Problems 15 through 24 deal with ODEs and techniques discussed in Section \(4.8 .\) Classify the following ordinary differential equations by determining whether they are linear, what their order is, whether they are homogeneous, and whether their coefficients are constant. (a) \((\sin x) y^{\prime \prime}+\cos x=0\). (f) \(\frac{d y}{d t}=\frac{1}{1+y}\) (b) \(y^{\prime \prime}+y^{2}=2 y^{\prime}\). (g) \(\frac{d y}{d x}=\frac{1}{1+x}\). (c) \(\frac{d^{3} y}{d t^{3}}+\frac{2 d y}{d t}=\sin y\). (h) \(\frac{d^{5} y}{d x^{5}}=x^{6}+5 x+6\) (d) \(\frac{d}{d t}\left(y^{2}+2 y\right)=y\). (i) \(t \frac{d y}{d t}+t y=1\). (e) \(\frac{d^{2} y}{d t^{2}}+2 \frac{d y}{d t}+3 y=e^{t}+e^{-t}\).

Short Answer

Expert verified
Each equation's classification is provided, noting their order, linearity, homogeneity, and if coefficients are constant or variable.

Step by step solution

01

Classify Equation (a)

The given equation is $$(\text{sin} x) y'' + \text{cos} x = 0$$First, identify its order: the highest derivative is \(y''\), making it a second-order equation. It’s nonlinear because it doesn’t fit the form of a linear ODE. The equation is homogeneous since the right-hand side equals zero, and it has variable coefficients with \text{sin} and \text{cos} as functions of \(x\).
02

Classify Equation (b)

The given equation is \(y'' + y^2 = 2y'\) Its order is determined by the highest derivative which is \(y''\), making it a second-order equation. The presence of the term \(y^2\) indicates it's nonlinear. It’s not homogeneous since the right-hand side isn't zero. This equation has constant coefficients in terms of the derivatives.
03

Classify Equation (c)

The given equation is $$\frac{d^3 y}{d t^3} + 2 \frac{d y}{d t} = \text{sin} y$$ The order is determined by the highest derivative, which is \(d^3 y/dt^3\), making it a third-order equation. It’s nonlinear because of the \text{sin} \(y\) term. It’s inhomogeneous since the right-hand side is not zero. This equation has constant coefficients in terms of the derivatives.
04

Classify Equation (d)

The given equation is $$\frac{d}{d t}(y^2 + 2y) = y$$ Expanding it gives $$2y y' + 2y' = y.$$ The highest derivative is \(y'\), making it a first-order equation. It's nonlinear due to the \(y^2\) term. It’s inhomogeneous because of the non-zero right-hand side, with constant coefficients.
05

Classify Equation (e)

The given equation is $$\frac{d^2 y}{d t^2} + 2 \frac{d y}{d t} + 3y = e^t + e^{-t}$$The highest derivative is \(d^2 y/dt^2\), making it a second-order equation. It’s linear because it fits the standard form of a linear ODE. It’s inhomogeneous since the right-hand side is not zero, with constant coefficients.
06

Classify Equation (f)

The given equation is $$\frac{d y}{d t} = \frac{1}{1+y}$$ The highest derivative is \(dy/dt\), making it a first-order equation. It’s nonlinear due to the fraction involving \(y\). It’s inhomogeneous with variable coefficients as a function of \(y\).
07

Classify Equation (g)

The given equation is $$\frac{d y}{d x} = \frac{1}{1+x}$$ The highest derivative is \(dy/dx\), making it a first-order equation. It’s linear because it can be written in the form of a linear ODE. It’s inhomogeneous as the right-hand side is not zero, with variable coefficients as a function of \(x\).
08

Classify Equation (h)

The given equation is $$\frac{d^5 y}{d x^5} = x^6 + 5x + 6$$ The highest derivative is \(d^5 y/dx^5\), making it a fifth-order equation. It’s linear because it fits the definition of a linear ODE. It’s inhomogeneous since the right-hand side is not zero, with variable coefficients as a function of \(x\).
09

Classify Equation (i)

The given equation is $$t \frac{d y}{d t} + ty = 1$$ The highest derivative is \(dy/dt\), making it a first-order equation. It’s linear because it can be written in the form of a linear ODE. It’s inhomogeneous with variable coefficients as a function of \(t\).

