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Chapter 3: Problem 51

From the solution \(y_{1}=e^{-4 x} \cos x\) we conclude that \(m_{1}=-4+i\) and \(m_{2}=-4-i\) are roots of the auxiliary equation. Hence another solution must be \(y_{2}=e^{-4 x} \sin x .\) Now dividing the polynomial \(m^{3}+6 m^{2}+m-34\) by \([m-(-4+i)][m-(-4-i)]=m^{2}+8 m+17\) gives \(m-2 .\) Therefore \(m_{3}=2\) is the third root of the auxiliary equation, and the general solution of the differential equation is $$y=c_{1} e^{-4 x} \cos x+c_{2} e^{-4 x} \sin x+c_{3} e^{2 x}$$

Short Answer

Expert verified
The general solution is \(y = c_1 e^{-4x} \cos x + c_2 e^{-4x} \sin x + c_3 e^{2x}\).

Step by step solution

01

Identify Given Solution Roots

The given solution suggests that \(y_1 = e^{-4x} \cos x\) implies that the roots of the auxiliary equation are \(m_1 = -4 + i\) and \(m_2 = -4 - i\). These are complex conjugates.
02

Find the Other Solution Based on Complex Roots

When roots are complex conjugates like \(m_1 = -4 + i\) and \(m_2 = -4 - i\), the differential equation solution includes a combination of \(e^{-4x} \sin x\). Thus, another solution is \(y_2 = e^{-4x} \sin x\).
03

Divide the Polynomial by Quadratic Factor

The polynomial to divide is \(m^3 + 6m^2 + m - 34\), and the factor we have is \(m^2 + 8m + 17\) derived from \([m-(-4+i)][m-(-4-i)]\). Dividing these two polynomials results in the linear factor \(m-2\).
04

Identify the Third Root

The quotient from division, \(m-2\), implies that the third root of the polynomial is \(m_3 = 2\). This corresponds to an exponential term in the solution.
05

Write the General Solution

Given all roots \(m_1 = -4 + i\), \(m_2 = -4 - i\), and \(m_3 = 2\), construct the general solution as \(y = c_1 e^{-4x} \cos x + c_2 e^{-4x} \sin x + c_3 e^{2x}\), 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.

Complex Roots
When solving differential equations, particularly linear differential equations with constant coefficients, complex roots can arise in the characteristic or auxiliary equation. Complex roots typically appear in conjugate pairs, such as \(m_1 = -4 + i\) and \(m_2 = -4 - i\). These roots occur when the discriminant of the characteristic equation is negative, leading to solutions that involve both exponential and trigonometric functions.
  • The presence of complex roots indicates oscillatory behavior in solutions.
  • The real part of the root, \(-4\), controls the exponential decay rate.
  • The imaginary part, \(\pm i\), introduces scillation, resulting in sine and cosine terms in the solution.
For instance, with the roots \(-4 \pm i\), the solutions can be expressed as \(y_1 = e^{-4x} \cos x\) and \(y_2 = e^{-4x} \sin x\). These solutions reflect the exponential decay modulated by oscillations, characteristic of complex conjugate roots.
Auxiliary Equation
The auxiliary equation plays a crucial role in finding solutions for linear differential equations. It's derived from the characteristic equation obtained by assuming solutions in an exponential form, such as \(e^{mx}\).
  • For a differential equation with constant coefficients, the auxiliary equation is a polynomial where the degree matches the order of the differential equation.
  • Solving the auxiliary equation involves finding roots, which can be real, repeated, or complex.
  • These roots inform the form of the solution to the differential equation.
In our exercise, the auxiliary equation is \(m^3 + 6m^2 + m - 34 = 0\). Solving this polynomial, we identify the roots \(-4 \pm i\) (complex) and \(2\) (real). These roots correspond to specific terms in the general solution based on their nature.
General Solution
The general solution of a differential equation encapsulates all possible solutions by incorporating its roots through characteristic terms. This solution is influenced by the nature of the roots found in the auxiliary equation.
  • For distinct complex conjugate roots, such as \(-4 \pm i\), the general solution includes terms like \(e^{-4x} \cos x\) and \(e^{-4x} \sin x\).
  • For real roots, like \(m=2\), the solution takes the form of an exponential function, such as \(e^{2x}\).
  • The arbitrary constants \(c_1, c_2, c_3\) allow for a wide range of specific solutions to satisfy initial or boundary conditions.
Thus, for the given differential equation, the general solution is formulated as \(y = c_1 e^{-4x} \cos x + c_2 e^{-4x} \sin x + c_3 e^{2x}\). This expression encompasses both the oscillatory decay due to complex roots and the growth associated with the real root.

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Most popular questions from this chapter

Using a CAS to solve the auxiliary equation \(6.11 \mathrm{m}^{3}+8.59 \mathrm{m}^{2}+7.93 \mathrm{m}+0.778=0\) we find \(m_{1}=-0.110241\) \(m_{2}=-0.647826+0.857532 i,\) and \(m_{3}=-0.647826-0.857532 i .\) The general solution is $$y=c_{1} e^{-0.110241 x}+e^{-0.647826 x}\left(c_{2} \cos 0.857532 x+c_{3} \sin 0.857532 x\right)$$

