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Chapter 5: Problem 8

In Exercises \(1-12\) find the general solution. $$ y^{\prime \prime}+y^{\prime}=0 $$

Short Answer

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Question: Find the general solution of the second-order homogeneous linear differential equation \(y'' + y' = 0\). Answer: The general solution is given by \(y(x) = -Ke^{-x} + C\), where \(K\) and \(C\) are arbitrary constants.

Step by step solution

01

Introduce the substitution

Let \(v = y'\), so \(v' = y''\). The given differential equation can now be written as: $$ v' + v = 0 $$ This is a first-order linear homogeneous differential equation.
02

Solve the first-order equation

The first-order linear homogeneous differential equation has the general solution: $$ v(x) = Ke^{-x}, $$ where \(K\) is an arbitrary constant.
03

Integrate to find the general solution of the original equation

From our substitution, we have \(y'(x) = v(x) = Ke^{-x}\). Integrate both sides to find the general solution for \(y(x)\): $$ y(x) = \int Ke^{-x} dx = -Ke^{-x} + C, $$ where \(C\) is another arbitrary constant.
04

Write the final general solution

The general solution of the original second-order homogeneous differential equation is given by: $$ y(x) = -Ke^{-x} + C $$ where \(K\) and \(C\) are arbitrary constants.

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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
A first-order linear differential equation is an equation of the form \( y' + Py = Q \), with \( P \) and \( Q \) as functions of \( x \). These equations can easily be solved using a standard method called the "integrating factor method."
This technique is useful because it lets us systematically solve the equation and find the function \( y(x) \) without guessing.
The integrating factor is derived from \( P \), specifically:
  • Calculate the integrating factor by finding \( \mu(x) = e^{\int P(x) \, dx} \).
  • Multiply through the equation by \( \mu(x) \) to make the left-hand side a perfect derivative.
  • Integrate both sides to find the general solution.
In our example, \( v' + v = 0 \) is already in the standard form, where \( P(x) = 1 \), making the integrating factor \( \mu(x) = e^{x} \). This leads to solutions that give insight into the behavior and characteristics of the function over the domain.
Homogeneous Differential Equations
Homogeneous differential equations, not to be confused with the term "homogeneous in the context of first-order equations," require that all terms can be expressed in terms of a single derivative.
Specifically, a differential equation is homogeneous if it can be written as all terms involving the dependent variable and its derivatives equal zero.
For our example, the original equation, \( y'' + y' = 0 \), is a second-order homogeneous differential equation.
Here's why it's homogeneous:
  • There are no constant or standalone functions of \( x \) on one side of the equation.
  • Both terms, \( y'' \) and \( y' \), include derivatives of the same unknown function \( y \).
These equations have particular solutions influenced only by the initial conditions or boundary values. Thus, solving homogeneous equations focuses heavily on finding these characteristic solutions.
Integration Techniques
Integration is a core process in solving differential equations. It helps to find the function when given its derivative.
After substituting and solving the first-order equation like \( v = Ke^{-x} \), integration brings us back to the original variable and its derivatives.
Consider the process:
  • Identify the form of the differential expression, which helps choose an appropriate integration method.
  • Apply basic integration rules for exponential or polynomial functions.
In our solution, to solve \( y'(x) = Ke^{-x} \), we integrate:\[\int Ke^{-x} \, dx = -Ke^{-x} + C\]The choice of integration technique will depend on the form of the integrand and could include basic rules, substitution, or parts if more complex. With \( K \) and \( C \) as arbitrary constants, the integration completes the solution of the differential equation.

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Most popular questions from this chapter

Find a fundamental set of solutions, given that \(y_{1}\) is a solution. \(x^{2}(\ln |x|)^{2} y^{\prime \prime}-(2 x \ln |x|) y^{\prime}+(2+\ln |x|) y=0 ; \quad y_{1}=\ln |x|\)

The nonlinear first order equation $$y^{\prime}+y^{2}+p(x) y+q(x)=0$$ is a Riccati equation. (See Exercise 2.4.55.) Assume that \(p\) and \(q\) are continuous. (a) Show that \(y\) is a solution of \((\mathrm{A})\) if and only if \(y=z^{\prime} / z,\) where $$z^{\prime \prime}+p(x) z^{\prime}+q(x) z=0$$ (b) Show that the general solution of \((\mathrm{A})\) is $$y=\frac{c_{1} z_{1}^{\prime}+c_{2} z_{2}^{\prime}}{c_{1} z_{1}+c_{2} z_{2}}$$ where \(\left\\{z_{1}, z_{2}\right\\}\) is a fundamental set of solutions of (B) and \(c_{1}\) and \(c_{2}\) are arbitrary constants. (c) Does the formula (C) imply that the first order equation (A) has a two- parameter family of solutions? Explain your answer.

Suppose \(f\) is continuous on an open interval that contains \(x_{0}=0 .\) Use variation of parameters to find a formula for the solution of the initial value problem $$ y^{\prime \prime}-y=f(x), \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}. $$

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ 4 x^{2} y^{\prime \prime}-4 x y^{\prime}+\left(4 x^{2}+3\right) y=x^{7 / 2} ; \quad y_{1}=\sqrt{x} \sin x, y_{2}=\sqrt{x} \cos x $$

Suppose the characteristic polynomial of \(a y^{\prime \prime}+b y^{\prime}+c y=0\) has complex conjugate roots \(\lambda \pm i \omega\). Use a method suggested by Exercise 22 to find a formula for the solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=0, \quad y\left(x_{0}\right)=k_{0}, \quad y^{\prime}\left(x_{0}\right)=k_{1} $$
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