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Chapter 7: Problem 2

Show that \(y=C e^{-2 x}+e^{x}\) is a solution of the differential equation \(y^{\prime}+2 y=3 e^{x}\) on \((-\infty, \infty)\).

Short Answer

Expert verified
We found the derivative of the given function, \(y'(x) = -2Ce^{-2x} + e^x\). After substituting \(y(x)\) and \(y'(x)\) into the differential equation, we simplified the equation and arrived at \(3e^x = 3e^x\), which holds for all values of \(x\) on \((-\infty, \infty)\). Thus, \(y(x) = Ce^{-2x} + e^x\) is a solution to the given differential equation \(y'(x) + 2y(x) = 3e^x\).

Step by step solution

01

Compute the derivative y'(x)

First, we will compute the \(y'(x)\), the derivative of the function \(y(x) = Ce^{-2x} + e^x\): Differentiation: \[y'(x) = \frac{d}{dx}[Ce^{-2x} + e^x]\] Now, apply the chain rule: \[y'(x) = C \cdot \frac{d}{dx}[e^{-2x}] + \frac{d}{dx}[e^x]\] Differentiate each term: \[y'(x) = -2Ce^{-2x} + e^x\] Now we have the derivative, we can proceed to check the given differential equation.
02

Substitute y(x) and y'(x) into the differential equation

Now, we will substitute the expressions for \(y(x)\) and \(y'(x)\) that we found in Step 1 into the given differential equation \(y'(x) + 2y(x) = 3e^x\): \[(-2Ce^{-2x} + e^x) + 2(Ce^{-2x} + e^x) = 3e^x\] Now, simplify the equation: \[(-2Ce^{-2x} + e^x) + 2Ce^{-2x} + 2e^x = 3e^x\] \[3e^x = 3e^x\]
03

Verify if the equation holds for all x

As we can see, the simplified equation holds for all values of \(x\) in the interval \((-\infty, \infty)\): \[3e^x = 3e^x\] Therefore, the function \(y(x) = Ce^{-2x} + e^x\) is indeed a solution of the given differential equation \(y'(x) + 2y(x) = 3e^x\) on the interval \((-\infty, \infty)\).

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Key Concepts

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

Chain Rule
When we tackle derivatives, the chain rule is a crucial tool that allows us to differentiate compositions of functions. It is used when a function is made up of one function inside of another, such as f(g(x)). The chain rule states that the derivative of f(g(x)) with respect to x is f'(g(x))*g'(x).

In the context of our example with the function y = Ce^{-2x} + e^x, we apply the chain rule to differentiate the term Ce^{-2x}. This term is composed of the functions f(u) = Ce^u where u = -2x is itself a function of x. So the derivative of u with respect to x is u' = -2, and we multiply this by the derivative of Ce^u with respect to u, which is simply Ce^u. Thus, the derivative with respect to x is -2Ce^{-2x}.
Exponential Functions
Exponential functions are a class of mathematical functions of the form f(x) = a^x, where a is a positive constant and x is the exponent. When the base a is the mathematical constant e (approximately equal to 2.71828), the function takes the special form of f(x) = e^x, which has unique properties and is often encountered in calculus and the natural sciences.

The derivative of f(x) = e^x is remarkable because it maintains its form; in other words, (e^x)' is again e^x. This property makes exponential functions particularly easy to differentiate and also essential in solving many differential equations. This is explicitly seen in our exercise where the term e^x is simply differentiated to itself, which significantly streamlines the process of solving the equation.
Derivative of Functions
The derivative represents the rate at which a function is changing at any given point and is a fundamental concept in calculus. For any function y = f(x), the derivative y'(x) or df/dx gives the slope of the tangent line to the curve of the function at any point x. This is crucial for understanding motion, growth, and change.

In the exercise, we find the derivative y' of the function y = Ce^{-2x} + e^x by first applying the chain rule and the properties of exponential functions, resulting in y'(x) = -2Ce^{-2x} + e^x. This derivative then allows us to substitute into the differential equation y' + 2y = 3e^x to verify that the function y is indeed a solution. Understanding how to take derivatives of various types of functions is key to analyzing and solving problems across many scientific and mathematical areas.

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Most popular questions from this chapter

Von Bertalanffy Growth Model The von Bertalanffy growth model is used to predict the length of commercial fish. The model is described by the differential equation $$ \frac{d x}{d t}=k(L-x) $$ where \(x(t)\) is the length of the fish at time \(t, k\) is a positive constant called the von Bertalanffy growth rate, and \(L\) is the maximum length of the fish. a. Find \(x(t)\) given that the length of the fish at \(t=0\) is \(x_{0}\). b. At the time the larvae hatch, the North Sea haddock are about \(0.4 \mathrm{~cm}\) long, and the average haddock grows to a length of \(10 \mathrm{~cm}\) after 1 year. Find an expression for the length of the North Sea haddock at time \(t\). c. Plot the graph of \(x\). Take \(L=100(\mathrm{~cm})\). d. On average, the haddock that are caught today are between \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\) long. What are the ages of the haddock that are caught?

The change from religious to lay teachers at Roman Catholic schools has been partly attributed to the decline in the number of women and men entering religious orders. The percentage of teachers who are lay teachers is given by $$ f(t)=\frac{98}{1+2.77 e^{-t}} \quad 0 \leq t \leq 4 $$ where \(t\) is measured in decades, with \(t=0\) corresponding to the beginning of 1960 . a. What percentage of teachers were lay teachers at the beginning of 1990 ? b. Find the year when the percentage of lay teachers was increasing most rapidly.

Solve the differential equation. $$ \frac{d y}{d x}+y \cot x=\cos x $$

In the Lotka-Volterra model it was assumed that an unlimited amount of food was available to the prey. In a situation in which there is a finite amount of natural resources available to the prey, the Lotka-Volterra model can be modified to reflect this situation. Consider the following system of differential equations: $$ \begin{array}{l} \frac{d x}{d t}=k x\left(1-\frac{x}{L}\right)-a x y \\ \frac{d y}{d t}=-b y+c x y \end{array} $$ where \(x(t)\) and \(y(t)\) represent the populations of prey and predators, respectively, and \(a, b, c, k\), and \(L\) are positive constants. a. Describe what happens to the prey population in the absence of predators. b. Describe what happens to the predator population in the absence of prey. c. Find all the equilibrium points and explain their significance.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If \(y_{1}\) is a solution of the homogeneous equation \(y^{\prime}+P y=0\) associated with the nonhomogeneous equation \(y^{\prime}+P y=f\) and \(y_{2}\) is a solution of the nonhomogeneous equation, then \(y=c y_{1}+y_{2}\) is a solution of the nonhomogeneous equation, where \(c\) is any constant.
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