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Chapter 1: Problem 8

8\. Find all solutions of the equation $$ y^{\prime}+2 y=b(x), \quad(-\infty1\).

Short Answer

Expert verified
Solve using the integrating factor method over defined regions of \( x \), combining solutions for different \( b(x) \).

Step by step solution

01

Understand the Given Differential Equation

We are given a first-order linear ordinary differential equation in the form \( y' + 2y = b(x) \). Here, \( b(x) \) is a piecewise function, which changes depending on the value of \( x \). Our goal is to solve this differential equation for \( y(x) \) over the entire range of \( x \).
02

Define the Function b(x)

The function \( b(x) \) is defined as \( b(x) = 1 - |x| \) for \( |x| \leq 1 \) and \( b(x) = 0 \) for \( |x| > 1 \). This means the function has a triangular shape, equaling 1 at \( x = 0 \), and linearly decreasing to 0 at \( x = \pm 1 \), and remains 0 beyond these points.
03

Solve for x such that |x| > 1

Since \( b(x) = 0 \) for \( |x| > 1 \), the differential equation becomes \( y' + 2y = 0 \). This is a homogeneous differential equation, and its solution is of the form \( y(x) = Ce^{-2x} \), where \( C \) is a constant determined by boundary conditions.
04

Solve for x such that |x| \leq 1

For \( |x| \leq 1 \), the equation is \( y' + 2y = 1 - |x| \). This is an inhomogeneous linear differential equation, and we solve it using the integrating factor method.
05

Determine the Integrating Factor

For an equation of the form \( y' + P(x)y = Q(x) \), the integrating factor is \( e^{\int P(x) \, dx} = e^{2x} \) in our case. Multiply the entire differential equation by the integrating factor, \( e^{2x}y' + 2e^{2x}y = e^{2x}(1 - |x|) \).
06

Solve the Transformed Equation

Notice the left-hand side is the derivative of \( e^{2x}y \). Integrating both sides, we get \( e^{2x}y = \int e^{2x}(1 - |x|) \, dx \). Evaluate the integral for the given range of \( x \).
07

Evaluate the Integral

The integral \( \int e^{2x}(1 - |x|) \, dx \) becomes separable. Solve this for both cases when \( x \geq 0 \) and \( x < 0 \) within the defined bounds. Simplify the expression to find \( y(x) \).
08

Combine Piecewise Solutions

Finally, the solution is \( y(x) = Ce^{-2x} \) for \( |x| > 1 \), and find the explicit form for the solution \( y(x) \) when \( |x| \leq 1 \) by solving the integral in the previous steps and applying boundary conditions. This gives a piecewise function for \( y(x) \), expressed separately over the defined domains.

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Key Concepts

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

First-order Linear Differential Equations
First-order linear differential equations are an essential type of ordinary differential equations (ODEs). They have the standard form: \[ y' + P(x) y = Q(x) \] where \( y' \) represents the derivative of \( y \) with respect to \( x \), and \( P(x) \) and \( Q(x) \) are functions of \( x \). These equations are linear because they involve no powers of \( y \) other than one.
  • **In homogeneous equations**, \( Q(x) = 0 \), and the equation simplifies.
  • **In inhomogeneous equations**, \( Q(x) \) is non-zero, requiring additional steps to solve.
In solving first-order linear ODEs, we often seek functions \( y(x) \) that satisfy the equation for all \( x \). Understanding these equations helps address various real-world problems, from population dynamics to electric circuits.
Piecewise Functions
Piecewise functions are versatile tools for modeling real-world situations where the behavior changes at certain points. The function defined in the exercise, \[ b(x)=1 - |x| \text{ for } |x| \leq 1 \] and \[ b(x)=0 \text{ for } |x|>1 \] is a perfect example of a piecewise function.
Piecewise functions commonly have various forms over different intervals:
  • **Continuous piecewise functions** follow a continuous path, except possibly for shifts at the switching points.
  • **Discontinuous piecewise functions** may jump or have gaps at the switching points.
In this exercise, the function \( b(x) \) takes on a triangular shape over \( |x| \leq 1 \), peaking at \( x = 0 \), and transitioning to zero for \( |x| > 1 \). Mastering piecewise functions is crucial for comprehensively solving differential equations with inputs that switch behaviors over different intervals.
Integrating Factor Method
The integrating factor method is a powerful technique for solving first-order linear differential equations. The idea is to multiply the entire equation by a specially chosen function, called the integrating factor, making it easier to solve.Consider the differential equation: \[ y' + P(x) y = Q(x) \] Here's how you can use the integrating factor:
  • **Find the integrating factor**: It is \( e^{\int P(x) \, dx} \), which often turns the left-hand side of the equation into the derivative of a product.
  • **Multiply the whole equation by this factor**: This transforms the equation nicely to \( \frac{d}{dx}(e^{\int P(x) \, dx} y) = e^{\int P(x) \, dx} Q(x) \).
  • **Integrate both sides**: This leads to an expression for \( y(x) \), emphasizing the relationships dictated by the equation.
Recognizing when to use the integrating factor method allows students to tackle more complex inhomogeneous linear equations effectively, making it an indispensable tool in the toolbox for solving differential equations.

