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Chapter 6: Problem 9

Losen Sie die folgenden Anfangswertprobleme: a) \(y^{\prime \prime}+4 y=0, \quad y(0)=2, y^{\prime}(0)=1\) b) \(y^{\prime \prime}+6 y^{\prime}+10 y=0, \quad y(0)=0, \quad y^{\prime}(0)=4\) c) \(y^{\prime \prime}+y^{\prime}=e^{-21}, \quad y(0)=0, \quad y^{\prime}(0)=1\) d) \(y^{\prime \prime}+2 y^{\prime}-3 y=2 t, \quad y(0)=1, \quad y^{\prime}(0)=0\)

Short Answer

Expert verified
a) \(y(t) = 2 \cos(2t) + \frac{1}{2} \sin(2t)\); b) \(y(t) = e^{-3 t} \frac{4}{3} \sin(t)\); c) \(y(t) = -1 + \frac{5}{3} e^{-t} - \frac{1}{3} e^{-2t}\); d) \(y(t) = e^{-3t} + \frac{5}{3} e^{t} - \frac{2}{3}t - 1\).

Step by step solution

01

Introduction

This problem requires solving four different initial value problems (IVPs). Each IVP involves a differential equation with given initial conditions.
02

Solve part (a) - Characteristic Equation

Given the differential equation \(y'' + 4y = 0\), find the characteristic equation: \(r^2 + 4 = 0\).
03

Find General Solution for part (a)

Solve the characteristic equation: \(r = \pm 2i\). Therefore, the general solution is \(y(t) = C_1 \cos(2t) + C_2 \sin(2t)\).
04

Apply Initial Conditions for part (a)

Use the initial conditions \(y(0)=2\) and \(y'(0)=1\) to find the constants \(C_1\) and \(C_2\): \(y(0) = C_1 = 2\), \(y'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t), \ y'(0) = 2C_2 = 1, \ C_2 = \frac{1}{2}\). Therefore, the solution is \(y(t) = 2 \cos(2t) + \frac{1}{2} \sin(2t)\).
05

Solve part (b) - Characteristic Equation

Given the differential equation \(y'' + 6y' + 10y = 0\), find the characteristic equation: \(r^2 + 6r + 10 = 0\).
06

Find General Solution for part (b)

Solve the characteristic equation: \(r = -3 \pm i\). Therefore, the general solution is: \(y(t) = e^{-3t}(C_1 \cos(t) + C_2 \sin(t))\).
07

Apply Initial Conditions for part (b)

Use the initial conditions \(y(0)=0\) and \(y'(0)=4\) to find the constants \(C_1\) and \(C_2\): \(y(0) = C_1 = 0\), \(y'(t) = e^{-3t}(-3C_1 \cos(t) - C_1 \sin(t) + 3C_2 \cos(t) + C_2 \sin(t)),\ y'(0) = 3C_2 = 4, \ C_2 = \frac{4}{3}\). Therefore, the solution is \(y(t) = e^{-3 t} \frac{4}{3} \sin(t)\).
08

Solve part (c) - Non-Homogeneous Equation

Given the differential equation \(y'' + y' = e^{-2t}\), first solve the homogeneous equation \(y_h'' + y_h' = 0\).
09

Find General Solution for Homogeneous part (c)

Solve the characteristic equation: \(r^2 + r = 0 \implies r(r+1)=0, \ r=0, -1\). Therefore, the homogeneous solution is \(y_h(t) = C_1 + C_2 e^{-t}\).
10

Find Particular Solution for part (c)

Use the method of undetermined coefficients to find a particular solution. Assume \(y_p = Ae^{-2t}\). Substitute and solve for \(A\): \((-4A + A)e^{-2t} = e^{-2t}, \ (-3A) = 1, \ A = -\frac{1}{3}\). Therefore, the particular solution is \(y_p = -\frac{1}{3}e^{-2t}\).
11

Apply Initial Conditions for part (c)

Combine solutions to get the general solution: \(y(t) = C_1 + C_2 e^{-t} - \frac{1}{3}e^{-2t}\). Use the initial conditions \(y(0)=0\) and \(y'(0)=1\) to find the constants \(C_1\) and \(C_2\): \(y(0) = C_1 + C_2 - \frac{1}{3} = 0, \ y'(0) = C_2 - \frac{2}{3} = 1, \ C_2 = \frac{5}{3}, \ C_1 = -1\). Therefore, the solution is \(y(t) = -1 + \frac{5}{3} e^{-t} - \frac{1}{3} e^{-2t}\).
12

Solve part (d) - Non-Homogeneous Equation

Given the differential equation \(y'' + 2y' - 3y = 2t\), first solve the homogeneous equation \(y_h'' + 2y_h' - 3y_h = 0\).
13

Find General Solution for Homogeneous part (d)

Solve the characteristic equation: \(r^2 + 2r - 3 = 0 \implies (r+3)(r-1) = 0, \ r = -3, 1\). Therefore, the homogeneous solution is \(y_h(t) = C_1 e^{-3t} + C_2 e^{t}\).
14

Find Particular Solution for part (d)

Use the method of undetermined coefficients to find a particular solution. Assume a polynomial form, e.g., \(y_p = At + B\). Substitute and solve for \(A\) and \(B\): \(A = -\frac{2}{3}\), thus the particular solution is \(y_p = -\frac{2}{3}t - 1\).
15

Apply Initial Conditions for part (d)

Combine solutions to get the general solution: \(y(t) = C_1 e^{-3t} + C_2 e^{t} - \frac{2}{3}t - 1\). Use the initial conditions \(y(0)=1\) and \(y'(0)=0\) to find the constants \(C_1\) and \(C_2\): \(y(0) = C_1 + C_2 - 1 = 1, \ y'(0) = -3 C_1 + C_2 - \frac{2}{3} = 0, \ C_1 = 1, \ C_2 = \frac{5}{3}\). Therefore, the solution is \(y(t) = e^{-3t} + \frac{5}{3} e^{t} - \frac{2}{3}t - 1\).

