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Chapter 3: Problem 21

From \(m^{3}+3 m^{2}+3 m+1=0\) we obtain \(m=-1, m=-1,\) and \(m=-1\) so that \\[ y=c_{1} e^{-x}+c_{2} x e^{-x}+c_{3} x^{2} e^{-x} \\]

Short Answer

Expert verified
The solution is \( y = c_1 e^{-x} + c_2 x e^{-x} + c_3 x^2 e^{-x} \).

Step by step solution

01

Identify the Differential Equation

The given polynomial equation is \(m^3 + 3m^2 + 3m + 1 = 0\). This equation is the characteristic equation of a third-order homogeneous differential equation with constant coefficients.
02

Solve the Characteristic Equation

The solutions of the characteristic equation \(m^3 + 3m^2 + 3m + 1 = 0\) are provided as \(m = -1, m = -1, m = -1\). This means all three roots are \(m = -1\), and they are repeated roots.
03

Formulate the General Solution

The general solution to a differential equation with repeated roots can be written using the formula for nth-order repeated roots. If the root \(m\) is repeated \(n\) times, the part of the solution corresponding to this is \(\sum_{k=0}^{n-1} c_k x^k e^{mx}\). Here, we have \(m = -1\) repeated 3 times.
04

Write the Final Solution

Based on our analysis, the general solution is given by: \( y = c_1 e^{-x} + c_2 x e^{-x} + c_3 x^2 e^{-x} \), where \(c_1, c_2,\) and \(c_3\) are arbitrary constants.

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Key Concepts

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

Characteristic Equation
The characteristic equation is a crucial concept when solving linear differential equations with constant coefficients. Think of it as a bridge between differential equations and algebra. Consider a third-order homogeneous differential equation:
  • The differential equation relates some function and its derivatives, typically represented as linear combinations.
  • "Characteristic" refers to how this equation helps identify the behavior or "characteristics" of the solution.
For the polynomial equation \(m^3 + 3m^2 + 3m + 1 = 0\), it is derived directly from the differential equation. Solving this polynomial tells us about the solutions to the differential equation. The roots of the characteristic equation are critical. They link directly to the terms in the general solution of the differential equation.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives. It has no standalone constant terms.
  • Takes the form \(a_n y^{(n)} + a_{n-1}y^{(n-1)} + \ldots + a_1y' + a_0y = 0\).
  • The term "homogeneous" indicates the absence of external non-zero terms, meaning it always equals zero without any constant or standalone function on the other side.
  • In our example, the solution is expressed with exponential functions \(e^{mx}\), signifying the natural exponential behavior of solutions.
These equations particularly pertain to linear systems, and the characteristic equation provides a method to solve them. The roots provide the base functions, which through linear combinations and possible modifications such as multiplying by \(x\), construct the complete solution.
Repeated Roots
Repeated roots occur in characteristic equations when a polynomial equation has multiple roots with the same value. This introduces a unique structure to the solution of the differential equation.
  • For a single root repeated \(n\) times, the solution is structured: \(y = (c_1 + c_2 x + \ldots + c_n x^{n-1})e^{rx}\), where \(r\) is the repeated root.
  • In the exemplified equation, root \(m = -1\) is repeated three times, necessitating the terms \(c_1 e^{-x} + c_2 x e^{-x} + c_3 x^2 e^{-x}\) in the solution.
This pattern reflects the idea that each repetition of a root requires an additional facet in the solution, specifically the multiplication by powers of \(x\). Repeated roots are an extension of the standard root solution, allowing for a proper spanning of the solution space.

