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State the order of the differential equation, and confirm that the functions in the given family are solutions. (a) \(2 \frac{d y}{d x}+y=x-1 ; y=c e^{-x / 2}+x-3\) (b) \(y^{\prime \prime}-y=0 ; \quad y=c_{1} e^{t}+c_{2} e^{-t}\)

Short Answer

Expert verified
(a) Order: 1, Confirmed; (b) Order: 2, Confirmed.

Step by step solution

01

Identify the Order of the Differential Equation (a)

The differential equation given is \(2 \frac{d y}{d x} + y = x - 1\). The highest derivative in this equation is \(\frac{d y}{d x}\), which is the first derivative, indicating that the order of the differential equation is 1.
02

Verify the Solution for Differential Equation (a)

To check if the function \(y = c e^{-x / 2} + x - 3\) is a solution to the differential equation, substitute \(y\) and its derivative \(\frac{d y}{d x}\) into the equation:First, find the derivative: \(\frac{d y}{d x} = -\frac{c}{2} e^{-x / 2} + 1\).Substitute into the equation:\[2 \left(-\frac{c}{2} e^{-x / 2} + 1 \right) + \left(c e^{-x / 2} + x - 3 \right) = x - 1\]Simplify the left-hand side:\[-c e^{-x / 2} + 2 + c e^{-x / 2} + x - 3 = x - 1\]This simplifies to:\[x - 1 = x - 1\]The original equation holds true, confirming that \(y = c e^{-x / 2} + x - 3\) is a solution.
03

Identify the Order of the Differential Equation (b)

The differential equation given is \(y'' - y = 0\). The highest derivative in this equation is the second derivative \(y''\), which indicates that the order of the differential equation is 2.
04

Verify the Solution for Differential Equation (b)

To verify if \(y = c_1 e^t + c_2 e^{-t}\) satisfies the differential equation \(y'' - y = 0\), first find the derivatives:\(y' = c_1 e^t - c_2 e^{-t}\)\(y'' = c_1 e^t + c_2 e^{-t}\)Substitute \(y\) and \(y''\) into the differential equation:\[c_1 e^t + c_2 e^{-t} - (c_1 e^t + c_2 e^{-t}) = 0\]Simplifying yields:\[0 = 0\]The equation holds, confirming that \(y = c_1 e^t + c_2 e^{-t}\) is a solution.

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Key Concepts

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

Ordinary Differential Equations
Differential equations play a crucial role in expressing the relationship between functions and their derivatives. An ordinary differential equation (ODE) involves derivatives of a function with respect to only one variable. They appear quite frequently in science and engineering to model physical and natural phenomena. ODEs can take on many forms, but they generally include a function, its derivatives, and an independent variable.
  • The order of an ODE is defined by the highest order of derivative present in the equation.
  • Solutions to ODEs can involve constants; these constants are usually determined by initial conditions or boundary conditions.
To grasp this concept, consider an ODE like "\(2 \frac{d y}{d x}+y=x-1\)". Here, the equation involves the first derivative, indicating it's a first-order differential equation. Meanwhile, the equation "\(y'' - y = 0\)" involves a second derivative which makes it a second-order equation. Understanding the order is key as it helps predict the behavior and the complexity of the solutions.
Solution Verification
Once an ODE is established, the next step is to verify potential solutions. This involves substituting the proposed solution back into the original differential equation to check whether it holds true. It’s quite like solving an algebra problem where you check your solution by inserting it back into the equation.
  • For the equation "\(2 \frac{d y}{d x} + y = x - 1\)", the proposed solution is "\(y = c e^{-x / 2} + x - 3\)". To verify, determine \(\frac{d y}{d x}\) and substitute back into the equation.
  • Simplify the expression carefully. All terms should cancel out correctly to give a true statement such as "\(x - 1 = x - 1\)".
  • This approach ensures that the functional form of the solution is a correct match for the differential equation.
Verification is crucial for ensuring that solutions not only theoretically satisfy the differential equation but also meet practical conditions, particularly for applications in the real world.
Order of Differential Equation
Understanding the order of a differential equation is imperative as it describes the highest derivative in the equation. The order directly influences the nature of the solution and the methods required to solve it.
  • A first-order differential equation may be easier to solve and often results in exponential solutions, as seen in scenarios involving simple exponential growth or decay.
  • Second-order differential equations are commonly found in systems exhibiting oscillatory behavior such as springs and circuits. They often require different techniques like the method of undetermined coefficients or variation of parameters to solve.
  • The further increase in order generally implies more complex systems and solutions. It might necessitate the application of advanced techniques, such as Laplace transforms or series solutions, particularly in engineering and physics applications.
By identifying and understanding the order, students can better select appropriate methods and interpret solutions in context, enhancing both their theoretical and practical understanding of the differential equation.

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Most popular questions from this chapter

A scientist wants to determine the half-life of a certain radioactive substance. She determines that in exactly 5 days a \(10.0\) -milligram sample of the substance decays to \(3.5\) milligrams. Based on these data, what is the half-life?

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A slope field of the form \(y^{\prime}=f(y)\) is said to be \(a u\) tonomous. (a) Explain why the tangent segments along any horizontal line will be parallel for an autonomous slope field. (b) The word autonomous means "independent." In what sense is an autonomous slope field independent? (c) Suppose that \(G(y)\) is an antiderivative of \(1 /[f(y)]\) and that \(C\) is a constant. Explain why any differentiable function defined implicitly by \(G(y)-x=C\) will be a solution to the equation \(y^{\prime}=f(y)\).

A rocket, fired upward from rest at time \(t=0\), has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k\), the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t\). It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation $$ m \frac{d v}{d t}=c k-m g $$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t\). (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in \(100 \mathrm{~s}\). Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) and \(\left.c=2500 \mathrm{~m} / \mathrm{s} .\right]\)

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