/*! 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 4 The auxiliary equation is \(m^{2... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 4

The auxiliary equation is \(m^{2}+1=0,\) so \(y_{c}=c_{1} \cos x+c_{2} \sin x\) and $$W=\left|\begin{array}{rr}\cos x & \sin x \\\\-\sin x & \cos x\end{array}\right|=1$$ Identifying \(f(x)=\sec x \tan x\) we obtain $$W=\left|\begin{array}{rr}\cos x & \sin x \\\\-\sin x & \cos x\end{array}\right|=1$$ $$\begin{array}{l}u_{1}^{\prime}=-\sin x(\sec x \tan x)=-\tan ^{2} x=1-\sec ^{2} x \\\u_{2}^{\prime}=\cos x(\sec x \tan x)=\tan x\end{array}$$ Then \(u_{1}=x-\tan x, u_{2}=-\ln |\cos x|,\) and $$\begin{aligned}y &=c_{1} \cos x+c_{2} \sin x+x \cos x-\sin x-\sin x \ln |\cos x| \\\&=c_{1} \cos x+c_{3} \sin x+x \cos x-\sin x \ln |\cos x|. \end{aligned}$$

Short Answer

Expert verified
General solution: \( y = c_1 \cos x + c_3 \sin x + x \cos x - \sin x \ln |\cos x| \).

Step by step solution

01

Identify Given Components

We start with the given differential equation and its characteristic (auxiliary) equation, which is \( m^2 + 1 = 0 \). This results in complex roots and the complementary solution for the homogeneous part is \( y_c = c_1 \cos x + c_2 \sin x \).
02

Evaluate the Wronskian

The Wronskian for the functions \( \cos x \) and \( \sin x \) is evaluated as \( W = \left|\begin{array}{rr}\cos x & \sin x \ -\sin x & \cos x \end{array}\right| = 1 \). This tells us the functions are linearly independent, which is necessary for solving the non-homogeneous part.
03

Identify the Non-Homogeneous Function

The function \( f(x) = \sec x \tan x \) is identified as the non-homogeneous part of the equation we want to solve. We need to find particular solutions corresponding to this function.
04

Compute Derivatives for Particular Solutions

We calculate the derivatives \( u_1' \) and \( u_2' \) using the method of variation of parameters: \[ u_1' = -\sin x (\sec x \tan x) = -\tan^2 x = 1 - \sec^2 x \] \[ u_2' = \cos x (\sec x \tan x) = \tan x \]
05

Integrate to Find Particular Solutions

The derivatives \( u_1' \) and \( u_2' \) are integrated to find \( u_1 \) and \( u_2 \): \[ u_1 = \int (1 - \sec^2 x) \, dx = x - \tan x \] \[ u_2 = \int \tan x \, dx = -\ln |\cos x| \]
06

Formulate the General Solution

We construct the general solution \( y \) by combining the complementary solution \( y_c \) and particular solution involving \( u_1 \) and \( u_2 \): \[ y = c_1 \cos x + c_2 \sin x + (x - \tan x) \cos x - \sin x \ln |\cos x| \] Simplifying terms, this becomes: \[ y = c_1 \cos x + c_3 \sin x + x \cos x - \sin x \ln |\cos x| \]

