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Chapter 1: Problem 13

Bestimme die allgemeine L枚sung der DGl \(3 y^{2} y^{\prime}-2 y^{3}=x+1\)

Short Answer

Expert verified
The general solution is given by the implicit equation \( y^{3} = 2 y^{3} x + \frac{x^{2}}{2} + x + C \).

Step by step solution

01

Rewrite the Differential Equation

Rewrite the given differential equation in a more familiar form. The given equation is:\[3 y^{2} y^{\text{\prime}} - 2 y^{3} = x + 1\]This can be rewritten as:\[3 y^{2} y^{\text{\prime}} = 2 y^{3} + x + 1\]
02

Separate the Variables

Separate the variables to one side of the equation. We need to isolate \(y\) and \(dy\) on one side, and \(x\) and \(dx\) on the other. Divide both sides by \(3 y^{2}\):\[y^{\text{\prime}} = \frac{2 y^{3} + x + 1}{3 y^{2}}\]Separate the variables:\[3 y^{2} dy = (2 y^{3} + x + 1) dx\]
03

Integrate Both Sides

Integrate both sides separately. First, integrate the left side with respect to \(y\):\[\int 3 y^{2} \, dy\]This results in:\[y^{3}\]Then, integrate the right side with respect to \(x\):\[\int (2 y^{3} + x + 1) \, dx\]Since \(2y^{3}\) is independent of \(x\), it remains a constant during integration:\[2 y^{3} x + \frac{x^{2}}{2} + x\]
04

Combine Results

Combine the integrated results into a single equation. This gives us:\[y^{3} = 2 y^{3} x + \frac{x^{2}}{2} + x + C\]Solve for the general solution. It is implicit and the exact form might vary based on simplifying or re-organizing overall constants.

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Key Concepts

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

Separation of Variables
Separation of variables is a method used to solve ordinary differential equations (ODEs) where the goal is to isolate each variable on different sides of the equation. In our given problem, we start with the equation:\[3 y^{2} y^{\text{\prime}} - 2 y^{3} = x + 1\]To apply separation of variables, we rewrite it as:\[3 y^{2} y^{\text{\prime}} = 2 y^{3} + x + 1\]By dividing both sides by \(3 y^{2}\), we get:\[y^{\text{\prime}} = \frac{2 y^{3} + x + 1}{3 y^{2}}\]Now, multiply both sides by \(dx\) and then divide both sides by \(y^{2}\) to get:\[3 y^{2} dy = (2 y^{3} + x + 1) dx\]This separates the variables effectively, allowing us to integrate both sides to find a solution.
Integration
Integration is the process of finding the integral of a function, which can be thought of as the opposite of differentiation. After separating the variables, we were left with:\[3 y^{2} dy = (2 y^{3} + x + 1) dx\]We then integrate both sides. First, we integrate the left side with respect to \(y\):\[\int 3 y^{2} \, dy = y^{3}\]Then, we integrate the right side with respect to \(x\). Since \(2y^{3}\) is a constant with respect to \(x\), we treat it as such during the integration:\[\int (2 y^{3} + x + 1) \, dx = 2 y^{3} x + \frac{x^{2}}{2} + x\]These integrations give us the expressions that we combine to form the solution.
General Solution
The general solution of a differential equation is the most comprehensive form that includes an arbitrary constant, representing all possible solutions. From our integrations, we obtained:\[y^{3} = 2 y^{3} x + \frac{x^{2}}{2} + x + C\]This equation contains an implicit solution for \(y\), meaning that \(y\) is not explicitly solved for in terms of \(x\). Here, \(C\) is the integration constant, which accounts for any initial conditions or specific solutions.
  • A general solution provides the overall behavior of the system described by the differential equation.
  • The presence of \(C\) means we can adjust it based on initial or boundary conditions to find specific solutions.
Our goal is to understand the relationship between \(y\) and \(x\), and the form we derived represents a broader class of solutions that fit different scenarios.

