/*! 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 35 From \(m^{3}+12 m^{2}+36 m=0\) w... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 35

From \(m^{3}+12 m^{2}+36 m=0\) we obtain \(m=0, m=-6,\) and \(m=-6\) so that \(y=c_{1}+c_{2} e^{-6 x}+c_{3} x e^{-6 x} .\) If \(y(0)=0, y^{\prime}(0)=1,\) and \(y^{\prime \prime}(0)=-7\) then $$c_{1}+c_{2}=0, \quad-6 c_{2}+c_{3}=1, \quad 36 c_{2}-12 c_{3}=-7$$ so \(c_{1}=5 / 36, c_{2}=-5 / 36, c_{3}=1 / 6,\) and \(y=\frac{5}{36}-\frac{5}{36} e^{-6 x}+\frac{1}{6} x e^{-6 x}\)

Short Answer

Expert verified
The constants are: \(c_1 = \frac{5}{36}, c_2 = -\frac{5}{36}, c_3 = \frac{1}{6}\). The solution is \(y = \frac{5}{36} - \frac{5}{36} e^{-6x} + \frac{1}{6} x e^{-6x}\).

Step by step solution

01

Identify roots of the polynomial

Given the polynomial \(m^3 + 12m^2 + 36m = 0\), we first need to factor it to find the roots. Notice that this is a cubic equation and can be factored as \(m(m^2 + 12m + 36) = 0\). The quadratic factor \(m^2 + 12m + 36\) can be written as \((m + 6)^2\), giving us three roots: \(m = 0, m = -6, m = -6\).
02

Write the general solution

Using the roots \(m = 0, m = -6\), and \(m = -6\), the general solution for the differential equation is given by \(y = c_1 + c_2 e^{-6x} + c_3 x e^{-6x}\), which accounts for the double root \(m = -6\).
03

Apply initial conditions to find constants

We have the conditions \(y(0) = 0\), \(y'(0) = 1\), and \(y''(0) = -7\). We substitute \(x = 0\) into the general solution and its derivatives to form the system of equations. For \(y(0) = 0\):\(c_1 + c_2 = 0\).For \(y'(0) = 1\), compute the derivative: \(y' = -6c_2 e^{-6x} + c_3 e^{-6x} - 6c_3 x e^{-6x}\) and substitute to get: \(-6c_2 + c_3 = 1\).For \(y''(0) = -7\), calculate the second derivative and substitute to find: \(36c_2 - 12c_3 = -7\).
04

Solve the system of equations

We have the system of equations:1. \(c_1 + c_2 = 0\)2. \(-6c_2 + c_3 = 1\)3. \(36c_2 - 12c_3 = -7\)Substitute \(c_1 = -c_2\) from equation 1 into the system. Solve equation 2 for \(c_3\): \(c_3 = 1 + 6c_2\). Substitute \(c_3\) into equation 3:\(36c_2 - 12(1 + 6c_2) = -7\) simplifies to:\(36c_2 - 12 - 72c_2 = -7\),\(-36c_2 = 5\),\(c_2 = -\frac{5}{36}\).Then \(c_1 = -(-\frac{5}{36}) = \frac{5}{36}\), and \(c_3 = 1 + 6(-\frac{5}{36}) = \frac{1}{6}\).
05

Write the final solution

With \(c_1 = \frac{5}{36}\), \(c_2 = -\frac{5}{36}\), and \(c_3 = \frac{1}{6}\), substitute these into the general solution:\[y = \frac{5}{36} - \frac{5}{36} e^{-6x} + \frac{1}{6} x e^{-6x}\]

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Key Concepts

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

Polynomial Roots
Understanding polynomial roots is crucial when dealing with differential equations. Polynomial roots are the values of the variable that make the polynomial equal to zero. In our exercise, we have the polynomial equation \(m^3 + 12m^2 + 36m = 0\). To find its roots, we first factor the polynomial. This particular equation can be factored as \(m(m^2 + 12m + 36)\).

The quadratic part, \(m^2 + 12m + 36\), can be further factored as \((m + 6)^2\). This gives us three roots: \(m = 0\), \(m = -6\), and \(m = -6\) (a double root).

Double roots occur when a factor is repeated. Recognizing these roots correctly is vital, as they determine the form of the solution to the differential equation.
Initial Conditions
Initial conditions are values given for the solution at a specific point. They help determine the specific solution from the general solution of a differential equation. In this problem, we have three initial conditions: \(y(0) = 0\), \(y'(0) = 1\), and \(y''(0) = -7\).

These conditions allow us to form a system of equations using the general solution and its derivatives. Initial conditions are useful because they ensure that our solution satisfies specific criteria at a particular point.

Applying these to our general solution鈥攁fter evaluating the function and its derivatives at \(x = 0\)鈥攔esults in three equations. Solving these equations yields the constants needed for our final specific solution.
General Solution
The general solution is the most comprehensive form of a solution to a differential equation. It involves arbitrary constants whose values are determined by initial conditions. Here, given the roots \(m = 0 \) and \(m = -6\) (with a double occurrence), the general solution takes the form:
  • \(y = c_1 + c_2 e^{-6x} + c_3 x e^{-6x}\)
The presence of a double root \(m = -6\) indicates that we need two terms related to this root in the solution. Hence, \(c_2 e^{-6x}\) and \(c_3 x e^{-6x}\) are included.

The constants \(c_1\), \(c_2\), and \(c_3\) need to be defined via initial conditions to transform this general expression into a particular solution.
System of Equations
A system of equations is a set of equations with multiple variables. Solving a system means finding values for the variables that satisfy all the equations simultaneously. In this exercise, the general solution and its derivatives lead to a system of three equations:
  • \(c_1 + c_2 = 0\)
  • \(-6c_2 + c_3 = 1\)
  • \(36c_2 - 12c_3 = -7\)
These equations arise from substituting the initial conditions into the respective expressions for \(y(0)\), \(y'(0)\), and \(y''(0)\).