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Key Concepts

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

Wronskian determinant
The Wronskian determinant is a vital tool in determining whether a set of solutions is linearly independent. In this context, the Wronskian for the functions \( \cos x \) and \( \sin x \) is computed using the determinant of the following matrix: \[ \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} = 1 \] A nonzero Wronskian indicates that the functions are linearly independent. This property is crucial as it ensures that the functions form a valid basis for the solution of the differential equation.
Linearly independent functions are essential when employing methods like the variation of parameters to solve non-homogeneous differential equations. If the Wronskian equals zero, the functions would be dependent, and we would need to find a different set of functions.
Complex roots of differential equations
When solving differential equations, we often encounter characteristic equations like \( m^2 + 1 = 0 \). This particular equation results in complex roots, specifically \( m = i \) and \( m = -i \). The presence of complex roots in the characteristic equation signifies that the solution will involve exponential and trigonometric functions.
Typically, for a complex root \( a + bi \), the complementary solution can be expressed as:
  • \( e^{ax} [c_1 \cos(bx) + c_2 \sin(bx)] \)
In our case, where \( a = 0 \) and \( b = 1 \), the complementary solution simplifies to:
  • \( c_1 \cos x + c_2 \sin x \)
Understanding how to handle complex roots is critical for forming the correct general solution to differential equations.
Particular solutions of differential equations
Finding a particular solution to a non-homogeneous differential equation often involves methods like undetermined coefficients or variation of parameters. Here, we use the variation of parameters technique. It begins by substituting dependent derivatives for each basis function. For example, we compute:
\[ u_1' = -\sin x(\sec x \tan x) = -\tan^2 x = 1 - \sec^2 x \]
\[ u_2' = \cos x(\sec x \tan x) = \tan x \]
Next, we integrate these expressions to find \( u_1 \) and \( u_2 \):
\[ u_1 = \int (1 - \sec^2 x) \, dx = x - \tan x \]
\[ u_2 = \int \tan x \, dx = -\ln |\cos x| \]
By combining these results with the complementary solution, we form the general solution:
\[ y = c_1 \cos x + c_3 \sin x + x \cos x - \sin x \ln |\cos x| \]
This method ensures that the particular solution appropriately complements the homogeneous solution when forming the general solution.

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Most popular questions from this chapter

In the case when \(\lambda=-\alpha^{2}<0,\) the solution of the differential equation is \(y=c_{1} \cosh \alpha x+c_{2} \sinh \alpha x .\) The condition \(y(0)=0\) gives \(c_{1}=0\) The condition \(y(1)-\frac{1}{2} y^{\prime}(1)=0 \quad\) applied to \(y=c_{2} \sinh \alpha x\) gives \(c_{2}\left(\sinh \alpha-\frac{1}{2} \alpha \cosh \alpha\right)=0 \quad\) or \(\tanh \alpha=\frac{1}{2} \alpha . \quad\) As can be seen from the figure, the graphs of \(y=\tanh x\) and \(y=\frac{1}{2} x\) intersect at a single point with approximate \(x\) -coordinate \(\alpha_{1}=1.915 .\) Thus, there is a single negative eigenvalue \(\lambda_{1}=-\alpha_{1}^{2} \approx-3.667\) and the corresponding eigenfuntion is \(y_{1}=\sinh 1.915 x\). For \(\lambda=0\) the only solution of the boundary-value problem is \(y=0\) For \(\lambda=\alpha^{2}>0\) the solution of the differential equation is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x .\) The condition \(y(0)=0\) gives \(c_{1}=0,\) so \(y=c_{2} \sin \alpha x .\) The condition \(y(1)-\frac{1}{2} y^{\prime}(1)=0\) gives \(c_{2}\left(\sin \alpha-\frac{1}{2} \alpha \cos \alpha\right)=0,\) so the eigenvalues are \(\lambda_{n}=\alpha_{n}^{2}\) when \(\alpha_{n}, n=2,3,4, \ldots,\) are the positive roots of \(\tan \alpha=\frac{1}{2} \alpha .\) Using a CAS we find that the first three values of \(\alpha\) are \(\alpha_{2}=4.27487, \alpha_{3}=7.59655,\) and \(\alpha_{4}=10.8127 .\) The first three eigenvalues are then \(\lambda_{2}=\alpha_{2}^{2}=18.2738, \lambda_{3}=\alpha_{3}^{2}=57.7075,\) and \(\lambda_{4}=\alpha_{4}^{2}=116.9139\) with corresponding eigenfunctions \(y_{2}=\sin 4.27487 x, y_{3}=\sin 7.59655 x,\) and \(y_{4}=\sin 10.8127 x\).

