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From \(0.75 x^{\prime \prime}+\beta x^{\prime}+6 x=0\) and \(\beta>3 \sqrt{2}\) we find that the roots of the auxiliary equation are \(m=-\frac{2}{3} \beta \pm \frac{2}{3} \sqrt{\beta^{2}-18}\) and \\[x=e^{-2 \beta t / 3}\left[c_{1} \cosh \frac{2}{3} \sqrt{\beta^{2}-18} t+c_{2} \sinh \frac{2}{3} \sqrt{\beta^{2}-18} t\right]\\] If \(x(0)=0\) and \(x^{\prime}(0)=-2\) then \(c_{1}=0\) and \(c_{2}=-3 / \sqrt{\beta^{2}-18}\).

Short Answer

Expert verified
\(c_1 = 0\) and \(c_2 = -\frac{3}{\sqrt{\beta^2-18}}\).

Step by step solution

01

Find the Characteristic Equation

For the differential equation \(0.75 x'' + \beta x' + 6x = 0\), we propose a solution of the form \(x = e^{mt}\). This leads to the characteristic equation \(0.75m^2 + \beta m + 6 = 0\).
02

Determine the Roots

Solve the characteristic equation \(0.75m^2 + \beta m + 6 = 0\) using the quadratic formula. The solutions are \(m = -\frac{\beta}{1.5} \pm \frac{1}{1.5} \sqrt{\beta^2 - 18}\). Simplifying gives the roots as \(m = -\frac{2}{3}\beta \pm \frac{2}{3} \sqrt{\beta^2 - 18}\).
03

Form the General Solution

With roots \(m_1\) and \(m_2\), the general solution is \(x(t) = e^{-\frac{2}{3}\beta t}(c_1 \cosh\left(\frac{2}{3}\sqrt{\beta^2-18} t\right) + c_2 \sinh\left(\frac{2}{3}\sqrt{\beta^2-18} t\right))\).
04

Apply Initial Conditions to Compute Constants

Apply the initial condition \(x(0) = 0\) to get \(c_1 \cosh(0) + c_2 \sinh(0) = 0\), resulting in \(c_1 = 0\). Next, for \(x'(0) = -2\), differentiate the general solution and substitute \(t = 0\) to get \(x'(0) = -\frac{2}{3}\beta c_1 + \frac{2}{3}\sqrt{\beta^2-18} c_2 = -2\). Substituting \(c_1 = 0\), solve for \(c_2\), resulting in \(c_2 = -\frac{3}{\sqrt{\beta^2-18}}\).
05

Verify the Solution

Plug back the values of \(c_1\) and \(c_2\) into the general solution and its derivative to ensure they satisfy both initial conditions.

