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Chapter 1: Problem 11

From \(y=e^{-x / 2}\) we obtain \(y^{\prime}=-\frac{1}{2} e^{-x / 2} .\) Then \(2 y^{\prime}+y=-e^{-x / 2}+e^{-x / 2}=0\)

Short Answer

Expert verified
The equation simplifies to 0.

Step by step solution

01

Given Functions

We begin with two functions: the original function, which is given as \( y = e^{-x/2} \), and its derivative, \( y' = -\frac{1}{2} e^{-x/2} \).
02

Substituting the Derivative

Substitute the derivative \( y' = -\frac{1}{2} e^{-x/2} \) into the given equation \( 2y' + y \). This yields \( 2(-\frac{1}{2} e^{-x/2}) + e^{-x/2} \).
03

Simplifying the Expression

Calculate \( 2(-\frac{1}{2} e^{-x/2}) = -e^{-x/2} \). Add this to \( e^{-x/2} \) from the original equation, resulting in \( -e^{-x/2} + e^{-x/2} \).
04

Get the Final Result

Realize that \( -e^{-x/2} + e^{-x/2} = 0 \). Thus, the equation simplifies to 0.

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Key Concepts

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

Exponential Functions
Exponential functions are a type of mathematical expressions where a constant base is raised to a variable exponent. In the given problem, the exponential function is expressed as \( y = e^{-x/2} \), where \( e \) is the base and \( -x/2 \) is the exponent. Exponential functions are important in mathematical modeling because they can easily capture phenomena involving growth or decay.

An exponential function has specific properties:
  • The rate of change is proportional to the value of the function.
  • The graph of an exponential function is a curve that can model rapid increases or decreases.
  • In calculus, they are often used to describe continuous growth or decay processes.
Understanding these properties allows us to manipulate and differentiate such functions, which is a core skill in calculus. For instance, in our exercise, we see the derivative of the exponential function being used in the solution process.
Differentiation
Differentiation is a core concept in calculus used to find the rate of change of a function. In the exercise provided, the differentiation of the function \( y = e^{-x/2} \) results in its derivative \( y' = -\frac{1}{2} e^{-x/2} \). This derivative tells us how the exponential function changes as \( x \) changes.

Key points to understand about differentiation include:
  • Derivatives represent the slope of the tangent line to the graph of the function.
  • For exponential functions of the form \( e^{u(x)} \), the derivative is found using the chain rule: \( d/dx (e^{u(x)}) = e^{u(x)} \, u'(x) \).
  • In our problem, the derivative reflects the rate at which the function \( y = e^{-x/2} \) decreases as \( x \) increases.
By calculating the derivative, we can substitute it back into the original equations to simplify and solve problems, just as shown in the solution steps.
Equation Simplification
Simplifying equations is an essential algebraic technique, enabling us to solve mathematical problems more easily. In the step-by-step solution provided, the simplification involves combining terms to get a final result of zero.

Here is a breakdown of the simplification process used in the exercise:
  • Substitute the value of the derivative \( y' = -\frac{1}{2} e^{-x/2} \) into the expression \( 2y' + y \).
  • Calculate \( 2(-\frac{1}{2} e^{-x/2}) = -e^{-x/2} \) which becomes one of our terms.
  • Add this to the original term \( e^{-x/2} \), resulting in \( -e^{-x/2} + e^{-x/2} \).
  • This simplification showcases the concept of canceling out terms to arrive at \( 0 \).
Effective simplification makes complex algebraic expressions more manageable, allowing students to better focus on the underlying calculus concepts without getting bogged down by complicated algebraic manipulations.

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Most popular questions from this chapter

Setting \(A^{\prime}(t)=-0.002\) and solving \(A^{\prime}(t)=-0.0004332 A(t)\) for \(A(t),\) we obtain $$A(t)=\frac{A^{\prime}(t)}{-0.0004332}=\frac{-0.002}{-0.0004332} \approx 4.6 \text { grams }.$$

The derivatives of the functions are \(\phi_{1}^{\prime}(x)=-x / \sqrt{25-x^{2}}\) and \(\phi_{2}^{\prime}(x)=x / \sqrt{25-x^{2}},\) neither of which is defined at \(x=\pm 5\).

Differentiating \(P=c_{1} e^{t} /\left(1+c_{1} e^{t}\right)\) we obtain \(\begin{aligned} \frac{d P}{d t} &=\frac{\left(1+c_{1} e^{t}\right) c_{1} e^{t}-c_{1} e^{t} \cdot c_{1} e^{t}}{\left(1+c_{1} e^{t}\right)^{2}}=\frac{c_{1} e^{t}}{1+c_{1} e^{t}} \frac{\left[\left(1+c_{1} e^{t}\right)-c_{1} e^{t}\right]}{1+c_{1} e^{t}} \\ &=\frac{c_{1} e^{t}}{1+c_{1} e^{t}}\left[1-\frac{c_{1} e^{t}}{1+c_{1} e^{t}}\right]=P(1-P) \end{aligned}\)

(a) \(\operatorname{since} \frac{d}{d x}\left(-\frac{1}{x+c}\right)=\frac{1}{(x+c)^{2}}=y^{2},\) we see that \(y=-\frac{1}{x+c}\) is a solution of the differential equation. (b) Solving \(y(0)=-1 / c=1\) we obtain \(c=-1\) and \(y=1 /(1-x) .\) Solving \(y(0)=-1 / c=-1\) we obtain \(c=1\) and \(y=-1 /(1+x) .\) Being sure to include \(x=0,\) we see that the interval of existence of \(y=1 /(1-x)\) is \((-\infty, 1),\) while the interval of existence of \(y=-1 /(1+x)\) is \((-1, \infty)\)

The differential equation \(y y^{\prime}-x y=0\) has normal form \(d y / d x=x .\) These are not equivalent because \(y=0\) is a solution of the first differential equation but not a solution of the second.
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