/*! 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 38 From \(m^{2}+4=0\) we obtain \(m... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 38

From \(m^{2}+4=0\) we obtain \(m=\pm 2 i\) so that \(y=c_{1} \cos 2 x+c_{2} \sin 2 x .\) If \(y(0)=0\) and \(y(\pi)=0\) then \(c_{1}=0\) and \(y=c_{2} \sin 2 x\).

Short Answer

Expert verified
The solution is \(y = c_2 \sin 2x\). \(c_1 = 0\) based on \(y(0) = 0\).

Step by step solution

01

Understand the Characteristic Equation

The given equation is \(m^2 + 4 = 0\). To solve for \(m\), we rearrange it to find the roots: \(m^2 = -4\). Taking the square root of both sides, we get \(m = \pm 2i\). This indicates a characteristic equation with complex roots.
02

Formulate the General Solution

With complex roots \(m = \pm 2i\), the general solution is given by \(y = c_1 \cos 2x + c_2 \sin 2x\). This solution represents a typical solution form when dealing with complex roots.
03

Apply Boundary Condition \(y(0) = 0\)

Substitute \(x = 0\) in the general solution: \(y(0) = c_1 \cos(0) + c_2 \sin(0) = c_1\). Given \(y(0) = 0\), we find that \(c_1 = 0\).
04

Simplify the Solution

Since \(c_1 = 0\), substitute this back into the general solution: \(y = c_2 \sin 2x\). This is the simplified form of the solution based on the boundary condition \(y(0) = 0\).
05

Apply Boundary Condition \(y(\pi) = 0\)

Now substitute \(x = \pi\) in the simplified solution: \(y(\pi) = c_2 \sin(2\pi) = 0\). This equation is satisfied for any \(c_2\) because \(\sin(2\pi) = 0\). Thus, \(c_2\) does not have to be zero, and \(y = c_2 \sin 2x\) will always satisfy \(y(\pi) = 0\).

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Key Concepts

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

Characteristic Equation
In differential equations, finding the characteristic equation is a key step when solving linear differential equations with constant coefficients.
This equation helps determine the nature of the solutions we can expect.
Let's break it down.
  • The characteristic equation is initially formed by replacing the derivatives in the differential equation with powers of a variable, usually denoted as \( m \).
  • For the example above, this results in the equation: \( m^2 + 4 = 0 \).
  • We solve this algebraic equation to find the roots, which would guide us to the general solution of the differential equation.
In this case, solving \( m^2 + 4 = 0 \) involves isolating \( m^2 \) giving \( m^2 = -4 \), and then taking the square root, resulting in the complex roots \( m = \pm 2i \).
Such roots imply that the solution will involve trigonometric functions.
Complex Roots
When we solve a characteristic equation and end up with complex roots, the form of our solution changes.
Complex roots occur when you have equations like \( m^2 = -4 \), giving us \( m = \pm 2i \). But what do these roots mean for our solutions?
  • Complex roots of the form \( a \pm bi \) translate to solutions involving sine and cosine functions.
  • Specifically, if we have roots \( \pm bi \), the general solution becomes \( y = c_1 \cos(bx) + c_2 \sin(bx) \).
In our scenario, with roots \( \pm 2i \), the general solution is \( y = c_1 \cos(2x) + c_2 \sin(2x) \).
The appearance of cosine and sine in the solutions is due to Euler's formula, which connects complex exponentials with these trigonometric functions.
General Solution
The term *general solution* in differential equations refers to the solution that encompasses all possible solutions to the given differential equation.
It is derived from the structure of the characteristic roots.
  • For real distinct roots, the form is straightforward with exponential solutions.
  • For repeated roots, the solution involves terms that include powers of \( x \).
  • For complex roots, as we had in the exercise above, the general solution takes the trigonometric form of \( y = c_1 \cos(bx) + c_2 \sin(bx) \).
This solution is then particularized by applying initial or boundary conditions.
In the example provided, these conditions were \( y(0) = 0 \) and \( y(\pi) = 0 \).
These conditions allow us to solve for specific constants, here leading to \( c_1 = 0 \), resulting in the specific solution \( y = c_2 \sin(2x) \).
By using these conditions, we match our general solution to the specific scenario outlined in the problem.

