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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) \(y^{\prime \prime}-y=0 ; y=c_{1} e^{t}+c_{2} e^{-t}\)

Short Answer

Expert verified
(a) Order 1; solution confirmed. (b) Order 2; solution confirmed.

Step by step solution

01

Identify the Order of the Differential Equation (a)

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

Identify the Order of the Differential Equation (b)

The given differential equation for part (b) is \(y'' - y = 0\). The highest derivative present in this equation is \(y''\), which is the second derivative. Hence, the order of the differential equation is 2.
03

Verify the Solution for Part (a)

We are given that \(y = c e^{-x/2} + x - 3\). Differentiate this function: \(\frac{d y}{d x} = -\frac{c}{2} e^{-x/2} + 1\). Substitute \(y\) and \(\frac{d y}{d x}\) into the original equation \(2 \frac{d y}{d x} + y = x - 1\): \[2(-\frac{c}{2} e^{-x/2} +1) + (c e^{-x/2} + x - 3) = x - 1\]. Simplifying, the expression is \[(-c e^{-x/2} +2 + c e^{-x/2} + x - 3) = x - 1\], which simplifies to \(x - 1 = x - 1\). Thus, the given family of functions satisfies the differential equation.
04

Verify the Solution for Part (b)

The given solution for part (b) is \(y = c_1 e^t + c_2 e^{-t}\). Compute the derivatives \(y' = c_1 e^t - c_2 e^{-t}\) and \(y'' = c_1 e^t + c_2 e^{-t}\). Substitute \(y\) and \(y''\) into the differential equation \(y'' - y = 0\): \[(c_1 e^t + c_2 e^{-t}) - (c_1 e^t + c_2 e^{-t}) = 0\]. This simplifies to \(0 = 0\), confirming that the function is indeed a solution of the differential equation.

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Key Concepts

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

Order of Differential Equation
The order of a differential equation is determined by its highest derivative term. It essentially tells us how many times we have differentiated a function before forming the equation. This concept is crucial as it indicates the level of complexity involved and helps categorize different kinds of differential equations.

For example, in our exercise, consider part (a) where the equation is given by: \(2 \frac{d y}{d x} + y = x - 1\). The highest derivative present is the first derivative \(\frac{d y}{d x}\). Therefore, this is a first-order differential equation.

Moving to part (b), we have the equation \(y'' - y = 0\). Here, the highest derivative is \(y''\), which is the second derivative. Consequently, this equation is second-order.

In practice:
  • Check the highest derivative in the equation.
  • The number of primes or the order of the differential operator indicates the order.
  • Order reflects the degree of the function's dependency on its derivative.
Verification of Solutions
Verification of solutions involves making sure that a function or set of functions satisfies a differential equation. This eliminates any guesswork by confirming that the proposed solutions fit the equation exactly.

Consider our solved examples. For part (a), the function \(y = c e^{-x / 2} + x - 3\) is proposed as a solution. By differentiating it and substituting back into the original differential equation, we determined that it satisfies the equation as it simplifies to an identity: \(x - 1 = x - 1\).

Similarly, for part (b), the function \(y = c_1 e^t + c_2 e^{-t}\) is tested by substituting its first and second derivatives into the equation \(y'' - y = 0\). Upon substitution, it results in \(0 = 0\), validating it as a real solution.

Key steps for verification:
  • Compute necessary derivatives of the proposed solution.
  • Substitute back into the differential equation.
  • Simplify the equation to see if both sides are equal.
High School Calculus
High school calculus lays the foundational knowledge of derivatives and integrals, which is essential for understanding differential equations. This stage introduces students to the concepts of limits, differentiation, and integration.

Here's how calculus connects to our topic:
  • **Differentiation:** This is the process of finding the derivative, or the rate at which a function is changing at any point. For differential equations, knowing how to differentiate is crucial, since these equations center around the derivatives of functions.
  • **Order and Degree:** Understanding derivatives helps in determining the order of a differential equation, which tells you how many times a function must be differentiated to arrive at the equation.
  • **Verification of Solutions:** Solving and verifying solutions for differential equations uses principles of differentiation to ensure functions meet the original equations.
By solidifying these concepts, students gain the analytical skills necessary to handle more advanced mathematical problems. It's important for students to see calculus not just as a set of rules but as a tool for understanding changes in physical and abstract systems.

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Most popular questions from this chapter

A student objects to Step 2 in the method of separation of variables because one side of the equation is integrated with respect to \(x\) while the other side is integrated with respect to \(y .\) Answer this student's objection. [Hint: Recall the method of integration by substitution.]

Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate five integral curves for the equation. $$y^{\prime}=\frac{y}{1+y^{2}}$$

(a) Suppose that a quantity \(y=y(t)\) changes in such a way that \(d y / d t=k \sqrt{y},\) where \(k>0 .\) Describe how \(y\) changes in words. (b) Suppose that a quantity \(y=y(t)\) changes in such a way that \(d y / d t=-k y^{3},\) where \(k>0 .\) Describe how \(y\) changes in words.

Determine whether the statement is true or false. Explain your answer. The equation $$ \left(\frac{d y}{d x}\right)^{2}=\frac{d y}{d x}+2 y $$ is an example of a second-order differential equation.

(a) There is a trick, called the Rule of 70 , that can be used to get a quick estimate of the doubling time or half-life of an exponential model. According to this rule, the doubling time or half-life is roughly 70 divided by the percentage growth or decay rate. For example, we showed in Example 5 that with a continued growth rate of \(1.08 \%\) per year the world population would double every 64 years. This result agrees with the Rule of \(70,\) since \(70 / 1.08 \approx 64.8\) Explain why this rule works. (b) Use the Rule of 70 to estimate the doubling time of a population that grows exponentially at a rate of \(1 \%\) per year. (c) Use the Rule of 70 to estimate the half-life of a population that decreases exponentially at a rate of \(3.5 \%\) per hour. (d) Use the Rule of 70 to estimate the growth rate that would be required for a population growing exponentially to double every 10 years.

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