For \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) we have $$y=c_{1} \cos \alpha x+c_{2} \sin \alpha x$$ Now \(y(-\pi)=y(\pi)=0\) implies \\[ \begin{array}{l} c_{1} \cos \alpha \pi-c_{2} \sin \alpha \pi=0 \\ c_{1} \cos \alpha \pi+c_{2} \sin \alpha \pi=0 \end{array} \\] This homogeneous system will have a nontrivial solution when \\[ \left|\begin{array}{rr} \cos \alpha \pi & -\sin \alpha \pi \\ \cos \alpha \pi & \sin \alpha \pi \end{array}\right|=2 \sin \alpha \pi \cos \alpha \pi=\sin 2 \alpha \pi=0 \\] Then \\[ 2 \alpha \pi=n \pi \quad \text { or } \quad \lambda=\alpha^{2}=\frac{n^{2}}{4} ; \quad n=1,2,3, \ldots \\] When \(n=2 k-1\) is odd, the eigenvalues are \((2 k-1)^{2} / 4 .\) since \(\cos (2 k-1) \pi / 2=0\) and \(\sin (2 k-1) \pi / 2 \neq 0\) we see from either equation in (1) that \(c_{2}=0 .\) Thus, the eigenfunctions corresponding to the eigenvalues \((2 k-1)^{2} / 4\) are \(y=\cos (2 k-1) x / 2\) for \(k=1,2,3, \ldots .\) Similarly, when \(n=2 k\) is even, the eigenvalues are \(k^{2}\) with corresponding eigenfunctions \(y=\sin k x\) for \(k=1,2,3, \ldots\)

We need to solve \(\left[1+\left(y^{\prime}\right)^{2}\right]^{3 / 2}=y^{\prime \prime} .\) Let \(u=y^{\prime}\) so that \(u^{\prime}=y^{\prime \prime} .\) The equation becomes \(\left(1+u^{2}\right)^{3 / 2}=u^{\prime}\) or \(\left(1+u^{2}\right)^{3 / 2}=d u / d x .\) Separating variables and using the substitution \(u=\tan \theta\) we have \\[ \begin{aligned} \frac{d u}{\left(1+u^{2}\right)^{3 / 2}}=d x & \Longrightarrow \int \frac{\sec ^{2} \theta}{\left(1+\tan ^{2} \theta\right)^{3 / 2}} d \theta=x \Longrightarrow \int \frac{\sec ^{2} \theta}{\sec ^{3} \theta} d \theta=x \\ & \Longrightarrow \int \cos \theta d \theta=x \Longrightarrow \sin \theta=x \Longrightarrow \frac{u}{\sqrt{1+u^{2}}}=x \\ & \Longrightarrow \frac{y^{\prime}}{\sqrt{1+\left(y^{\prime}\right)^{2}}}=x \Longrightarrow\left(y^{\prime}\right)^{2}=x^{2}\left[1+\left(y^{\prime}\right)^{2}\right]=\frac{x^{2}}{1-x^{2}} \\\ & \Longrightarrow y^{\prime}=\frac{x}{\sqrt{1-x^{2}}}(\text { for } x>0) \Longrightarrow y=-\sqrt{1-x^{2}} \end{aligned} \\]

Using a CAS to solve the auxiliary equation \(m^{3}-6 m^{2}+2 m+1\) we find \(m_{1}=-0.270534\) \(m_{2}=0.658675,\) and \(m_{3}=5.61186 .\) The general solution is $$y=c_{1} e^{-0.270534 x}+c_{2} e^{0.658675 x}+c_{3} e^{5.61186 x}$$

(a) The weight of \(x\) feet of the chain is \(2 x,\) so the corresponding mass is \(m=2 x / 32=x / 16 .\) The only force acting on the chain is the weight of the portion of the chain hanging over the edge of the platform. Thus, by Newton's second law, \(\frac{d}{d t}(m v)=\frac{d}{d t}\left(\frac{x}{16} v\right)=\frac{1}{16}\left(x \frac{d v}{d t}+v \frac{d x}{d t}\right)=\frac{1}{16}\left(x \frac{d v}{d t}+v^{2}\right)=2 x\) and \(x d v / d t+v^{2}=32 x .\) Now, by the Chain Rule, \(d v / d t=(d v / d x)(d x / d t)=v d v / d x,\) so \(x v d v / d x+v^{2}=32 x\) (b) We separate variables and write the differential equation as \(\left(v^{2}-32 x\right) d x+x v d v=0 .\) This is not an exact form, but \(\mu(x)=x\) is an integrating factor. Multiplying by \(x\) we get \(\left(x v^{2}-32 x^{2}\right) d x+x^{2} v d v=0 .\) This form is the total differential of \(u=\frac{1}{2} x^{2} v^{2}-\frac{32}{3} x^{3},\) so an implicit solution is \(\frac{1}{2} x^{2} v^{2}-\frac{32}{3} x^{3}=c .\) Letting \(x=3\) and \(v=0\) we find \(c=-288 .\) Solving for \(v\) we get \(\frac{d x}{d t}=v=\frac{8 \sqrt{x^{3}-27}}{\sqrt{3} x}, \quad 3 \leq x \leq 8\) (c) Separating variables and integrating we obtain \(\frac{x}{\sqrt{x^{3}-27}} d x=\frac{8}{\sqrt{3}} d t \quad\) and \(\quad \int_{3}^{x} \frac{s}{\sqrt{s^{3}-27}} d s=\frac{8}{\sqrt{3}} t+c\) since \(x=3\) when \(t=0,\) we see that \(c=0\) and \\[ t=\frac{\sqrt{3}}{8} \int_{3}^{x} \frac{s}{\sqrt{s^{3}-27}} d s \\] We want to find \(t\) when \(x=7 .\) Using a CAS we find \(t(7)=0.576\) seconds.
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