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The differential equation \(y^{\prime}=3 x^{2} y-\cos (x) y^{\prime \prime}\) is linear.

Short Answer

Expert verified
Yes, the differential equation is linear.

Step by step solution

01

Understanding Linear Differential Equations

A linear differential equation is one where the dependent variable and its derivatives appear to the power of one, and no products of the dependent variable and its derivatives occur. Additionally, a linear equation must not have powers or functions of dependent variable without being multiplied by independent variable terms.
02

Analyzing Given Equation

The given equation is \( y^{\prime} = 3x^{2}y - \cos(x)y^{\prime\prime} \). We need to check whether the equation is linear with respect to the dependent variable \( y \) and its derivatives.
03

Identifying Terms and Checking Linearity

The term \( 3x^{2}y \) is linear in \( y \) because \( y \) appears to the first power and is multiplied by a function of \( x \) only. The term \( \cos(x)y^{\prime\prime} \) involves the second derivative of \( y \), but it is multiplied by a function of \( x \) only, which aligns with linearity conditions of higher-order derivatives.
04

Final Verification

Each derivative of \( y \) in the differential equation is multiplied only by expressions involving \( x \) (whether constants or functions of \( x \)). Since no products of derivatives or powers other than one are present, the equation satisfies the conditions of linearity.

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Key Concepts

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

First-order Derivatives
First-order derivatives are fundamental in the study of differential equations. They refer to the rate at which a dependent variable, typically denoted as \( y \), changes with respect to an independent variable, often represented as \( x \). Simply put, the first-order derivative of \( y \) with respect to \( x \), is the slope of the tangent line to the curve at any point on the graph of \( y \).

In mathematical notation, the first-order derivative is represented by \( y' \) or \( \frac{dy}{dx} \). This value depicts how swiftly or slowly \( y \) is changing as \( x \) varies. Understanding first-order derivatives is crucial because they appear in almost all differential equations, whether linear or non-linear, describing a wide range of phenomena, from motion dynamics to population growth.
  • The expression \( y' = 3x^2 y \) in the original equation demonstrates a first-order derivative.
  • This represents how the function \( y \) changes in response to \( x \), scaled by \( 3x^2 \).

The first-order derivative is essential in verifying the linearity of an equation since it needs to appear to the power of one. In the given problem, this aspect is preserved, ensuring an essential property of linear differential equations.
Second-order Derivatives
Second-order derivatives explore the change that occurs to the rate of change of a function. In other words, they measure how the first derivative itself changes as the independent variable varies. This can be visualized as the curvature or concavity of a graph at a point, indicating whether the graph is bending upwards or downwards.

In formulaic terms, the second-order derivative of a function \( y \) with respect to \( x \) is denoted by \( y'' \) or \( \frac{d^2y}{dx^2} \). These derivatives gain importance in solving higher-order differential equations where the behavior of a system is influenced not just by its current state but also by how the state is changing.
  • The term \( -\cos(x)y'' \) in the given equation represents a second-order derivative.
  • This expresses how the curvature of \( y \) is affected by the function \( \cos(x) \), contingent only on \( x \).

Second-order derivatives must also adhere to the linearity conditions — they appear with no powers other than one and without multiplicative terms from other derivatives. The original problem meets these conditions, which is a key reason the equation is classified as linear.
Linearity Conditions
Linearity conditions are integral for characterizing a differential equation as linear. For a differential equation to be linear, each term involving the dependent variable \( y \) or its derivatives needs to meet specific criteria:
  • Each derivative must appear to the power of one, not squared or multiplied by another derivative.
  • The dependent variable \( y \) and its derivatives should only be multiplied by functions or constants of \( x \), the independent variable.
  • No terms should combine different derivatives, such as products like \( y \cdot y' \) or higher powers like \( (y')^2 \).

In the provided differential equation, \( y' = 3x^2 y - \cos(x)y'' \), the conditions of linearity are thoroughly satisfied. Each derivative is appropriately managed:
  • \( y' \) is multiplied only by a function of \( x \), adhering to the linearity criteria.
  • \( y'' \) likewise fits the linearity condition because it is also multiplied explicitly by a function of \( x \), \( \cos(x) \).
  • There are no mixed or squared terms of the derivatives, aligning with the linear equation principles.

Understanding and verifying these conditions ensure that the equation behaves predictably within systems it models. This aspect allows us to apply certain mathematical techniques and solutions patterns unique to linear differential equations. In this particular exercise, recognizing the linearity properties confirms the equation's qualification as a linear equation.

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