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Chapter 5: Problem 2

Gegeben ist die Differentialgleichung \(y^{\prime \prime}-4 y^{\prime}-5 y=0 .\) Zeigen Sie, da \(B\) diese Gleichung die allgemeine L?sung \(y=C_{1} \cdot \mathrm{e}^{5 x}+C_{2} \cdot \mathrm{e}^{-x}\) besitzt \(\left(C_{1}, C_{2} \in \mathbb{R}\right)\).

Short Answer

Expert verified
The general solution is \(y = C_1 e^{5x} + C_2 e^{-x}\).

Step by step solution

01

- Find the characteristic equation

For the differential equation \(y'' - 4y' - 5y = 0\), assume a solution of the form \(y = e^{rx}\). Substitute \(y\), \(y'\) and \(y''\) into the equation to get the characteristic equation: \(r^2e^{rx} - 4re^{rx} - 5e^{rx} = 0\).
02

- Simplify the characteristic equation

Factor out \(e^{rx}\), which is never zero, to obtain the simpler characteristic equation \(r^2 - 4r - 5 = 0\).
03

- Solve the characteristic equation

Solve the quadratic equation \(r^2 - 4r - 5 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). Here, \(a = 1\), \(b = -4\), and \(c = -5\). Substituting these values gives \(r = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}\).
04

- Find the roots

The solutions to the quadratic equation are \(r_1 = 5\) and \(r_2 = -1\).
05

- Write the general solution

For the roots \(r_1 = 5\) and \(r_2 = -1\), the general solution to the differential equation takes the form \(y = C_1 e^{r_1 x} + C_2 e^{r_2 x} = C_1 e^{5x} + C_2 e^{-x}\), where \(C_1\) and \(C_2\) are arbitrary constants.

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Key Concepts

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

Characteristic Equation
When dealing with second-order linear differential equations such as the given differential equation \(y'' - 4y' - 5y = 0\), the first step is to find the characteristic equation. This is done by assuming a solution of the form \(y = e^{rx}\).
By substituting \(y = e^{rx}\), \(y' = re^{rx}\), and \(y''=r^2e^{rx}\) into the differential equation, we get:
\[r^2e^{rx} - 4re^{rx} - 5e^{rx} = 0\]
Factoring out \(e^{rx}\), which can never be zero, we simplify to get the characteristic equation:
\[r^2 - 4r - 5 = 0\].
The characteristic equation is essential because its solutions will help us determine the general solution of the differential equation.
Quadratic Formula
Solving the characteristic equation involves using the quadratic formula. In our case, the characteristic equation is \(r^2 - 4r - 5 = 0\).
The quadratic formula is given by:
\[r = \frac{-b \textpm \sqrt{b^2 - 4ac}}{2a}\]
Here, \a = 1, b = -4\, and \c = -5\. Substituting these values into the quadratic formula, we get:
\[r = \frac{4 \textpm \sqrt{16 + 20}}{2} = \frac{4 \textpm \sqrt{36}}{2} = \frac{4 \textpm 6}{2} \]
which simplifies to:
\[r_1 = 5\ and \r_2 = -1\].
Using the quadratic formula allows us to find the roots of the characteristic equation, which are crucial for determining the form of the general solution.
General Solution
For a second-order linear differential equation with constant coefficients, the general solution depends on the roots of the characteristic equation. In our case, the roots are \(r_1 = 5\) and \(r_2 = -1\).
Therefore, the general solution of the differential equation is given by:
\[y = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]
where \(C_1\) and \(C_2\) are arbitrary constants.
Substituting the values of \(r_1\) and \(r_2\), we get:
\[y = C_1 e^{5x} + C_2 e^{-x}\]
This form covers all possible solutions to the differential equation. The constants \(C_1\) and \(C_2\) can be determined if initial conditions are provided.
Roots of Equation
The roots of the characteristic equation provide critical information about the behavior of solutions to the differential equation. In the equation \(r^2 - 4r - 5 = 0\), we found the roots to be \(r_1 = 5\) and \(r_2 = -1\).
These roots are distinct real numbers, which means that the general solution will be of the form:
\[y = C_1 e^{5x} + C_2 e^{-x}\].
Different types of roots lead to different forms of solutions:
  • If the roots are repeated (e.g., both roots are the same), the solution would include a term \(x e^{rx}\).
  • If the roots are complex, the solution would involve sine and cosine functions.

For our current problem, since the roots are real and distinct, the solution is simply a linear combination of the exponential functions corresponding to these roots.

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Most popular questions from this chapter

Die folgenden Anfangswertprobleme beschreiben mechanische Schwingungen im aperiodischen Grenz fall. Wie lauten die L?sungen? a) \(2 \vec{x}+10 \dot{x}+12,5 x=0, \quad x(0)=5, \quad \dot{x}(0)=1\) b) \(\quad \ddot{x}+\dot{x}+0,25 x=0, \quad x(0)=1, \quad \dot{x}(0)=-1\)

Lösen Sie das Anfangswertproblem $$ \dot{\mathbf{x}}=\left(\begin{array}{ll} 1 & 4 \\ 1 & 1 \end{array}\right) \mathbf{x}+\left(\begin{array}{r} -1 \\ 2 \end{array}\right) \mathrm{e}^{t}, \quad \mathbf{x}(0)=\left(\begin{array}{c} -0,5 \\ 0 \end{array}\right) $$ a) durch „Aufsuchen einer partikulären Lösung", b) nach dem Einsetzungs- oder Eliminationsverfahren.

Ein K?rper wird zur Zeit \(t=0\) aus der H?he \(s_{0}=10 \mathrm{~m}\) mit der Anfangsgeschwindigkeit \(v_{0}=30 \mathrm{~m} / \mathrm{s}\) senkrecht nach oben geworien (sog. senkrechter Wurf). Bestimmen Sie das Weg-Zeit-Gesetz \(s=s(t)\) und das Geschwindigkeit-Zeit-Ge\(\operatorname{setz} v=v(t)\). Anleitung: Die Bewegung genügt der Differentialgleichung \(\ddot{s}=-g(g:\) Erdbeschleunigung: vgl. hierzu auch das einführende Beispiel in Abschnitt V.1.1).

Ein Stromkreis mit einem zeitabhängigen ohmschen Widerstand werde durch die Differentialgleichung 1. Ordnung $$ \frac{d i}{d t}+(2 \cdot \sin t) \cdot i=\sin (2 t) \quad(t \geqslant 0) $$ beschrieben. Ermitteln Sie den zeitlichen Verlauf der Stromstärke \(i\) durch Variation der Konstanten für den Anfangswert \(l(0)=0\).

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) \(\ddot{x}+7 \dot{x}+12 x=0, \quad x(0)-5, \quad \dot{x}(0)=0\)
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