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Chapter 2: Problem 32

For \(y^{\prime}+y=f(x)\) an integrating factor is \(e^{x}\) so that $$y e^{x}=\left\\{\begin{array}{ll} e^{x}+c_{1}, & 0 \leq x \leq 1 \\ -e^{x}+c_{2}, & x>1 \end{array}\right.$$ If \(y(0)=1\) then \(c_{1}=0\) and for continuity we must have \(c_{2}=2 e\) so that $$y=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq 1 \\ 2 e^{1-x}-1, & x>1 \end{array}\right.$$

Short Answer

Expert verified
The piecewise solution for the differential equation is \( y=1 \) for \( 0\leq x \leq 1 \) and \( y=2e^{1-x}-1 \) for \( x>1 \).

Step by step solution

01

Identify the Differential Equation

The given differential equation is \( y^{\prime} + y = f(x) \). We are tasked to solve for \( y \) using the integrating factor method.
02

Determine the Integrating Factor

We are given that the integrating factor is \( e^{x} \). The purpose of using the integrating factor is to make the left-hand side of the differential equation into the derivative of a product. This helps us integrate both sides easily.
03

Multiply Equation by the Integrating Factor

Multiply each term of the differential equation \( y^{\prime} + y = f(x) \) by the integrating factor \( e^x \). This transforms the equation into \( e^x y^{\prime} + e^x y = e^x f(x) \).
04

Express Left-Hand Side as a Single Derivative

The left-hand side is now \( \frac{d}{dx}(e^x y) \), since \( \frac{d}{dx}(e^x y) = e^x y^{\prime} + e^x y \). Hence, the equation becomes \( \frac{d}{dx}(e^x y) = e^x f(x) \).
05

Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^x y) \, dx = \int e^x f(x) \, dx \]The left-hand side integration results in \( e^x y \), and the solution depends on the form \( \int e^x f(x) \, dx \), provided in the piecewise defined equation.
06

Use Initial Condition \( y(0)=1 \)

For \( 0 \leq x \leq 1 \), the equation is \( e^x y = e^x + c_1 \). Given \( y(0) = 1 \), substitute into the equation: \( e^0 \cdot 1 = e^0 + c_1 \), which simplifies to \( 1 = 1 + c_1 \). Thus, \( c_1 = 0 \).
07

Check Continuity at \( x=1 \) and Find \( c_2 \)

Continuity at \( x=1 \) requires that \( y \) is continuous across the boundary conditions. - For \( 0 \leq x \leq 1 \), \( y = 1 \) at \( x = 1 \).- For \( x > 1 \), \( e^x y = -e^x + c_2 \), leading to \( y = 2e^{1-x} - 1 \) for \( x = 1 \).Equating the functions at \( x=1 \) gives \( 2e^{1-1} - 1 = 1 \), means \( 2e - 1 = 1 \), implying \( c_2 = 2e \).
08

Write the Piecewise Solution

The solution of the differential equation is:\[ y = \begin{cases} 1, & 0 \leq x \leq 1 \ 2e^{1-x} - 1, & x > 1 \end{cases} \]

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Key Concepts

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

Integrating Factor
The integrating factor is a crucial tool in solving first-order linear differential equations. It's used to simplify these equations and make them easier to solve. The integrating factor is typically denoted as \( \mu(x) \), and it transforms a linear differential equation into a form that can be easily integrated. For an equation of the form \( y' + P(x)y = Q(x) \), the integrating factor is obtained as \( \mu(x) = e^{\int P(x) \, dx} \).

In this exercise, we have that \( P(x) = 1 \), therefore, the integrating factor becomes \( e^x \). By multiplying the entire differential equation \( y' + y = f(x) \) by \( e^x \), the left-hand side becomes a single derivative, \( \frac{d}{dx}(e^x y) \). This conversion enables us to integrate both sides of the equation seamlessly to find \( y \).
  • Step: Multiply the equation by the integrating factor \( e^x \).
  • Result: The equation simplifies to a total derivative, enabling easy integration.
Piecewise Functions
Piecewise functions are used to define a function by different expressions for different intervals of the domain. They are especially useful in mathematics to model situations where a function's behavior changes over certain ranges.

In the given problem, the solution to the differential equation is expressed as a piecewise function:\[y = \begin{cases} 1, & 0 \leq x \leq 1 \2e^{1-x} - 1, & x > 1 \\end{cases}\]

This piecewise function reflects how the behavior and continuity of \( y \) change at \( x=1 \). The first part, \( y = 1 \), holds for \( 0 \leq x \leq 1 \), while the second part governs \( y \) for \( x > 1 \), ensuring the continuation and proper solution of the differential equation across these segments.
  • Part 1: \( y = 1 \) for interval \( 0 \leq x \leq 1 \).
  • Part 2: \( y = 2e^{1-x} - 1 \) for \( x > 1 \).
Initial Condition
An initial condition is a value that defines the specific solution of a differential equation from a family of solutions. It provides a reference point鈥攗sually given as \( y(x_0) = y_0 \)鈥攖o determine a unique solution by eliminating arbitrary constants.