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Key Concepts

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

Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They can model various real-world phenomena like population growth, physics systems, and economics. The goal is to determine the function that satisfies the equation. In simple terms, if you think of an equation involving just numbers and basic operations, a differential equation involves rates of change and the objects they describe. To solve these equations, we often need techniques beyond algebra, especially when initial conditions are specified.
Characteristic Equation
When solving linear homogeneous differential equations with constant coefficients, the characteristic equation is a crucial tool. It's derived by assuming solutions in the form of exponential functions, which translates the differential equation into a polynomial equation involving the exponents (often denoted by 'r'). For example, for the differential equation \(y'' + 4y = 0\), translating this into a characteristic equation gives us \(r^2 + 4 = 0\). Solving this polynomial provides the roots, which then help us form the general solution to the differential equation. The nature of the roots (real or complex) determines the form of the solution.
Non-Homogeneous Equations
A non-homogeneous differential equation has terms that do not solely depend on the function and its derivatives but also contains standalone functions or constants. For example, \(y'' + 2y' - 3y = 2t\) is non-homogeneous because of the 2t term on the right-hand side. Solving these involves two main steps: First, solve the corresponding homogeneous equation (by setting the non-homogeneous term to zero). Second, find a particular solution that accounts for the non-homogeneous part. The overall solution is the sum of the homogeneous solution and the particular solution.
Method of Undetermined Coefficients
The method of undetermined coefficients is used to solve non-homogeneous linear differential equations. This involves guessing the form of the particular solution based on the non-homogeneous term and determining the unknown coefficients that make this guess satisfy the equation. For instance, if our equation is \(y'' + y' = e^{-2t}\), we might guess a particular solution of the form \(y_p = Ae^{-2t}\). Substituting \(y_p\) into the differential equation allows us to solve for the coefficient \(A\). The method is systematic and works well when the non-homogeneous term is a simple function, like a polynomial, exponential, or sine/cosine.

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Most popular questions from this chapter

Gegeben ist das Funktionenpaar $$ f(t)=\sin (\omega t) \propto F(s)=\frac{\omega}{s^{2}+\omega^{2}} $$ Bestimmen Sie hieraus unter Verwendung des Ableitungssatzes für Bildfunktionen die Laplace-Transformierte von \(f_{1}(t)=t \cdot \sin (\omega t)\) und \(f_{2}(t)=t^{2} \cdot \sin (\omega t)\).

Berechnen Sie die Laplace-Transformierte von \(f(t)=\cos (\omega t)\) unter Berücksichtigung der folgenden Beriehungen: $$ \cos (\omega t)=\frac{\mathrm{e}^{j \omega t}+\mathrm{e}^{-j \omega t}}{2} \quad \text { und } \quad \mathscr{E}\left\\{\mathrm{e}^{a \prime}\right\\}=\frac{1}{s-a}. $$

Bestimmen Sie unter Verwendung des Dämpfungssatzes sowie der jeweils angegebenen Laplace-Transformierten die Bildfunktionen der folgenden ,gedämpften" Originalfunktionen: a) \(f(t)=e^{3 t} \cdot \cos (2 t), \quad F(s)=\mathscr{L}\\{\cos (2 t)\\}=\frac{s}{s^{2}+4}\) b) \(f(t)=A \cdot \mathrm{e}^{-\delta t} \cdot \sin (\omega t), \quad F(s)=\mathscr{L}\left\\{\sin (\omega t\\}=\frac{\omega}{s^{2}+\omega^{2}}\right.\) c) \(f(t)=2^{3 t}, \quad F(s)=\mathscr{L}\\{1\\}=\frac{1}{s}\)

Die Funktion \(f(t)=\sin (\omega t)\) ist eine Lösung der Schwingungsgleichung \(f^{\prime \prime}=-\omega^{2} \cdot f\). Bestimmen Sie hieraus die zugeh?rige Bildfunktion \(F(s)\). indem Sie die Laplace-Transformation auf diese Differentialgleichung anwenden und dabei den Ableitungssatz fir Originalfunktionen verwenden.

Bestimmen Sie unter Verwendung des Ahalichkeitssatzes und des entsprechenden Verschiebungssatzes die Laplace-Transformierte von \(\sin (\omega t+\varphi)\) für \(\varphi>0 .\) Anleitung: Gehen Sie von der Funktion \(f(t)=\sin t\) aus und verschieben Sie diese zunächst um die Strecke \(\varphi\) nach links. AnschlieBend wird die verschobene Kurve mit der Gleichung \(g(t)=f(t+\varphi)=\sin (t+\varphi)\) der Áhnlichkeitstransformation \(t \rightarrow \omega t\) unterworfen. Ferner ist \(\mathscr{Y}\\{\sin t\\}=\frac{1}{s^{2}+1}\).
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