frac{d^{2} y}{d x^{2}} = \cos x \). This is a second-order differential equation, and it can be expressed in standard linear form as \( y'' - \cos x = 0 \) where \( P(x) = 0 \), and has constant coefficients. Thus, equation (d) is a linear equation.
06
Conclusion on Linear Equations
Equations (a), (c), and (d) are linear equations, while equation (b) is not.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equations
A first-order differential equation involves derivatives of a function with respect to one variable, often written as \( \frac{dy}{dx} \). The hallmark of a first-order **linear** differential equation is that it can be expressed in the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] Here, \( P(x) \) and \( Q(x) \) are functions solely of the independent variable \( x \). This form ensures that derivatives and the unknown function \( y \) itself are not multiplied or divided by one another.
- **Example**: \( \frac{dy}{dx} + \frac{y}{x} = \ln x \) is a first-order linear equation because it's in the correct form.
- The simplicity in the form is what makes solving these equations generally more accessible, using methods like integrating factors.
Higher-Order Linear Differential Equations
When a differential equation involves higher derivatives, such as second or third derivatives of a function, it becomes a higher-order differential equation. A linear form of such an equation can be complex. However, it always takes the general form: \[ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \, ext{...} \, + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x) \] Each term contains coefficients \( a_i(x) \) which are functions of \( x \). For example, in the equation \( \frac{d^{2} y}{d x^{2}} = \cos x \), it translates to: \[ y'' - \cos x = 0 \]
- This is a second-order linear differential equation.
- Even though \( \cos x \) doesn鈥檛 multiply any derivative, it functions as the forcing term \( g(x) \).
Standard Form of Linear Equations
Any linear differential equation must adhere to a predictable form to qualify as a linear equation. The standard form ensures that each occurrence of the unknown function and its derivatives adhere strictly to linearity. For first-order equations, we are familiar with \( \frac{dy}{dx} + P(x)y = Q(x) \). For higher-order equations, the general form is: \[ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \, \ldots \, + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x) \]
- The terms \( y^{(i)} \) cannot involve nonlinear combinations such as products or powers of \( y \).
- Linearity requires that each term is a first-degree term with respect to the dependent variable \( y \) and its derivatives.
Continuous Functions
In the context of linear differential equations, continuous functions, such as \( P(x) \) and \( Q(x) \) in first-order differential equations or \( a_i(x) \) in higher-order ones, play a key role. **Why Continuity Matters:**
- **Existence and Uniqueness:** The solution to a linear differential equation is guaranteed to exist and be unique in some interval if the coefficients involved are continuous.
- **Intermediary Steps:** Many methods for solving these equations, like integrating factors or the method of undetermined coefficients, assume continuity to ensure the operations behave as expected (e.g., integrating over intervals).