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Most popular questions from this chapter

(a) From \(\frac{1}{4} x^{\prime \prime}+x^{\prime}+5 x=0, x(0)=1 / 2,\) and \(x^{\prime}(0)=1\) we obtain \(x=e^{-2 t}\left(\frac{1}{2} \cos 4 t+\frac{1}{2} \sin 4 t\right)\) (b) \(x=\frac{1}{\sqrt{2}} e^{-2 t} \sin \left(4 t+\frac{\pi}{4}\right)\) (c) If \(x=0\) then \(4 t+\pi / 4=\pi, 2 \pi, 3 \pi, \ldots\) so that the times heading downward are \(t=(7+8 n) \pi / 16\) for \(n=0,1,2, \dots\)

From \(\left(D^{2}+5\right) x-2 y=0\) and \(-2 x+\left(D^{2}+2\right) y=0\) we obtain \(y=\frac{1}{2}\left(D^{2}+5\right) x, D^{2} y=\frac{1}{2}\left(D^{4}+5 D^{2}\right) x,\) and \(\left(D^{2}+1\right)\left(D^{2}+6\right) x=0 .\) The solution is $$\begin{aligned} &x=c_{1} \cos t+c_{2} \sin t+c_{3} \cos \sqrt{6} t+c_{4} \sin \sqrt{6} t\\\ &y=2 c_{1} \cos t+2 c_{2} \sin t-\frac{1}{2} c_{3} \cos \sqrt{6} t-\frac{1}{2} c_{4} \sin \sqrt{6} t \end{aligned}.$$

Define \(y=u(x) e^{-2 x}\) so $$y^{\prime}=-2 u e^{-2 x}+u^{\prime} e^{-2 x}, \quad y^{\prime \prime}=u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}+4 u e^{-2 x}$$ and $$y^{\prime \prime}-4 y=e^{-2 x} u^{\prime \prime}-4 e^{-2 x} u^{\prime}=0 \quad \text { or } \quad u^{\prime \prime}-4 u^{\prime}=0$$ If \(w=u^{\prime}\) we obtain the linear first-order equation \(w^{\prime}-4 w=0\) which has the integrating factor \(e^{-4 \int d x}=e^{-4 x} .\) Now $$\frac{d}{d x}\left[e^{-4 x} w\right]=0 \quad \text { gives } \quad e^{-4 x} w=c$$ Therefore \(w=u^{\prime}=c e^{4 x}\) and \(u=c_{1} e^{4 x} .\) A second solution is \(y_{2}=e^{-2 x} e^{4 x}=e^{2 x} .\) We see by observation that a particular solution is \(y_{p}=-1 / 2 .\) The general solution is \\[ y=c_{1} e^{-2 x}+c_{2} e^{2 x}-\frac{1}{2} \\]

If the constant \(-c_{1}^{2}\) is used instead of \(c_{1}^{2},\) then, using partial fractions \\[ y=-\int \frac{d x}{x^{2}-c_{1}^{2}}=-\frac{1}{2 c_{1}} \int\left(\frac{1}{x-c_{1}}-\frac{1}{x+c_{1}}\right) d x=\frac{1}{2 c_{1}} \ln \left|\frac{x+c_{1}}{x-c_{1}}\right|+c_{2} \\] Alternatively, the inverse hyperbolic tangent can be used.

For \(\lambda=\alpha^{2}>0\) the general solution is \(y=c_{1} \cos \sqrt{\alpha} x+c_{2} \sin \sqrt{\alpha} x .\) Setting \(y(0)=0\) we find \(c_{1}=0,\) so that \(y=c_{2} \sin \sqrt{\alpha} x .\) The boundary condition \(y(1)+y^{\prime}(1)=0\) implies \\[ c_{2} \sin \sqrt{\alpha}+c_{2} \sqrt{\alpha} \cos \sqrt{\alpha}=0. \\] Taking \(c_{2} \neq 0,\) this equation is equivalent to \(\tan \sqrt{\alpha}=-\sqrt{\alpha} .\) Thus, the eigenvalues are \(\lambda_{n}=\alpha_{n}^{2}=x_{n}^{2}\) \(n=1,2,3, \ldots,\) where the \(x_{n}\) are the consecutive positive roots of \(\tan \sqrt{\alpha}=-\sqrt{\alpha}\).
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