In this exercise, the given initial condition is \( y(0) = 1 \). This is used to find the constant \( c_1 \) in the piecewise function. By substituting \( x = 0 \) into the equation for \( y \), we get \( e^0 \, y = e^0 + c_1 \), leading to \( 1 = 1 + c_1 \) and thus \( c_1 = 0 \).

Having a correct initial condition ensures the solution obtained reflects the actual requirement given in the problem statement and also maintains the continuity of the piecewise function at \( x = 1 \).
  • Initial condition: \( y(0) = 1 \).
  • Result: Solving for \( c_1 \) gives \( c_1 = 0 \).

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Most popular questions from this chapter

For \(y^{\prime}+2 y=f(x)\) an integrating factor is \(e^{2 x}\) so that $$y e^{2 x}=\left\\{\begin{array}{ll} \frac{1}{2} e^{2 x}+c_{1}, & 0 \leq x \leq 3 \\ c_{2}, & x>3 \end{array}\right.$$ If \(y(0)=0\) then \(c_{1}=-1 / 2\) and for continuity we must have \(c_{2}=\frac{1}{2} e^{6}-\frac{1}{2}\) so that $$y=\left\\{\begin{array}{ll} \frac{1}{2}\left(1-e^{-2 x}\right), & 0 \leq x \leq 3 \\ \frac{1}{2}\left(e^{6}-1\right) e^{-2 x}, & x>3 \end{array}\right.$$

Assume \(R d q / d t+(1 / C) q=E(t), R=200, C=10^{-4},\) and \(E(t)=100\) so that \(q=1 / 100+c e^{-50 t} .\) If \(q(0)=0\) then \(c=-1 / 100\) and \(i=\frac{1}{2} e^{-50 t}\).

(a) \(\mathrm{By}\) inspection \(y=x\) and \(y=-x\) are solutions of the differential equation and not members of the family \(y=x \sin \left(\ln x+c_{2}\right).\) (b) Letting \(x=5\) and \(y=0\) in \(\sin ^{-1}(y / x)=\ln x+c_{2}\) we get \(\sin ^{-1} 0=\ln 5+c\) or \(c=-\ln 5 .\) Then \(\sin ^{-1}(y / x)=\ln x-\ln 5=\ln (x / 5) .\) Because the range of the arcsine function is \([-\pi / 2, \pi / 2]\) we must have $$\begin{array}{c} -\frac{\pi}{2} \leq \ln \frac{x}{5} \leq \frac{\pi}{2} \\ e^{-\pi / 2} \leq \frac{x}{5} \leq e^{\pi / 2} \\ 5 e^{-\pi / 2} \leq x \leq 5 e^{\pi / 2}. \end{array}$$ The interval of definition of the solution is approximately [1.04, 24.05].

Assume \(L d i / d t+R i=E(t), E(t)=E_{0} \sin \omega t,\) and \(i(0)=i_{0}\) so that $$i=\frac{E_{0} R}{L^{2} \omega^{2}+R^{2}} \sin \omega t-\frac{E_{0} L \omega}{L^{2} \omega^{2}+R^{2}} \cos \omega t+c e^{-R t / L}$$ since \(i(0)=i_{0}\) we obtain \(c=i_{0}+\frac{E_{0} L \omega}{L^{2} \omega^{2}+R^{2}}\)

(a) The equilibrium solutions \(y(x)=2\) and \(y(x)=-2\) satisfy the initial conditions \(y(0)=2\) and \(y(0)=-2\) respectively. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(1+c e^{4 x}\right) /\left(1-c e^{4 x}\right)\) we obtain $$1=2 \frac{1+c e}{1-c e}, \quad 1-c e=2+2 c e, \quad-1=3 c e, \quad \text { and } \quad c=-\frac{1}{3 e}$$. (b) Separating variables and integrating yields \\[ \begin{aligned} \frac{1}{4} \ln |y-2|-\frac{1}{4} \ln |y+2|+\ln c_{1} &=x \\ \ln |y-2|-\ln |y+2|+\ln c &=4 x \\ \ln \left|\frac{c(y-2)}{y+2}\right| &=4 x \\ c \frac{y-2}{y+2} &=e^{4 x}.\end{aligned}\\] Solving for \(y\) we get \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right) .\) The initial condition \(y(0)=-2\) implies \(2(c+1) /(c-1)=-2\) which yields \(c=0\) and \(y(x)=-2 .\) The initial condition \(y(0)=2\) does not correspond to a value of \(c,\) and it must simply be recognized that \(y(x)=2\) is a solution of the initial-value problem. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right)\) leads to \(c=-3 e .\) Thus, a solution of the initial-value problem is $$y=2 \frac{-3 e+e^{4 x}}{-3 e-e^{4 x}}=2 \frac{3-e^{4 x-1}}{3+e^{4 x-1}}.$$
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