Chapter 4: Problem 208
Are the following differential equations linear? Explain your reasoning. $$ \frac{d y}{d x}=x^{2} y+\sin x $$
Short Answer
Expert verified
The differential equation is linear because it matches the standard linear form and \( P(x) \) and \( Q(x) \) are functions of \( x \) only.
Step by step solution
01
Identify the General Form of a Linear Differential Equation
A linear first-order differential equation has the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \) only. Identify these components from the given equation.
02
Rewrite the Given Equation to Compare with the Linear Form
The given equation is \( \frac{dy}{dx} = x^2 y + \sin x \). Rewrite this as \( \frac{dy}{dx} - x^2 y = \sin x \) to match the linear form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \( P(x) = -x^2 \) and \( Q(x) = \sin x \).
03
Check if P(x) and Q(x) Depend on y or Its Derivatives
For the equation to be linear, \( P(x) \) and \( Q(x) \) must not involve \( y \) or its derivatives. In this case, both \( P(x) = -x^2 \) and \( Q(x) = \sin x \) are independent of \( y \).
04
Conclusion: Determine the Linearity of the Equation
The given differential equation \( \frac{dy}{dx} = x^2 y + \sin x \) is linear because it fits the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \) and neither \( P(x) \) nor \( Q(x) \) depend on \( y \).
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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 is a type of equation involving derivatives of a function and the function itself. Specifically, it contains the first derivative, such as \( \frac{dy}{dx} \), but not higher order derivatives. This simplicity focuses on the relationship between a function and its rate of change. When solving these equations, our goal is often to express the function in terms of an independent variable, frequently denoted by \( x \). First-order differential equations take many forms and can be linear or non-linear.
- Linear first-order differential equations follow the standard format: \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) depend only on \( x \).
- Non-linear equations might involve \( y^2, \sin(y) \), or other more complex interactions between \( y \) and its derivatives.
- Understanding the classification—linear or non-linear—helps us decide the best techniques for finding solutions.
Linearity in Differential Equations
Understanding linearity is critical when dealing with differential equations. A differential equation is linear if it can be expressed in the standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \) only.For linear equations:
- The coefficient of \( y \) (i.e., \( P(x) \)) should not involve \( y \) or any functions of \( y \).
- The term \( Q(x) \) should also not be a function of \( y \).
Functions of x
In the context of differential equations, understanding the role of \( x \)-dependent functions is crucial. Functions like \( P(x) \) and \( Q(x) \) in the linear first-order form \( \frac{dy}{dx} + P(x)y = Q(x) \) are solely functions of the independent variable \( x \). Their independence from \( y \) or its derivatives is what allows the equation to be classified as linear.Here's a closer look at these functions:
- \( P(x) \): This function multiplies the dependent variable, \( y \), and influences how quickly \( y \) changes with \( x \). In our example, \( P(x) = -x^2 \), shows that the rate of change scales with the square of \( x \).
- \( Q(x) \): This function represents the non-homogeneous part of the differential equation, determining the equation's driving force independent of \( y \). Here, \( Q(x) = \sin x \), highlighting periodic influences from \( x \).