/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 From \(m^{3}+3 m^{2}-4 m-12=0\) ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 18

From \(m^{3}+3 m^{2}-4 m-12=0\) we obtain \(m=-2, m=2,\) and \(m=-3\) so that \\[ y=c_{1} e^{-2 x}+c_{2} e^{2 x}+c_{3} e^{-3 x} \\]

Short Answer

Expert verified
The general solution is \(y = c_{1} e^{-2x} + c_{2} e^{2x} + c_{3} e^{-3x}\).

Step by step solution

01

Identify Roots of the Polynomial

Given the cubic equation \(m^3 + 3m^2 - 4m - 12 = 0\), we already have the roots: \(m = -2, m = 2, m = -3\). These roots represent the solutions to the polynomial equation, and each root corresponds to an exponential term in a general solution of a differential equation.
02

Form the General Solution

Using the roots from the polynomial, the general solution is constructed using the formula \[y = c_{1} e^{m_{1} x} + c_{2} e^{m_{2} x} + c_{3} e^{m_{3} x}\] where \(m_1, m_2, m_3\) are the roots. For our roots, \(m_1 = -2\), \(m_2 = 2\), and \(m_3 = -3\). Thus, the solution becomes: \[y = c_{1} e^{-2x} + c_{2} e^{2x} + c_{3} e^{-3x}\]
03

Confirm Each Term Corresponds to a Root

Verify that each exponential term in the general solution corresponds to one of the roots of the polynomial. The term \(e^{-2x}\) corresponds to \(m = -2\), \(e^{2x}\) to \(m = 2\), and \(e^{-3x}\) to \(m = -3\). This ensures that the general solution is correctly derived from the differential equation that would have given the original polynomial as its characteristic equation.

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Key Concepts

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

Cubic Equations
A cubic equation is a type of polynomial equation of the third degree. It takes the general form: \[ ax^3 + bx^2 + cx + d = 0 \] where \(a\), \(b\), \(c\), and \(d\) are constants, with \(a eq 0\). These equations are fundamental in mathematics because they can represent complex, nonlinear relationships. They are called 'cubic' because the highest power of the variable \(x\) is three.
  • Finding the roots of a cubic equation is crucial, as they indicate the values of \(x\) that satisfy the equation.
  • The roots can be real or complex numbers, and there can be as many as three real roots.
Cubic equations often arise in various mathematical contexts, including differential equations, as their roots are directly linked to the solutions of such equations.
Characteristic Equation
The characteristic equation is a critical concept when dealing with linear differential equations, particularly those with constant coefficients. This is usually a polynomial equation derived from the differential equation, and its solutions (roots) play a crucial role in determining the general solution.
  • For a homogeneous linear differential equation with constant coefficients, the characteristic equation is obtained by replacing the derivative operator with a variable, often \(m\).
  • The roots of the characteristic equation can be real, repeated, or complex, and each type of root influences the form of the solution differently.
In the exercise you've seen, the characteristic equation \(m^3 + 3m^2 - 4m - 12 = 0\) helps us determine the form of the exponential terms that comprise the solution to the differential equation.
General Solution of Differential Equations
When solving differential equations, the general solution is the collection of all possible solutions, representing a family of functions that will satisfy the equation. For linear differential equations with constant coefficients, like the ones we discuss here, this solution is particularly elegant.
  • The general solution often takes the form of linear combinations of exponential functions, each associated with a distinct root of the characteristic equation.
  • For an equation with order \(n\), there can be \(n\) linearly independent solutions, leading to \(n\) terms in the general solution.
Each term in the general solution represents a specific behavior dictated by the corresponding root. For example, a root \(m_i\) results in the term \(c_i e^{m_i x}\), where \(c_i\) is a constant determined by initial conditions or boundary values. This process allows for a systematic way to capture all potential behaviors of the system described by the differential equation.

