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Chapter 5: Problem 8

L?sen Sie die folgenden Schwingungsprobleme (aperiodische Schwingungen): a) \(\vec{x}+6 \dot{x}+5 x=0, \quad x(0)=10, \quad \dot{x}(0)=2\) b) \(\vec{x}+\dot{x}+0,16 x=0, \quad x(0)=2, \quad \dot{x}(0)=-4\) c) \(\ddot{x}+7 \dot{x}+12 x=0, \quad x(0)-5, \quad \dot{x}(0)=0\)

Short Answer

Expert verified
Solve the characteristic equation for each part, use initial conditions to find constants and write the specific solution.

Step by step solution

01

Understand the form of the differential equations

Each of the given differential equations is a second-order linear homogeneous differential equation with constant coefficients. The general form of such an equation is \( \frac{d^2 x}{dt^2} + a \frac{dx}{dt} + b x = 0 \). Identify the coefficients for each part.
02

Solve the characteristic equation for each problem

For each equation, solve the corresponding characteristic equation of the form \( r^2 + ar + b = 0 \) to find the roots \(r_1\) and \(r_2\).
03

Determine the general solution

Using the roots found from the characteristic equation in Step 2, write the general solution: If the roots are real and distinct, the solution is \( x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \). If the roots are complex, use the appropriate form based on whether they are real, repeated, or complex conjugates.
04

Apply initial conditions to find specific constants

Substitute the initial conditions \( x(0) = x_0 \) and \( \frac{dx}{dt}(0) = v_0 \) into the general solution from Step 3 to solve for the constants \( C_1 \) and \( C_2 \).
05

Write the specific solution for each problem

Using the values of \( C_1 \) and \( C_2 \) found in Step 4, write the specific solution for each given initial condition problem.

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Key Concepts

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

linear homogeneous differential equations
In this equation, \(a\) and \(b\) are constants that describe the system's properties. The term \frac{d^2 x}{dt^2} represents the acceleration,___frac{dx}{dt} represents the velocity, and \(x\) represents the position. Because there are no additional functions or terms involved, we call it homogeneous.
characteristic equation
The next step in solving these types of differential equations is to form what's called the characteristic equation. This comes from assuming a solution of the form \( x(t) = Ce^{rt} \), where \(C\) is a constant and \(r\) is the rate of growth or decay. By substituting this assumed solution into the differential equation, we get: ewline___frac{d^2 Ce^{rt}}{dt^2} + a\frac{d Ce^{rt}}{dt} + b Ce^{rt} = 0 onumber. ewlineFactoring out \(e^{rt}\) (which is never zero), we end up with: ewliner^2 + ar + b = 0 onumber, which is our characteristic equation.
general solution
The general solution of the second-order differential ensures it covers all possible behaviors related to our system. Depending on the nature of roots of the characteristic equation, the general solution can have different forms:
    If the roots are real and distinct, the solution would be+x(t)=C1e^{r1t}+C2e^{r2t}<. ul> If the roots are complex, we don't have separately real exponentials, use: x(t)=e^{rt}(C cos(wt)+sin(wt)) which accounts for using Euler's formula. Repeated roots result in the general solution of:x(t)=(C1+C2 e^{rt}.

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Most popular questions from this chapter

Das Anfangswertproblem $$ \ddot{x}+4 \dot{x}+29 x=0, \quad x(0)=1, \quad \dot{x}(0)=v(0)=-2 $$ beschreibt eine ged?mpfte (mechanische) Schwingung \((x:\) Auslenkung; \(v=\dot{x}:\) Geschwindigkeit). Wie groB sind Auslenkung und Geschwindigkeit zur Zeit \(t=0,1\) a) bei exakter Lösung, b) bei näherungsweiser Lösung der Differentialgleichung nach dem Runge- KuttaVerfahren 4. Ordnung (Schrittweite: \(h=\Delta t=0,05\) ).

Die Aufladung eines Kondensators der Kapazit?t \(C\) über einen ohmschen Widerstand \(R\) auf die Endspannung \(u_{0}\) erfolgt nach dem Exponentialgesetz $$ u_{C}(t)=u_{0}\left(1-\mathrm{e}^{-\frac{t}{R C}}\right) \quad(t \geqslant 0) $$ Zeigen Sie, daB diese Funktion eine (partikul?re) Lôsung der Differentialgleichung 1. Ordnung $$ R C \frac{d u_{C}}{d t}+u_{C}=u_{0} $$ ist, die diesen Einschaltworgang beschreibt (sog. RC-Glied, Bild V-63).

Bestimmen Sie die allgemeinen L?sungen der folgenden inhomogenen linearen Differentialgleichungen 2. Ordnung: a) \(y^{\prime \prime}+2 y^{\prime}-3 y=3 x^{2}-4 x\) b) \(y^{\prime \prime}-y=x^{3}-2 x^{2}-4\) c) \(\ddot{x}-2 \dot{x}+x=\mathrm{e}^{2 t}\) d) \(y^{\prime \prime}-2 y^{\prime}-3 y=-2 \cdot \mathrm{e}^{3 x}\) e) \(\quad \bar{x}+10 \dot{x}+25 x=3 \cdot \cos (5 t)\) f) \(y^{\prime \prime}+10 y^{\prime}-24 y=2 x^{2}-6 x\) g) \(\quad \vec{x}-x=t \cdot \sin t\) h) \(y^{\prime \prime}+12 y^{\prime}+36 y=3 \cdot \mathrm{e}^{-6 x}\) i) \(y^{\prime \prime}+4 y=10 \cdot \sin (2 x)+2 x^{2}-x+e^{-x}\) j) \(y^{\prime \prime}+2 y^{\prime}+y=x^{2} \cdot e^{x}+x-\cos x\)

Lösen Sie die folgenden Anfangswertprobleme: a) \(y^{\prime}+4 y=x^{3}-x, \quad y(1)=2\) b) \(y^{\prime}-y=\mathrm{e}^{x}, \quad y(0)=1\) c) \(y^{\prime}+3 y=-\cos x, \quad y(0)=5\)

Wie lauten die allgemeinen Lösungen der folgenden homogenen linearen Differentialgleichungen \(1 .\) Ordnung mit konstanten Koeffizienten? a) \(y^{\prime}+4 y=0\) b) \(\quad 2 y^{\prime}+4 y=0\) c) \(\quad-3 y^{\prime}=8 y\) d) \(a y^{\prime}-b y=0\) c) \(\quad \vec{n}=-\lambda n\) f) \(-3 y^{\prime}+18 y=0\) g) \(L \frac{d i}{d t}+R i=0\) h) \(\quad 2 \frac{d y}{d x}+18 y=0\) i) \(3 y^{\prime}-5 a y=0\) j) \(\quad T u+u=0\)
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