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Chapter 6: Problem 3

Lösen Sie die folgenden linearen Differentialgleichungen 1. Ordnung mit konstanten Koefizienten: a) \(y^{\prime}-3 y=4 t \cdot \mathrm{e}^{\prime}\) b) \(y^{\prime}-4 y=5 \cdot \sin t\)

Short Answer

Expert verified
Solutions: (a) \(y = Ce^{3t} + (At + B)e^t\); (b) \(y = De^{4t} + A \sin{t} + B \cos{t}\).

Step by step solution

01

Homogeneous Solution for part (a)

To find the homogeneous solution, solve the equation without the non-homogeneous part: \(y' - 3y = 0\). Solve it as a separable equation by setting \(y_h = Ce^{3t}\).
02

Particular Solution for part (a)

Assume a particular solution to the non-homogeneous equation \(y' - 3y = 4t e^t\). Try a solution of the form \(y_p = (At + B)e^t\). Differentiate and substitute back into the equation, solving for A and B.
03

Complete Solution for part (a)

Combine the homogeneous and particular solutions: \(y = y_h + y_p\). The complete solution is \(y = Ce^{3t} + (At + B)e^t\). Substitute found A and B values to get the final form.
04

Homogeneous Solution for part (b)

Solve the homogeneous equation \(y' - 4y = 0\) similarly. Set \(y_h = De^{4t}\).
05

Particular Solution for part (b)

Assume a particular solution to the non-homogeneous equation \(y' - 4y = 5 \sin{t}\). Try a solution of the form \(y_p = A \sin{t} + B \cos{t}\). Differentiate and substitute back into the equation, then solve for A and B.
06

Complete Solution for part (b)

Combine the homogeneous and particular solutions: \(y = y_h + y_p\). The complete solution is \(y = De^{4t} + A \sin{t} + B \cos{t}\).

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Key Concepts

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

First-Order Differential Equations
A first-order differential equation involves the first derivative of the function but no higher derivatives. Such equations commonly appear in both physics and engineering. The general form of a first-order linear differential equation is: \[y' + p(t)y = g(t)\]

Here, \(y'\) is the first derivative of \(y\) with respect to \(t\), \(p(t)\) is a function of \(t\), and \(g(t)\) is another given function.

They can be solved by finding both homogeneous and particular solutions, which we will discuss next.
Homogeneous Solutions
A homogeneous differential equation is one where \(g(t) = 0\). Solving it involves finding the solution of \( y' + p(t)y = 0 \). This can often be tackled by using separation of variables or an integrating factor.

For example, consider the equation \(y' - 3y = 0\). This can be solved by separation of variables:
  • Separate terms: \(\frac{dy}{y} = 3 dt\)
  • Integrate both sides: \(\text{ln}|y| = 3t + C\)
  • Solve for \(y\): \(y_h = Ce^{3t}\)
The result is the homogeneous solution.
Particular Solutions
A particular solution is found by solving the non-homogeneous equation. The general form looks like \( y' + p(t)y = g(t) \). Here, we guess a form for \( y_p \) based on the function \( g(t) \) and then determine the coefficients by substituting back into the original equation.

For the example \( y' - 3y = 4t e^t \), we might guess \( y_p = (At + B)e^t \). Differentiate and substitute back:
  • Differentiate: \(y_p' = (A + At + B)e^t\)
  • Substitute back: \((A + At + B)e^t - 3(At + B)e^t = 4t e^t \)
Solve for \( A \) and \( B \) to find the particular solution.
Non-Homogeneous Equations
A non-homogeneous differential equation includes a non-zero function on the right side: \( y' + p(t)y = g(t) \). Solving it involves two main steps:

  • Find the homogeneous solution by assuming \( g(t) = 0 \).
  • Find the particular solution for the given \( g(t) \).
The overall solution is the sum of the homogeneous and particular solutions.

For instance, with \( y' - 4y = 5 \sin(t) \):
  • Homogeneous part: \( y_h = De^{4t} \)
  • Particular part: Assume \( y_p = A \sin(t) + B \cos(t) \). Differentiate and solve.
Combine them to get the full solution.
Separable Equations
A separable differential equation is a special case where the equation can be written in the form: \( \frac{dy}{dx} = g(x)h(y) \).

The strategy here is to rearrange the equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other:
  • Rewrite: \( \frac{1}{h(y)} dy = g(x) dx \)
  • Integrate both sides: \( \int \frac{1}{h(y)} dy = \int g(x) dx \)
  • Solve the resulting equation for \( y \).
This method works well for many simple forms of equations.

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Most popular questions from this chapter

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 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) \(f(t)=2^{3 t}, \quad F(s)=\mathscr{L}\\{1\\}=\frac{1}{s}\)

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

Berechnen Sie mit Hilfe des entsprechenden Grenzwertsatzes aus der gegebenen Bildfunktion \(F(s)\) den Endwert \(f(\infty)\) der zugeh?rigen Originalfunktion \(f(t):\) a) \(F(s)=\frac{1}{s(s+4)}\) b) \(F(s)=\frac{\mathrm{e}^{s}-s-1}{s^{2}}\)

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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