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Chapter 3: Problem 19

Find the general solutions of the differential equations in Problems. \(y^{(3)}+y^{\prime \prime}-y^{\prime}-y=0\)

Short Answer

Expert verified
The general solution is \( y(x) = C_1e^x + C_2e^{-x} + C_3xe^{-x} \).

Step by step solution

01

Identify and write the auxiliary equation

The given differential equation is a linear homogeneous differential equation with constant coefficients: \( y^{(3)} + y'' - y' - y = 0 \). The auxiliary equation is found by substituting \( y = e^{rx} \) into the differential equation, which yields the characteristic polynomial: \( r^3 + r^2 - r - 1 = 0 \).
02

Factor the characteristic polynomial

To solve \( r^3 + r^2 - r - 1 = 0 \), try to find rational roots using the Rational Root Theorem. Testing possible roots, we find that \( r = 1 \) is a root. Thus, the polynomial can be factored as \((r - 1)(r^2 + 2r + 1) = 0\).
03

Solve the quadratic equation

The quadratic \( r^2 + 2r + 1 = 0 \) can be factored further as \((r + 1)^2 = 0\), which means \( r = -1 \) is a root of multiplicity 2.
04

Write the general solution

The roots of the characteristic equation are \( r = 1, -1, -1 \). Therefore, the general solution to the differential equation is \( y(x) = C_1e^x + C_2e^{-x} + C_3xe^{-x} \), 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
In the context of solving differential equations, the **characteristic equation** is fundamental for finding solutions to linear homogeneous differential equations with constant coefficients. When given such a differential equation, like the example here, the characteristic equation is derived by assuming a solution of the form \( y = e^{rx} \). By substituting this assumed solution back into the differential equation, we transform it into a polynomial. This polynomial is known as the characteristic equation.
  • For example, the given differential equation is \( y^{(3)} + y'' - y' - y = 0 \).
  • By substituting \( y = e^{rx} \), we transform it into a characteristic equation: \( r^3 + r^2 - r - 1 = 0 \).
The roots of this polynomial give insight into the nature of the solution of the differential equation.
Auxiliary Equation
The **auxiliary equation** in differential equations is often synonymous with the characteristic equation in the context of linear differential equations. After forming the characteristic polynomial by assuming \( y = e^{rx} \), the next step is to solve this characteristic polynomial, which is often referred to as the auxiliary equation.
  • The process starts with identifying the polynomial, such as \( r^3 + r^2 - r - 1 = 0 \).
  • This polynomial is then solved to find its roots, which are essential to determine the general solution of the differential equation.
Finding the roots can involve various techniques, including factoring, applying the Rational Root Theorem, or using the quadratic formula for simpler cases.
General Solution
The **general solution** of a differential equation is a broad solution encompassing all possible specific solutions. It's a key part of solving differential equations as it involves determining expressions that satisfy the original differential equation.
  • Once the roots of the characteristic or auxiliary equation are known, they are used to construct the general solution.
  • In our problem, the roots are \( r = 1, -1, -1 \).
Thus, the general solution to the differential equation is \( y(x) = C_1e^x + C_2e^{-x} + C_3xe^{-x} \), with \( C_1, C_2, \) and \( C_3 \) being arbitrary constants indicating the homogeneous nature of the solution.
Homogeneous Differential Equation
A **homogeneous differential equation** is one where every term is a function of the unknown variable and its derivatives, typically set equal to zero. This property simplifies the process of finding solutions, as the equation represents a system in equilibrium. In the given example, the equation \( y^{(3)} + y'' - y' - y = 0 \) is homogeneous.
  • This is because all terms involve the function \( y \) and its derivatives, with no additional constants or functions of \( x \).
  • This form allows us to use techniques like substitution and the characteristic polynomial to find solutions.
Overall, the homogeneous nature assures that solutions can often be expressed using exponential functions, shaped by the roots of the characteristic equation.

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Most popular questions from this chapter

As indicated by the cart-with-flywheel example discussed in this section, an unbalanced rotating machine part typically results in a force having amplitude proportional to the square of the frequency \(\omega\). (a) Show that the amplitude of the steady periodic solution of the differential equation $$ m x^{\prime \prime}+c x^{\prime}+k x=m A \omega^{2} \cos \omega t $$ (with a forcing term similar to that in Eq. (17)) is given by $$ C(\omega)=\frac{m A \omega^{2}}{\sqrt{\left(k-m \omega^{2}\right)^{2}+(c \omega)^{2}}}. $$ (b) Suppose that \(c^{2}<2 m k .\) Show that the maximum amplitude occurs at the frequency \(\omega_{m}\) given by $$ \omega_{m}=\sqrt{\frac{k}{m}\left(\frac{2 m k}{2 m k-c^{2}}\right)} . $$ Thus the resonance frequency in this case is larger (in contrast with the result of Problem 27 ) than the natural frequency \(\omega_{0}=\sqrt{k / m} .\) (Suggestion: Maximize the square of \(C .\) )

In Problems, one solution of the differential equation is given. Find the general solution. $$ 6 y^{(4)}+5 y^{(3)}+25 y^{\prime \prime}+20 y^{\prime}+4 y=0 ; y=\cos 2 x $$

In Problems 13 through 20, a third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions. $$ \begin{aligned} &y^{(3)}-6 y^{\prime \prime}+11 y^{\prime}-6 y=0 ; y(0)=0, y^{\prime}(0)=0, y^{\prime \prime}(0)=3 ; \\ &y_{1}=e^{x}, y_{2}=e^{2 x}, y_{3}=e^{3 x} \end{aligned} $$

Find and plot both the steady periodic solution \(x_{\mathrm{sp}}(t)=C \cos (\omega t-\alpha)\) of the given differential equation and the actual solution \(x(t)=x_{\mathrm{sp}}(t)+\) \(x_{\mathrm{tr}}(t)\) that satisfies the given initial conditions. $$ x^{\prime \prime}+2 x^{\prime}+26 x=600 \cos 10 t ; x(0)=10, x^{\prime}(0)=0 $$

Consider the eigenvalue problem $$ y^{\prime \prime}+2 y^{\prime}+\lambda y=0 ; \quad y(0)=0, \quad y^{\prime}(1)=0 $$ Show that the eigenvalues are all positive and that the \(n\) th positive eigenvalue is \(\lambda_{n}=\alpha_{n}^{2}+1\) with associated eigenfunction \(y_{n}(x)=e^{-x} \sin \alpha_{n} x\), where \(\alpha_{n}\) is the \(n\) th positive root of \(\tan z=z\)
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