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Most popular questions from this chapter

(a) When \(T(x)=x^{2}\) the given differential equation is the Cauchy-Euler equation \\[ x^{2} y^{\prime \prime}+2 x y^{\prime}+\rho \omega^{2} y=0 \\] The solutions of the auxiliary equation \\[ m(m-1)+2 m+\rho \omega^{2}=m^{2}+m+\rho \omega^{2}=0 \\] are \\[ m_{1}=-\frac{1}{2}-\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} i, \quad m_{2}=-\frac{1}{2}+\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} \\] when \(\rho \omega^{2}>0.25 .\) Thus \\[ y=c_{1} x^{-1 / 2} \cos (\lambda \ln x)+c_{2} x^{-1 / 2} \sin (\lambda \ln x) \\] where \(\lambda=\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} .\) Applying \(y(1)=0\) gives \(c_{1}=0\) and consequently \\[ y=c_{2} x^{-1 / 2} \sin (\lambda \ln x) \\]. The condition \(y(e)=0\) requires \(c_{2} e^{-1 / 2} \sin \lambda=0 . \quad\) We obtain a nontrivial solution when \(\lambda_{n}=n \pi\) \(n=1,2,3, \ldots .\) But $$\lambda_{n}=\frac{1}{2} \sqrt{4 \rho \omega_{n}^{2}-1}=n \pi$$ Solving for \(\omega_{n}\) gives \\[ \omega_{n}=\frac{1}{2} \sqrt{\left(4 n^{2} \pi^{2}+1\right) / \rho} \\]. The corresponding solutions are \\[ y_{n}(x)=c_{2} x^{-1 / 2} \sin (n \pi \ln x) \\]. b.

For \(\lambda=\alpha^{4}, \alpha>0,\) the general solution of the boundary-value problem \\[ y^{(4)}-\lambda y=0, \quad y(0)=0, y^{\prime \prime}(0)=0, y(1)=0, y^{\prime \prime}(1)=0 \\] is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} \cosh \alpha x+c_{4} \sinh \alpha x\). The boundary conditions \(y(0)=0, y^{\prime \prime}(0)=0\) give \(c_{1}+c_{3}=0\) and \(-c_{1} \alpha^{2}+c_{3} \alpha^{2}=0,\) from which we conclude \(c_{1}=c_{3}=0 .\) Thus, \(y=c_{2} \sin \alpha x+c_{4} \sinh \alpha x .\) The boundary conditions \(y(1)=0, y^{\prime \prime}(1)=0\) then give \\[ \begin{aligned} c_{2} \sin \alpha+c_{4} \sinh \alpha &=0 \\ -c_{2} \alpha^{2} \sin \alpha+c_{4} \alpha^{2} \sinh \alpha &=0 \end{aligned} \\]. In order to have nonzero solutions of this system, we must have the determinant of the coefficients equal zero, that is, \\[ \left|\begin{array}{cc} \sin \alpha & \sinh \alpha \\ -\alpha^{2} \sin \alpha & \alpha^{2} \sinh \alpha \end{array}\right|=0 \quad \text { or } \quad 2 \alpha^{2} \sinh \alpha \sin \alpha=0 \\] But since \(\alpha>0,\) the only way that this is satisfied is to have \(\sin \alpha=0\) or \(\alpha=n \pi .\) The system is then satisfied by choosing \(c_{2} \neq 0, c_{4}=0,\) and \(\alpha=n \pi .\) The eigenvalues and corresponding eigenfunctions are then \\[ \lambda_{n}=\alpha^{4}=(n \pi)^{4}, n=1,2,3, \ldots \quad \text { and } \quad y=\sin n \pi x \\].

For large values of \(t\) the differential equation is approximated by \(x^{\prime \prime}=0 .\) The solution of this equation is the linear function \(x=c_{1} t+c_{2} .\) Thus, for large time, the restoring force will have decayed to the point where the spring is incapable of returning the mass, and the spring will simply keep on stretching.

From \(k_{1}=40\) and \(k_{2}=120\) we compute the effective spring constant \(k=4(40)(120) / 160=120 .\) Now, \(m=20 / 32\) so \(k / m=120(32) / 20=192\) and \(x^{\prime \prime}+192 x=0 . \quad\) Using \(x(0)=0\) and \(x^{\prime}(0)=2\) we obtain \(x(t)=\frac{\sqrt{3}}{12} \sin 8 \sqrt{3} t\).

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