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In Exercises \(3-6,\) determine whether the differential equation is linear. Explain your reasoning. $$2 x y-y^{\prime} \ln x=y$$

Short Answer

Expert verified
The differential equation \(2xy - y'\ln x = y\) is not linear.

Step by step solution

01

Identify the differential equation

The differential equation given is \(2xy - y'\ln x = y\).
02

Check for linearity

Linearity of a differential equation means the equation is a polynomial in the dependent variable and its derivatives, where the power is 1, coefficients are constants and the term involving dependent variable is separable from terms involving its derivatives. Looking at the given differential equation, one can see that it doesn't follow these conditions. The variable factor \(\ln x\) which multiplies with the derivative of y breaks the linearity rules of constant coefficients and separable terms.
03

Conclusion

The given differential equation is not linear because it doesn't fulfill the properties required to be considered as linear according to the definition.

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Key Concepts

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

Linearity
In the context of differential equations, linearity is a fundamental property that helps determine the equation's complexity and the methods used for its solution. A linear differential equation has the following characteristics:
  • The dependent variable and its derivatives appear only to the first power. This means there are no squares, cubes, or any other powers of the dependent variable or its derivatives.
  • The coefficients of the dependent variable and its derivatives are constants or functions only of the independent variable, not the dependent variable.
  • The dependent variable and its derivatives are not multiplied together.

In the equation from the exercise, we see a term \(y' \ln x\). Here, \(\ln x\) is a variable factor and thus violates the condition of having constant coefficients, indicating the equation is not linear.
Dependent Variable
In differential equations, the dependent variable is the unknown function we are trying to solve for. It is called ‘dependent’ because its value depends on the independent variable(s). For example, in most cases, if \(y\) is the dependent variable, it will be expressed as a function of \(x\), the independent variable, like \(y = f(x)\).

A key point in linear differential equations is that the dependent variable and all its derivatives appear in the first degree only. This impacts how we evaluate linearity, as higher powers of the dependent variable or its derivatives would mean the equation is non-linear. In our exercise, \(y\) is the dependent variable, and although it appears to the first power, the presence of variable coefficients makes the equation non-linear.
Derivatives
Derivatives represent the rate of change of the dependent variable concerning the independent variable. They are central to differential equations, as these equations describe relationships involving quantities and their rates of change.

In a linear differential equation, any derivative of the dependent variable is linear (first degree) and not raised to any power other than one. This means terms like \(y'\ln{x}\) are problematic when determining linearity because the differential \(y'\) is multiplied by a variable term, \(\ln{x}\). This violates the consistency needed for linear derivatives, thereby contributing to non-linearity in the given problem.
Coefficients
Coefficients in differential equations are the values or functions that multiply the dependent variable and its derivatives. For a differential equation to be linear, these coefficients should either be constants or functions of the independent variable only.

Having constant coefficients simplifies the equation and aids in straightforward analytical solutions. However, when coefficients depend on variables other than the independent one, like \(\ln x\) in our equation, it disrupts the linearity. It forces terms involving derivatives to behave non-linearly, complicating the solution process. Hence, understanding the nature and role of coefficients is critical when analyzing the linearity of differential equations.

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Most popular questions from this chapter

In Exercises 65 and \(66,\) determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. \(y^{\prime}+x \sqrt{y}=x^{2}\) is a first-order linear differential equation.

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Investment Growth A large corporation starts at time \(t=0\) to invest part of its receipts continuously at a rate of \(P\) dollars per year in a fund for future corporate expansion. Assume that the fund earns \(r\) percent interest per year compounded continuously. So, the rate of growth of the amount \(A\) in the fund is given by \(d A / d t=r A+P,\) where \(A=0\) when \(t=0 .\) Solve this differential equation for \(A\) as a function of \(t .\)

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