We have \(y_{1}^{\prime}=1\) and \(y_{1}^{\prime \prime}=0,\) so \(x y_{1}^{\prime \prime}-x y_{1}^{\prime}+y_{1}=0-x+x=0\) and \(y_{1}(x)=x\) is a solution of the differential equation. Letting \(y=u(x) y_{1}(x)=x u(x)\) we get $$y^{\prime}=x u^{\prime}(x)+u(x) \text { and } y^{\prime \prime}=x u^{\prime \prime}(x)+2 u^{\prime}(x)$$ Then \(x y^{\prime \prime}-x y^{\prime}+y=x^{2} u^{\prime \prime}+2 x u^{\prime}-x^{2} u^{\prime}-x u+x u=x^{2} u^{\prime \prime}-\left(x^{2}-2 x\right) u^{\prime}=0 .\) If we make the substitution \(w=u^{\prime},\) the linear first-order differential equation becomes \(x^{2} w^{\prime}-\left(x^{2}-x\right) w=0,\) which is separable: $$\begin{aligned}\frac{d w}{d x} &=\left(1-\frac{1}{x}\right) w \\\\\frac{d w}{w} &=\left(1-\frac{1}{x}\right) d x \\\\\ln w &=x-\ln x+c \\\w &=c_{1} \frac{e^{x}}{x.}\end{aligned}$$ Then \(u^{\prime}=c_{1} e^{x} / x\) and \(u=c_{1} \int e^{x} d x / x .\) To integrate \(e^{x} / x\) we use the series representation for \(e^{x} .\) Thus, a second solution is $$\begin{aligned}y_{2}=x u(x) &=c_{1} x \int \frac{e^{x}}{x} d x \\\&=c_{1} x \int \frac{1}{x}\left(1+x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\cdots\right) d x \\\&=c_{1} x \int\left(\frac{1}{x}+1+\frac{1}{2 !} x+\frac{1}{3 !} x^{2}+\cdots\right) d x \\\&=c_{1} x\left(\ln x+x+\frac{1}{2(2 !)} x^{2}+\frac{1}{3(3 !)} x^{3}+\cdots\right) \\\&=c_{1}\left(x \ln x+x^{2}+\frac{1}{2(2 !)} x^{3}+\frac{1}{3(3 !)} x^{4}+\cdots\right)\end{aligned}$$ An interval of definition is probably \((0, \infty)\) because of the \(\ln x\) term.

Let \(m\) be the mass and \(k_{1}\) and \(k_{2}\) the spring constants. Then \(k=4 k_{1} k_{2} /\left(k_{1}+k_{2}\right)\) is the effective spring constant of the system. since the initial mass stretches one spring \(\frac{1}{3}\) foot and another spring \(\frac{1}{2}\) foot, using \(F=k s,\) we have \(\frac{1}{3} k_{1}=\frac{1}{2} k_{2}\) or \(2 k_{1}=3 k_{2}\). The given period of the combined system is \(2 \pi / \omega=\pi / 15,\) so \(\omega=30 .\) since a mass weighing 8 pounds is \(\frac{1}{4}\) slug, we have from \(w^{2}=k / m\) \\[30^{2}=\frac{k}{1 / 4}=4 k \quad \text { or } \quad k=225\\] We now have the system of equations \\[\begin{aligned}\frac{4 k_{1} k_{2}}{k_{1}+k_{2}} &=225 \\\2 k_{1} &=3 k_{2}\end{aligned}\\] Solving the second equation for \(k_{1}\) and substituting in the first equation, we obtain \\[\frac{4\left(3 k_{2} / 2\right) k_{2}}{3 k_{2} / 2+k_{2}}=\frac{12 k_{2}^{2}}{5 k_{2}}=\frac{12 k_{2}}{5}=225\\] Thus, \(k_{2}=375 / 4\) and \(k_{1}=1125 / 8 .\) Finally, the weight of the first mass is \\[32 m=\frac{k_{1}}{3}=\frac{1125 / 8}{3}=\frac{375}{8} \approx 46.88 \mathrm{lb}\\].