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Most popular questions from this chapter

Wie lautet die L枚sung des Anfangswertproblems $$ y^{\prime \prime}=y^{\prime} \cdot(y+1), \quad y(0)=0, \quad y^{\prime}(0)=\frac{1}{2} ? $$

Bestimme die allgemeinen L枚sungen der DGln (a) \(x y^{\prime}=y \cdot \ln \left|\frac{y}{x}\right|(x>0)\); (b) \(y^{\prime}=(2 x+2 y-1)^{2}\)

Gegeben sei die DGl $$ y^{\prime}(x)-k y(x)=0, \quad k \in \mathbb{R} $$ (a) Zeige, daB \(y(x)=C \mathrm{e}^{k x}(C \in \mathbb{R}\) beliebig) der DGl gen眉gt. (b) Weise nach, daB jede weitere L枚sung \(\tilde{y}(x)\) der DGl notwendig die in (a) auftretende Gestalt besitzt: Anleitung: Betrachte den Quotienten \(\frac{\tilde{y}(x)}{\mathrm{e}^{k x}}\)

Best?tige durch nachrechnen: \(\left[\mathrm{e}^{-t} \cos t, \mathrm{e}^{-t} \sin t\right]^{T}, t \in \mathbb{R}\), ist eine L枚sung des Systems $$ \left\\{\begin{array}{l} \dot{x}_{1}=-x_{1}-x_{2} \\ \dot{x}_{2}=x_{1}-x_{2} \end{array}\right. $$ Welcher Orbit entspricht dieser L枚sung?

Sei \(\varphi(t)\) die zu den Anfangsbedingungen \(\varphi(0)=0, \varphi^{\prime}(0)=\varphi_{1}>0\) geh枚rende L枚sung der Pendelgleichung $$ m l \varphi^{\prime \prime}(t)=-m g \sin \varphi(t) $$ und \(E=\frac{m}{2} l^{2} \varphi_{1}^{2}\) die zugeh?rige (kinetische) Energie des Pendels zum Zeitpunkt \(t=0 . Z \mathrm{Zu}\) einem beliebigen Zeitpunkt \(t\) ergibt sich \(E\) als Summe aus kinetischer und potentieller Energie der Pendelmasse: $$ E=\frac{m}{2} l^{2}\left[\varphi^{\prime}(t)\right]^{2}+m g l[1-\cos \varphi(t)] $$ (s. Abschn. 1.3.3, Anwendung II). Beweise unter der Voraussetzung, daB die Gesamtenergie gr枚Ber ist als die maximale potentielle Energie (d.h. \(E>2 m g l\) ): (a) Die L枚sung \(\varphi(t)\) der Pendelgleichung existiert f眉r alle \(t\) und ist eine ungerade Funktion. Ferner gilt \(\varphi^{\prime}(t)>0\) f眉r alle \(t\) und \(\lim _{t \rightarrow \pm \infty} \varphi(t)=\pm \infty\) (b) Es gibt genau eine reelle Zahl \(\tau\) mit \(\varphi(\tau)=2 \pi\) (benutze (a)). Welchen Wert besitzt \(\varphi^{\prime}(\tau)\) ? (c) F眉r alle \(t\) gilt: \(\varphi(t+\tau)=\varphi(t)+2 \pi\). Anleitung: Zeige, daB sowohl \(\varphi_{0}(t):=\varphi(t+\tau)\) als auch \(\psi_{0}(t):=\varphi(t)+2 \pi\) der Pendelgleichung mit gemeinsamen Anfangsdaten gen眉gt. (d) Die Bewegung $$ \left[\begin{array}{c} x(t) \\ y(t) \end{array}\right]=\left[\begin{array}{l} l \cos \varphi(t) \\ l \sin \varphi(t) \end{array}\right] $$ der Pendelmasse ist periodisch mit der Periode \(\tau\) (benutze (c)). Gib einen Integralausdruck f眉r \(\tau\) an.
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