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Key Concepts

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

Linear vs Nonlinear ODE
Ordinary Differential Equations (ODEs) can be classified into linear and nonlinear based on the structure of the equation. If all the terms involving the dependent variable and its derivatives are linear, then the ODE is linear. This means that the dependent variable and its derivatives appear only to the first power and are not multiplied or divided by each other.
For example, the equation \(\frac{d^2 y}{d t^2} + 2 \frac{d y}{d t} + 3y = e^t + e^{-t}\) is linear because it fits the form y'' + a(t)y' + b(t)y = g(t), where a(t) and b(t) are functions of t, and g(t) is a nonzero function, making it inhomogeneous but still linear.

In contrast, if the dependent variable or its derivatives appear to any power other than one or if they are multiplied together, the ODE is nonlinear. For instance, the equation \y'' + y^2 = 2y'\ is nonlinear due to the y^2 term. Such nonlinearities often make solving the ODEs more complex.
Order of ODE
The order of an ordinary differential equation is the highest derivative of the dependent variable that appears in the equation. Identifying the order is crucial as it determines the formulation and methods used for solving the ODE.
For example, the equation \(\frac{d^3 y}{d t^3} + 2 \frac{d y}{d t} = \text{sin} y\) is a third-order ODE because the highest derivative here is \( \frac{d^3 y}{d t^3}\).

First-order ODEs, like \( t \frac{d y}{d t} + t y = 1 \), involve only the first derivative \( \frac{d y}{d t} \). Second-order ODEs, such as \( (\text{sin} x) y'' + \text{cos} x = 0 \), include the second derivative \( y'' \). By classifying the order, we can apply the correct method to find the solution.
Homogeneous vs Inhomogeneous ODE
An ODE is homogeneous if every term is a function of the dependent variable and its derivatives, and the equation equals zero on the right-hand side. For example, \( (\text{sin} x) y'' + \text{cos} x = 0 \) is a homogeneous ODE because every term involves the dependent variable or its derivatives, and the right-hand side is zero.
Homogeneous ODEs can be simpler to solve as they often allow for eigenvalue methods and other proportional techniques.

In contrast, an inhomogeneous ODE has a nonzero right-hand side, typically involving terms that do not depend on the dependent variable or its derivatives. For example, \( y'' + 2y' + 3y = e^t + e^{-t} \) is inhomogeneous due to the \( e^t + e^{-t} \) term. Solving inhomogeneous ODEs often requires finding a particular solution to the nonhomogeneous part followed by solving the associated homogeneous equation.
Constant vs Variable Coefficients
Classification of an ODE can also depend on whether the coefficients of the terms involving the dependent variable and its derivatives are constant or variable.
**Constant Coefficients:** An ODE has constant coefficients if the coefficients are constants (numbers). For instance, \( y'' + 2y' + 3y = e^t + e^{-t} \) has constant coefficients (2 and 3). These ODEs are often simpler to solve, with solutions involving exponential functions, sine, and cosine functions.

**Variable Coefficients:** An ODE has variable coefficients if the coefficients are functions of the independent variable. For example, \( (\text{sin} x) y'' + \text{cos} x = 0 \) includes the coefficients \( \text{sin} x \) and \( \text{cos} x \), which are functions of x. Variable coefficient ODEs are typically more complex, and their solutions might require special methods or approximations, such as series solutions or numerical methods.

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Most popular questions from this chapter

In this question we deal with a number of variants of the chemostat. (a) How would you expect the model to differ if there were two growth-limiting nutrients? (b) Suppose that at high densities bacteria start secreting a chemical that inhibits their own growth. How would you model this situation? (c) In certain cases two (or more) bacterial species are kept in the same chemostat and compete for a common nutrient. Suggest a model for such competition experiments.

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