Solving this system involves methods like substitution or elimination to find the values of \(c_1\), \(c_2\), and \(c_3\). Each equation provides essential information that helps isolate the constants, resulting in the particular solution of the differential equation.

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Most popular questions from this chapter

(a) Let \((r, \theta)\) denote the polar coordinates of the destroyer \(S_{1}\). When \(S_{1}\) travels the 6 miles from (9,0) to (3,0) it stands to reason, since \(S_{2}\) travels half as fast as \(S_{1}\), that the polar coordinates of \(S_{2}\) are \(\left(3, \theta_{2}\right),\) where \(\theta_{2}\) is unknown. In other words, the distances of the ships from (0,0) are the same and \(r(t)=15 t\) then gives the radial distance of both ships. This is necessary if \(S_{1}\) is to intercept \(S_{2}\) (b) The differential of arc length in polar coordinates is \((d s)^{2}=(r d \theta)^{2}+(d r)^{2},\) so that \\[ \left(\frac{d s}{d t}\right)^{2}=r^{2}\left(\frac{d \theta}{d t}\right)^{2}+\left(\frac{d r}{d t}\right)^{2} \\] Using \(d s / d t=30\) and \(d r / d t=15\) then gives \\[ \begin{aligned} 900 &=225 t^{2}\left(\frac{d \theta}{d t}\right)^{2}+225 \\ 675 &=225 t^{2}\left(\frac{d \theta}{d t}\right)^{2} \\ \frac{d \theta}{d t} &=\frac{\sqrt{3}}{t} \\ \theta(t) &=\sqrt{3} \ln t+c=\sqrt{3} \ln \frac{r}{15}+c \end{aligned} \\] When \(r=3, \theta=0,\) so \(c=-\sqrt{3} \ln \frac{1}{5}\) and \\[ \theta(t)=\sqrt{3}\left(\ln \frac{r}{15}-\ln \frac{1}{5}\right)=\sqrt{3} \ln \frac{r}{3} \\] Thus \(r=3 e^{\theta / \sqrt{3}},\) whose graph is a logarithmic spiral. (c) The time for \(S_{1}\) to go from (9,0) to \((3,0)=\frac{1}{5}\) hour. Now \(S_{1}\) must intercept the path of \(S_{2}\) for some angle \(\beta,\) where \(0<\beta<2 \pi .\) At the time of interception \(t_{2}\) we have \(15 t_{2}=3 e^{\beta / \sqrt{3}}\) or \(t=\frac{1}{5} e^{\beta / \sqrt{3}} .\) The total time is then \\[ t=\frac{1}{5}+\frac{1}{5} e^{\beta / \sqrt{3}}<\frac{1}{5}\left(1+e^{2 \pi / \sqrt{3}}\right) \\].

(a) When \(T(x)=x^{2}\) the given differential equation is the Cauchy-Euler equation \\[ x^{2} y^{\prime \prime}+2 x y^{\prime}+\rho \omega^{2} y=0 \\] The solutions of the auxiliary equation \\[ m(m-1)+2 m+\rho \omega^{2}=m^{2}+m+\rho \omega^{2}=0 \\] are \\[ m_{1}=-\frac{1}{2}-\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} i, \quad m_{2}=-\frac{1}{2}+\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} \\] when \(\rho \omega^{2}>0.25 .\) Thus \\[ y=c_{1} x^{-1 / 2} \cos (\lambda \ln x)+c_{2} x^{-1 / 2} \sin (\lambda \ln x) \\] where \(\lambda=\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} .\) Applying \(y(1)=0\) gives \(c_{1}=0\) and consequently \\[ y=c_{2} x^{-1 / 2} \sin (\lambda \ln x) \\]. The condition \(y(e)=0\) requires \(c_{2} e^{-1 / 2} \sin \lambda=0 . \quad\) We obtain a nontrivial solution when \(\lambda_{n}=n \pi\) \(n=1,2,3, \ldots .\) But $$\lambda_{n}=\frac{1}{2} \sqrt{4 \rho \omega_{n}^{2}-1}=n \pi$$ Solving for \(\omega_{n}\) gives \\[ \omega_{n}=\frac{1}{2} \sqrt{\left(4 n^{2} \pi^{2}+1\right) / \rho} \\]. The corresponding solutions are \\[ y_{n}(x)=c_{2} x^{-1 / 2} \sin (n \pi \ln x) \\]. b.

If \(1-i\) is a root of the auxiliary equation then so is \(1+i,\) and the auxiliary equation is $$(m-2)[m-(1+i)][m-(1-i)]=m^{3}-4 m^{2}+6 m-4=0.$$ We need \(m^{3}-4 m^{2}+6 m-4\) to have the form \(m(m-1)(m-2)+b m(m-1)+c m+d .\) Expanding this last expression and equating coefficients we get \(b=-1, c=3,\) and \(d=-4 .\) Thus, the differential equation is $$x^{3} y^{\prime \prime \prime}-x^{2} y^{\prime \prime}+3 x y^{\prime}-4 y=0.$$

$$\text { since } x=\frac{\sqrt{85}}{4} \sin (4 t-0.219)-\frac{\sqrt{17}}{2} e^{-2 t} \sin (4 t-2.897), \text { the amplitude approaches } \sqrt{85} / 4 \text { as } t \rightarrow \infty$$.

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 $$y=c_{1} e^{-4 x} \cos x+c_{2} e^{-4 x} \sin x+c_{3} e^{2 x}$$
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