(a) The general solution of \\[ \frac{d^{2} \theta}{d t^{2}}+\theta=0 \\] is \(\theta(t)=c_{1} \cos t+c_{2} \sin t .\) From \(\theta(0)=\pi / 12\) and \(\theta^{\prime}(0)=-1 / 3\) we find \\[ \theta(t)=(\pi / 12) \cos t-(1 / 3) \sin t \\] Setting \(\theta(t)=0\) we have \(\tan t=\pi / 4\) which implies \(t_{1}=\tan ^{-1}(\pi / 4) \approx 0.66577\) (b) We set \(\theta(t)=\theta(0)+\theta^{\prime}(0) t+\frac{1}{2} \theta^{\prime \prime}(0) t^{2}+\frac{1}{6} \theta^{\prime \prime \prime}(0) t^{3}+\cdots\) and use \(\theta^{\prime \prime}(t)=-\sin \theta(t)\) together with \(\theta(0)=\pi / 12\) and \(\theta^{\prime}(0)=-1 / 3 .\) Then \\[ \theta^{\prime \prime}(0)=-\sin (\pi / 12)=-\sqrt{2}(\sqrt{3}-1) / 4 \\] and \\[ \theta^{\prime \prime \prime}(0)=-\cos \theta(0) \cdot \theta^{\prime}(0)=-\cos (\pi / 12)(-1 / 3)=\sqrt{2}(\sqrt{3}+1) / 12 \\] Thus \\[ \theta(t)=\frac{\pi}{12}-\frac{1}{3} t-\frac{\sqrt{2}(\sqrt{3}-1)}{8} t^{2}+\frac{\sqrt{2}(\sqrt{3}+1)}{72} t^{3}+\cdots \\] (c) Setting \(\pi / 12-t / 3=0\) we obtain \(t_{1}=\pi / 4 \approx 0.785398\) (d) Setting \\[ \frac{\pi}{12}-\frac{1}{3} t-\frac{\sqrt{2}(\sqrt{3}-1)}{8} t^{2}=0 \\] and using the positive root we obtain \(t_{1} \approx 0.63088\) (e) Setting \\[ \frac{\pi}{12}-\frac{1}{3} t-\frac{\sqrt{2}(\sqrt{3}-1)}{8} t^{2}+\frac{\sqrt{2}(\sqrt{3}+1)}{72} t^{3}=0 \\] we find with the help of a CAS that \(t_{1} \approx 0.661973\) is the first positive root. (f) From the output we see that \(y(t)\) is an interpolating function on the interval \(0 \leq t \leq 5,\) whose graph is shown. The positive root of \(y(t)=0\) near \(t=1\) is \(t_{1}=0.666404\) (g) To find the next two positive roots we change the interval used in NDSolve and Plot from \(\\{\mathbf{t}, \mathbf{0}, \mathbf{5}\\}\) to \(\\{\mathbf{t}, \mathbf{0}, \mathbf{1 0}\\} .\) We see from the graph that the second and third positive roots are near 4 and \(7,\) respectively. Replacing \(\\{\mathbf{t}, \mathbf{1}\\}\) in FindRoot with \(\\{\mathbf{t}, \mathbf{4}\\}\) and then \(\\{\mathbf{t}, \mathbf{7}\\}\) we obtain \(t_{2}=3.84411\) and \(t_{3}=7.0218\)

From \(x^{\prime \prime}+4 x=-5 \sin 2 t+3 \cos 2 t, x(0)=-1,\) and \(x^{\prime}(0)=1\) we obtain \(x_{c}=c_{1} \cos 2 t+c_{2} \sin 2 t, x_{p}=\) \(\frac{3}{4} t \sin 2 t+\frac{5}{4} t \cos 2 t,\) and \\[x=-\cos 2 t-\frac{1}{8} \sin 2 t+\frac{3}{4} t \sin 2 t+\frac{5}{4} t \cos 2 t\\]
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