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Most popular questions from this chapter

The auxiliary equation is \(m^{3}+m=m\left(m^{2}+1\right)=0,\) so \(y_{c}=c_{1}+c_{2} \cos x+c_{3} \sin x\) and $$W=\left|\begin{array}{rrr}1 & \cos x & \sin x \\\0 & -\sin x & \cos x \\ 0 & -\cos x & -\sin x\end{array}\right|=1$$ Identifying \(f(x)=\tan x\) we obtain $$u_{1}^{\prime}=W_{1}=\left|\begin{array}{ccr}0 & \cos x & \sin x \\\0 & -\sin x & \cos x \\ \tan x & -\cos x & -\sin x\end{array}\right|=\tan x$$ $$u_{2}^{\prime}=W_{2}=\left|\begin{array}{rrr}1 & 0 & \sin x \\\0 & 0 & \cos x \\\0 & \tan x & -\sin x\end{array}\right|=-\sin x$$ $$u_{3}^{\prime}=W_{3}=\left|\begin{array}{ccc}1 & \cos x & 0 \\\0 & -\sin x & 0 \\\0 & -\cos x & \tan x\end{array}\right|=-\sin x \tan x=\frac{\cos ^{2} x-1}{\cos x}=\cos x-\sec x$$ Then $$\begin{array}{l}u_{1}=-\ln |\cos x| \\\u_{2}=\cos x \\\u_{3}=\sin x-\ln |\sec x+\tan x| \end{array}$$ and $$y=c_{1}+c_{2} \cos x+c_{3} \sin x-\ln |\cos x|+\cos ^{2} x$$ $$+\sin ^{2} x-\sin x \ln |\sec x+\tan x|$$ $$=c_{4}+c_{2} \cos x+c_{3} \sin x-\ln |\cos x|-\sin x \ln |\sec x+\tan x|$$ for \(-\pi / 2

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 \(f(x)=\sec ^{2} x\) we obtain $$\begin{aligned}&u_{1}^{\prime}=-\frac{\sin x}{\cos ^{2} x}\\\&u_{2}^{\prime}=\sec x\end{aligned}$$ Then $$\begin{aligned}&u_{1}=-\frac{1}{\cos x}=-\sec x\\\&u_{2}=\ln |\sec x+\tan x|\end{aligned}$$ and $$\begin{aligned}y &=c_{1} \cos x+c_{2} \sin x-\cos x \sec x+\sin x \ln |\sec x+\tan x| \\\&=c_{1} \cos x+c_{2} \sin x-1+\sin x \ln |\sec x+\tan x|.\end{aligned}$$

From \((D+5) x+y=0\) and \(4 x-(D+1) y=0\) we obtain \(y=-(D+5) x\) so that \(D y=-\left(D^{2}+5 D\right) x .\) Then \(4 x+\left(D^{2}+5 D\right) x+(D+5) x=0\) and \((D+3)^{2} x=0 .\) Thus $$\begin{aligned} &x=c_{1} e^{-3 t}+c_{2} t e^{-3 t}\\\ &y=-\left(2 c_{1}+c_{2}\right) e^{-3 t}-2 c_{2} t e^{-3 t} .\end{aligned}$$ Using \(x(1)=0\) and \(y(1)=1\) we obtain $$\begin{array}{c}c_{1} e^{-3}+c_{2} e^{-3}=0 \\\\-\left(2 c_{1}+c_{2}\right) e^{-3}-2 c_{2} e^{-3}=1\end{array}$$ Or $$\begin{array}{c}c_{1}+c_{2}=0 \\\2 c_{1}+3 c_{2}=-e^{3}.\end{array}$$ Thus \(c_{1}=e^{3}\) and \(c_{2}=-e^{3} .\) The solution of the initial value problem is $$\begin{aligned}&x=e^{-3 t+3}-t e^{-3 t+3}\\\&y=-e^{-3 t+3}+2 t e^{-3 t+3}.\end{aligned}$$

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}\\].

Write the equation in the form $$y^{\prime \prime}+\frac{1}{x} y^{\prime}+\left(1-\frac{1}{4 x^{2}}\right) y=x^{-1 / 2}$$ and identify \(f(x)=x^{-1 / 2} .\) From \(y_{1}=x^{-1 / 2} \cos x\) and \(y_{2}=x^{-1 / 2} \sin x\) we compute $$W\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc} x^{-1 / 2} \cos x & x^{-1 / 2} \sin x \\\\-x^{-1 / 2} \sin x-\frac{1}{2} x^{-3 / 2} \cos x & x^{-1 / 2} \cos x-\frac{1}{2} x^{-3 / 2} \sin x\end{array}\right|=\frac{1}{x}$$ Now $$u_{1}^{\prime}=-\sin x \quad \text { so } \quad u_{1}=\cos x,$$ and $$u_{2}^{\prime}=\cos x \quad \text { so } \quad u_{2}=\sin x.$$ Thus a particular solution is $$y_{p}=x^{-1 / 2} \cos ^{2} x+x^{-1 / 2} \sin ^{2} x,$$ and the general solution is $$\begin{aligned}y &=c_{1} x^{-1 / 2} \cos x+c_{2} x^{-1 / 2} \sin x+x^{-1 / 2} \cos ^{2} x+x^{-1 / 2} \sin ^{2} x \\\&=c_{1} x^{-1 / 2} \cos x+c_{2} x^{-1 / 2} \sin x+x^{-1 / 2}.\end{aligned}$$
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