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Most popular questions from this chapter

Let \(\phi\) be the solution of \(y^{\prime}+i y=x\) such that \(\phi(0)=2\). Find \(\phi(\pi)\).

5\. Consider the equation $$ L y^{\prime}+R y=E e^{i \omega x} $$ where \(L, R, E, \omega\) are positive constants. (a) Compute the solution \(\phi\) which satisfies \(\phi(0)=0\). (b) Using the differential equation show that \(\phi_{1}=\operatorname{Re} \phi\) satisfies $$ L y^{\prime}+R y=E \cos \omega x $$ Compute \(\phi_{1}\). (c) Using the differential equation show that \(\phi_{2}=\operatorname{Im} \phi\) satisfies $$ L y^{\prime}+R y=E \sin \omega x $$ Compute \(\phi_{2}\).

Verify that the following are solutions of the differential equations given: (a) \(\phi(x)=e^{-\sin x}\), for \(y^{\prime}+(\cos x) y=0\) (b) \(\phi(x)=\sin x-1\), for \(y^{\prime}+(\cos x) y=\sin x \cos x\) (c) \(\phi(x)=1\), for \(y^{\prime \prime}-y^{\prime}=0\) (d) \(\phi(x)=e^{x}\), for \(y^{\prime \prime}-y^{\prime}=0\) (e) \(\phi(x)=c_{1}+c_{2} e^{x}\), for \(y^{\prime \prime}-y^{\prime}=0,\left(c_{1}, c_{2}\right.\) any constants) (f) \(\phi(x)=\sin 2 x\), for \(y^{\prime \prime}+4 y=0\) (g) \(\phi(x)=e^{2 i x}\), for \(y^{\prime \prime}+4 y=0\) (h) \(\phi(x)=c_{1} \cos k x+c_{2} \sin k x\), for \(y^{\prime \prime}+k^{2} y=0,\left(k\right.\) a positive constant, and \(c_{1}, c_{2}\) any constants)

Find all solutions of the following equations: (a) \(y^{\prime}-2 y=1\) (b) \(y^{\prime}+y=e^{x}\) (c) \(y^{\prime}-2 y=x^{2}+x\) (d) \(3 y^{\prime}+y=2 e^{-x}\) (e) \(y^{\prime}+3 y=e^{i x}\)

7\. Consider the equation $$ y^{\prime}+a y=b(x) $$ where \(a\) is a constant, and \(b\) is a continuous function on \(0 \leqq x<\infty\), satisfying there \(|b(x)| \leqq k\), where \(k\) is some positive number. (a) Find the solution \(\phi\) satisfying \(\phi(0)=0\). (b) If Re \(a \neq 0\), show that this solution satisfies $$ |\phi(x)| \leqq \frac{k}{\operatorname{Re} a}\left[1-e^{-(\operatorname{Re} a) x}\right] $$ (c) Show that the right side of the inequality in (b) is the solution of $$ y^{\prime}+(\operatorname{Re} a) y=k, \quad(\operatorname{Re} a \neq 0) $$ whose graph passes through the origin.
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