/*! 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 1 From \(m^{2}+3 m+2=0\) we find \... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 1

From \(m^{2}+3 m+2=0\) we find \(m_{1}=-1\) and \(m_{2}=-2 .\) Then \(y_{c}=c_{1} e^{-x}+c_{2} e^{-2 x}\) and we assume \(y_{p}=A\) Substituting into the differential equation we obtain \(2 A=6 .\) Then \(A=3, y_{p}=3\) and \(y=c_{1} e^{-x}+c_{2} e^{-2 x}+3\).

Short Answer

Expert verified
The general solution is \(y = c_1 e^{-x} + c_2 e^{-2x} + 3\).

Step by step solution

01

Recognize the Type of Equation

The given equation is a homogeneous second-order differential equation with constant coefficients. The general solution consists of a complementary function (denoted as \(y_c\)) and a particular solution (denoted as \(y_p\)).
02

Solve the Auxiliary Equation

The characteristic equation for the differential equation is given by \(m^2 + 3m + 2 = 0\). Factoring this gives \((m+1)(m+2)=0\), leading to roots \(m_1 = -1\) and \(m_2 = -2\).
03

Complementary Solution

Using the roots from the auxiliary equation, the complementary solution is \(y_c = c_1 e^{-x} + c_2 e^{-2x}\), where \(c_1\) and \(c_2\) are constants.
04

Assume a Particular Solution

Assume a particular solution of the form \(y_p = A\), where \(A\) is a constant. This is because the non-homogeneous part is a constant (i.e., 6).
05

Substitute for Particular Solution

Substitute \(y_p = A\) into the differential equation. This gives \(2A = 6\). Solve for \(A\) by dividing both sides by 2, resulting in \(A = 3\).
06

General Solution

The general solution of the differential equation is the sum of the complementary and particular solutions: \(y = c_1 e^{-x} + c_2 e^{-2x} + 3\).

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Key Concepts

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

Understanding Constant Coefficients in Differential Equations
When dealing with differential equations, especially second-order, the term "constant coefficients" refers to when the coefficients of the derivatives are constants. In the equation \(m^2 + 3m + 2 = 0\), the coefficients are simply 1, 3, and 2. These do not change with the variable (e.g., \(x\) or \(t\)), which makes the equation easier to solve. Constant coefficients allow us to employ various powerful methods, such as using the characteristic equation to find solutions.
This is particularly straightforward compared to variable coefficients, where the coefficients could be functions of the variable. With constant coefficients, once we have the linear differential equation, we can quickly move to solve it by using algebraic techniques.
Unveiling the Characteristic Equation
The characteristic equation is pivotal in finding solutions to linear differential equations with constant coefficients. For a second-order equation such as \(m^2 + 3m + 2 = 0\), this equation is derived from replacing the derivatives in the differential equation with powers of \(m\).
The characteristic equation's roots, which can be obtained by factoring or using the quadratic formula, guide you to the solution structure of the differential equation. In our example, solving \((m+1)(m+2)=0\) gives roots \(m_1 = -1\) and \(m_2 = -2\). These roots are crucial because they dictate the form of the complementary solution and help identify the behavior of the system described by the equation.
Finding the Complementary Solution
Once you have the roots from the characteristic equation, you're ready to construct the complementary solution. This solution, \(y_c\), forms part of the general solution to the differential equation. For distinct real roots like \(-1\) and \(-2\), the complementary solution is expressed as a sum of exponential terms. Specifically, \(y_c = c_1 e^{-x} + c_2 e^{-2x}\), where \(c_1\) and \(c_2\) are arbitrary constants.
The complementary solution represents the homogeneous part of the differential equation's solution, the part that would solve the equation if there were no external forces or inputs. This will always involve terms related to the roots of the characteristic equation, reflecting natural modes of the system's behavior.
Determining the Particular Solution
The particular solution, \(y_p\), accounts for any non-homogeneous part of the differential equation. In cases where the independent term is a constant, like 6 in our example, a natural assumption is \(y_p = A\), where \(A\) is a constant.
To find \(A\), substitute \(y_p = A\) into the differential equation and solve for \(A\). For our equation, substituting gives \(2A = 6\), leading us to \(A = 3\). The particular solution ties the entire equation to the specific external influences or inputs it experiences. Adding this to the complementary solution results in the general solution of the form \(y = c_1 e^{-x} + c_2 e^{-2x} + 3\), which covers both natural and forced behaviors of the system.

