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Chapter 1: Problem 1

In Problems 1-10 state whether the given differential equation is linear or nonlinear. Give the order of each equation. 1\. \((1-x) y^{\prime \prime}-4 x y^{\prime}+5 y=\cos x\)

Short Answer

Expert verified
The equation is linear and of order 2.

Step by step solution

01

Identify the Type of Differential Equation

Look at the given differential equation \((1-x) y'' - 4x y' + 5y = \cos x\). A differential equation is linear if it can be written in the form \(a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + ... + a_1(x) y' + a_0(x) y = g(x)\). In our case, each term involving \(y\), \(y'\), or \(y''\) has coefficients that are functions of \(x\), i.e., \(1-x\), \(-4x\), and \(5\). The equation is linear because it fits this form and there are no terms like \(y^2\) or \((y')^2\) which would indicate nonlinearity.
02

Determine the Order of the Differential Equation

The order of a differential equation is the highest derivative that appears in the equation. In the equation \((1-x) y'' - 4x y' + 5y = \cos x\), the highest derivative present is \(y''\), which is the second derivative. Therefore, the order of the differential equation is 2.

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Key Concepts

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

Linear Differential Equation
A linear differential equation is a type of equation involving derivatives of a function that adds up terms linearly. To identify one, you check whether the equation can be put into a standard linear form:
  • \[a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \, \ldots \, + a_1(x) y' + a_0(x) y = g(x)\]
Here, the coefficients \(a_n(x)\) are functions of \(x\), and \(g(x)\) is the non-homogeneous part of the equation. If there are no nonlinear terms such as \(y^2\) or \((y')^2\), then the equation is linear.
For instance, in the equation \((1-x) y'' - 4x y' + 5y = \cos x\), every term involving \(y\) or its derivatives has a coefficient that is only dependent on \(x\). Thus, it classifies as linear.
Order of a Differential Equation
The order of a differential equation is determined by the highest derivative present within the equation. The prominence of this concept is crucial as it provides insight into the equation's complexity and the possible methods of solution.
To find the order, simply look at each derivative term in the equation.
  • For instance, consider the equation \((1-x) y'' - 4x y' + 5y = \cos x\).
    The terms are \(y''\), which is the second derivative of \(y\), \(y'\), and \(y\) without a derivative. The highest among these derivatives is \(y''\).

  • This indicates that the equation is of the second order, making it more complex compared to first-order differential equations.
Second-Order Differential Equation
Second-order differential equations involve derivatives of the second degree as their highest order of derivative. These equations frequently appear in various fields like engineering, physics, and more due to their utility in modeling a wide range of systems.
When solving a second-order differential equation, typical tasks include looking for a solution that satisfies the equation based on initial conditions that can relate to physical scenarios such as initial position or velocity in mechanical problems.
  • They often take the form of \(a(x) y'' + b(x) y' + c(x) y = g(x)\) where \(y''\) suggests it is of the second order.

  • In our equation \((1-x) y'' - 4x y' + 5y = \cos x\), the presence of \(y''\) confirms that it is indeed second-order.
Understanding such equations is key to analyzing dynamic systems.

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Most popular questions from this chapter

Theorem \(1.1\) guarantees that there is only one solution of the differential equation \(y^{\prime}=3 y^{43} \cos x\) passing through any specified point \(\left(x_{0}, y_{0}\right)\) in the \(x y\)-plane. The interval of existence of a solution depends on the initial condition \(y\left(x_{0}\right)=y_{0}\). Use the one-parameter family of solutions \(y=1 /(c-\sin x)^{3}\) to find a solution satisfying \(y(x)=1 / 8\). Find a solution satisfying \(y(\pi)=8\). Use these two solutions as a basis for discussing the following questions. From the given family of solutions, when do you think the interval of existence of the initialvalue problem is a finite interval? An infinite interval?

Determine a region of the \(x y\)-plane for which the given differential equation would have a unique solution through a point \(\left(x_{0}, y_{0}\right)\) in the region. $$ \frac{d y}{d x}=x^{3} \cos y $$

\(x^{2} y^{\prime \prime}-3 x y^{4}+4 y=0 ; \quad y=x^{2}+x^{2} \ln x, x>0\)

After a mass \(m\) is attached to a spring, it stretches the spring \(s\) units and then hangs at rest in the equilibrium position, as shown in Figure \(1.19(b)\). After the spring/mass system has been set in motion, let \(x(t)\) denote the directed distance of the mass beyond the equilibrium position. Assume that the downward direction is positive and that the motion takes place in a vertical straight line through the center of gravity of the mass. Assume, too, that the only forces acting on the system are the weight \(m g\) of the mass and the restoring force of the stretched spring, which, by Hooke's law, is proportional to its total elongation. Determine a differential equation for the displacement \(x(t)\) for any time \(t\).

There exists an interval centered at 2 on which the unique solution of the initial-value problem \(y^{\prime}=(y-1)^{3}, y(2)=1\) is \(y=1\).
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