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Key Concepts

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

Characteristic Equation
In the context of solving differential equations, the characteristic equation is a core concept. For the given differential equation \[0.75 x'' + \beta x' + 6x = 0,\]we start by assuming a solution of the type\[x = e^{mt}.\]This assumption allows us to transform the original equation into what we call the characteristic equation:\[0.75m^2 + \beta m + 6 = 0.\]This equation is crucial as it transforms the problem from a differential equation into an algebraic one, making it easier to handle. Finding the characteristic equation is a vital step because its roots give us insight into the form of the general solution. It essentially tells us how the solution behaves over time, including whether it oscillates or decays.
Roots of a Polynomial
Once we have the characteristic equation \[0.75m^2 + \beta m + 6 = 0,\]our next task is to solve it for its roots. This involves applying the quadratic formula:\[m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]where in our specific context, \(a = 0.75\), \(b = \beta\), and \(c = 6\). When you plug these values into the formula, you get:\[m = -\frac{2}{3}\beta \pm \frac{2}{3} \sqrt{\beta^2 - 18}.\]These roots determine the form of our solution. If the term under the square root (\[\beta^2 - 18\]) is positive, the roots are real and distinct. This affects the usage of hyperbolic functions in the solution. Understanding roots is key, as they provide the fundamental building blocks for constructing the full solution.
Initial Conditions
With our general solution constructed using the roots of the characteristic equation, \[x(t) = e^{-\frac{2}{3}\beta t}\left(c_1 \cosh\left(\frac{2}{3}\sqrt{\beta^2-18} t\right) + c_2 \sinh\left(\frac{2}{3}\sqrt{\beta^2-18} t\right)\right),\]we need specific values to solve for the constants \(c_1\) and \(c_2\). This is where initial conditions come into play. For this problem, we are given:
  • \(x(0) = 0\)
  • \(x'(0) = -2\)
Plugging \(t = 0\) into the general solution and its derivative gives equations that allow us to isolate \(c_1\) and \(c_2\). Initial conditions are essential because they tailor the general solution to a specific scenario. Without them, the solution remains too vague, providing only a general behavior rather than a precise answer.
General Solution
By combining the roots of the characteristic equation with our hyperbolic functions, we form the general solution. It captures all possible behaviors of the system described by the differential equation:\[x(t) = e^{-\frac{2}{3}\beta t}\left(c_1 \cosh\left(\frac{2}{3}\sqrt{\beta^2-18} t\right) + c_2 \sinh\left(\frac{2}{3}\sqrt{\beta^2-18} t\right)\right).\]The presence of \(\cosh\) and \(\sinh\) suggests that these functions handle scenarios where the roots are real, which usually leads to exponential growth or decay rather than oscillations. In this way, the general solution offers a comprehensive overview of potential system behaviors before specific initial conditions are applied. By using this framework, we can accommodate a wide range of scenarios just by tweaking \(c_1\) and \(c_2\) based on how the system starts.

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Most popular questions from this chapter

From \(m^{2}+1=0\) we obtain \(m=\pm i\) so that \(y=c_{1} \cos x+c_{2} \sin x\) and \(y^{\prime}=-c_{1} \sin x+c_{2} \cos x .\) From \(y^{\prime}(0)=c_{1}(0)+c_{2}(1)=c_{2}=0\) and \(y^{\prime}(\pi / 2)=-c_{1}(1)=0\) we find \(c_{1}=c_{2}=0 .\) A solution of the boundary-value problem is \(y=0\).

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 \(y_{4}=\sin 10.8127 x\).

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 the differential equation is \(y=c_{1} \cos 4 x+c_{2} \sin 4 x .\) From \(y_{0}=y(0)=c_{1}\) we see that \(y=y_{0} \cos 4 x+c_{2} \sin 4 x .\) From \(y_{1}=y(\pi / 2)=y_{0}\) we see that any solution must satisfy \(y_{0}=y_{1} .\) We also see that when \(y_{0}=y_{1}, y=y_{0} \cos 4 x+c_{2} \sin 4 x\) is a solution of the boundary-value problem for any choice of \(c_{2}\). Thus, the boundary-value problem does not have a unique solution for any choice of \(y_{0}\) and \(y_{1}\). (b) Whenever \(y_{0}=y_{1}\) there are infinitely many solutions. (c) When \(y_{0} \neq y_{1}\) there will be no solutions. (d) The boundary-value problem will have the trivial solution when \(y_{0}=y_{1}=0 .\) This solution will not be unique.

Using the double-angle formula for the cosine, we have $$\sin x \cos 2 x=\sin x\left(\cos ^{2} x-\sin ^{2} x\right)=\sin x\left(1-2 \sin ^{2} x\right)=\sin x-2 \sin ^{3} x$$ since \(\sin x\) is a solution of the related homogeneous differential equation we look for a particular solution of the form \(y_{p}=A x \sin x+B x \cos x+C \sin ^{3} x .\) Substituting into the differential equation we obtain $$2 A \cos x+(6 C-2 B) \sin x-8 C \sin ^{3} x=\sin x-2 \sin ^{3} x$$ Equating coefficients we find \(A=0, C=\frac{1}{4},\) and \(B=\frac{1}{4} .\) Thus, a particular solution is $$y_{p}=\frac{1}{4} x \cos x+\frac{1}{4} \sin ^{3} x$$

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