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Most popular questions from this chapter

The thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form \\[ \begin{array}{c} y(x)=y(0)+y^{\prime}(0) x+\frac{1}{2 !} y^{\prime \prime}(0) x^{2}+\frac{1}{3 !} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{4 !} y^{(4)}(0) x^{4} \\ +\frac{1}{5 !} y^{(5)}(0) x^{5}+\frac{1}{6 !} y^{(6)}(0) x^{6} \end{array} \\] From \(y^{\prime \prime}(x)=e^{y}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=e^{y} y^{\prime} \\ y^{(4)}(x) &=e^{y}\left(y^{\prime}\right)^{2}+e^{y} y^{\prime \prime} \\ y^{(5)}(x) &=e^{y}\left(y^{\prime}\right)^{3}+3 e^{y} y^{\prime} y^{\prime \prime}+e^{y} y^{\prime \prime \prime} \\ y^{(6)}(x) &=e^{y}\left(y^{\prime}\right)^{4}+6 e^{y}\left(y^{\prime}\right)^{2} y^{\prime \prime}+3 e^{y}\left(y^{\prime \prime}\right)^{2}+4 e^{y} y^{\prime} y^{\prime \prime \prime}+e^{y} y^{(4)} \end{aligned} \\] Using \(y(0)=0\) and \(y^{\prime}(0)=-1\) we find \\[ y^{\prime \prime}(0)=1, \quad y^{\prime \prime \prime}(0)=-1, \quad y^{(4)}(0)=2, \quad y^{(5)}(0)=-5, \quad y^{(6)}(0)=16 \\] An approximate solution is \\[ y(x)=-x+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}+\frac{1}{12} x^{4}+\frac{1}{24} x^{5}+\frac{1}{45} x^{6} .\\]

Solving \(\frac{1}{2} q^{\prime \prime}+20 q^{\prime}+1000 q=0\) we obtain \(q_{c}(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The steady- state charge has the form \(q_{p}(t)=A \sin 60 t+B \cos 60 t+C \sin 40 t+D \cos 40 t .\) Substituting into the differential equation we find $$\begin{aligned}(-1600 A-2400 B) & \sin 60 t+(2400 A-1600 B) \cos 60 t \\ +&(400 C-1600 D) \sin 40 t+(1600 C+400 D) \cos 40 t\end{aligned}$$ $$=200 \sin 60 t+400 \cos 40 t$$ Equating coefficients we obtain \(A=-1 / 26, B=-3 / 52, C=4 / 17,\) and \(D=1 / 17 .\) The steady-state charge is \\[q_{p}(t)=-\frac{1}{26} \sin 60 t-\frac{3}{52} \cos 60 t+\frac{4}{17} \sin 40 t+\frac{1}{17} \cos 40 t\\] and the steady-state current is \\[i_{p}(t)=-\frac{30}{13} \cos 60 t+\frac{45}{13} \sin 60 t+\frac{160}{17} \cos 40 t-\frac{40}{17} \sin 40 t\\].

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.

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

(a) The auxiliary equation is \(m^{2}-64 / L=0\) which has roots \(\pm 8 / \sqrt{L}\). Thus, the general solution of the differential equation is \(x=c_{1} \cosh (8 t / \sqrt{L})+c_{2} \sinh (8 t / \sqrt{L})\) (b) Setting \(x(0)=x_{0}\) and \(x^{\prime}(0)=0\) we have \(c_{1}=x_{0}, 8 c_{2} / \sqrt{L}=0 .\) Solving for \(c_{1}\) and \(c_{2}\) we get \(c_{1}=x_{0}\) and \(c_{2}=0,\) so \(x(t)=x_{0} \cosh (8 t / \sqrt{L})\) (c) When \(L=20\) and \(x_{0}=1, x(t)=\cosh (4 t / \sqrt{5}) .\) The chain will last touch the peg when \(x(t)=10\) Solving \(x(t)=10\) for \(t\) we get \(t_{1}=\frac{1}{4} \sqrt{5} \cosh ^{-1} 10 \approx 1.67326 .\) The velocity of the chain at this instant is \(x^{\prime}\left(t_{1}\right)=12 \sqrt{11 / 5} \approx 17.7989 \mathrm{ft} / \mathrm{s}\).
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