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Chapter 11: Problem 69

Are the statements true or false? Give an explanation for your answer. The only solution to the differential equation \(d y / d x=\) \(3 y^{2 / 3}\) passing through the point (0,0) is \(y=x^{3}\)

Short Answer

Expert verified
True, the only solution passing through (0,0) is \(y=x^3\).

Step by step solution

01

Verify the Point through Proposed Solution

To determine if the proposed solution is indeed the only solution, start by checking if it satisfies the differential equation and the point (0,0). Substitute \( x = 0 \) into \( y = x^3 \) to find \( y = 0^3 = 0 \). Therefore, the point \( (0,0) \) satisfies the proposed solution.
02

Differentiate the Proposed Solution

Differentiate the proposed function \( y = x^3 \) with respect to \( x \). The derivative \( \frac{dy}{dx} = 3x^2 \).
03

Substitute Back into the Differential Equation

Substitute \( y = x^3 \) into the given differential equation \( \frac{dy}{dx} = 3y^{2/3} \). This becomes \( 3x^2 = 3(x^3)^{2/3} = 3x^2 \), which holds true. This shows that \( y = x^3 \) is indeed a solution to the differential equation.
04

Consider Possible Solutions with Arbitrary Constants

The differential equation \( \frac{dy}{dx} = 3y^{2/3} \) is separable. Rewriting it, we have \( dy = 3y^{2/3}dx \). Integrating both sides gives \( \int rac{1}{y^{2/3}} \, dy = \int 3 \, dx \), leading to \( y^{1/3} = x + C \), or \( y = (x + C)^3 \), where \( C \) is an integration constant.
05

Apply Initial Condition to General Solution

Apply the initial condition \( (0,0) \) to find the value of \( C \): Substitute \( x = 0 \) and \( y = 0 \) into \( y = (x + C)^3 \), resulting in \( 0 = C^3 \). Hence, \( C = 0 \), which gives \( y = x^3 \).
06

Conclusion: Determine Uniqueness of Solution

Given that \( y = (x + C)^3 \) leads to \( C = 0 \) when passing through \( (0,0) \), the only possible solution in the context of this initial condition is \( y = x^3 \). The differential equation and initial condition uniquely determine this solution.

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Key Concepts

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

Initial Value Problem
An initial value problem in differential equations deals with finding a specific solution that satisfies both a differential equation and a given set of initial conditions. In our context, the initial condition is a point, (0,0), through which the solution to the differential equation must pass.

To solve such a problem, one must:
  • Identify a differential equation.
  • Find the general solution, which may contain arbitrary constants.
  • Use the initial condition to solve for the constants, thus deriving the unique solution.
The key point here is that the initial condition is crucial. It helps determine the exact form of the solution that not only solves the differential equation but also aligns perfectly with the given initial value. Without an initial condition, multiple solutions might exist.
Separable Equations
Separable equations are a special type of differential equations that can be separated into two integrals; one involving only the dependent variable (like \(y\)) and the other involving only the independent variable (like \(x\)).

For our differential equation \(\frac{dy}{dx} = 3y^{2/3}\), the process of separation involves rearranging the terms so that each variable is on a different side of the equation:
  • First, write it as \(dy = 3y^{2/3}dx\).
  • The equation can be separated to form \(\int \frac{1}{y^{2/3}} \, dy = \int 3 \, dx\).
  • Solving these integrals leads to \(y^{1/3} = x + C\).
The essence of separable equations is that it allows us to integrate each side independently, simplifying the process of finding solutions to certain differential equations.
Uniqueness of Solutions
In the realm of differential equations, the uniqueness of solutions is an important concept.

Given an initial value problem, a unique solution exists when a certain combination of conditions is fulfilled:
  • The differential equation is valid over a region containing the initial point.
  • Any assumptions used, such as continuity and differentiability, are met in this region.
For this specific exercise, after solving the separable equation and applying the initial condition \((0,0)\), we derived the equation \(y = (x+C)^3\), where \(C\) is a constant. Using the initial condition, we found \(C = 0\), leading us back to \(y = x^3\).

Because this solution satisfies both the differential equation and the initial condition, it is the unique solution. The uniqueness is confirmed by the fact that no other value of \(C\) is possible under the given initial condition.

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Most popular questions from this chapter

Analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-x-\frac{y}{3}\right)\\\&\frac{d y}{d t}=y\left(1-y-\frac{x}{2}\right)\end{aligned}$$

Morphine is often used as a pain-relieving drug. The half-life of morphine in the body is 2 hours. Suppose morphine is administered to a patient intravenously at a rate of \(2.5 \mathrm{mg}\) per hour, and the rate at which the morphine is eliminated is proportional to the amount present. (a) Use the half-life to show that, to three decimal places, the constant of proportionality for the rate at which morphine leaves the body (in mg/hour) is \(k=-0.347\) (b) Write a differential equation for the quantity, \(Q,\) of morphine in the blood after \(t\) hours. (c) Use the differential equation to find the equilibrium solution. (This is the long-term amount of morphine in the body, once the system has stabilized.)

The differential equation \(d Q / d t=-0.15 Q+25\) represents the quantity of a drug in the body if the drug is metabolized at a continuous rate of \(15 \%\) per day and an IV line is delivering the drug at a constant rate of \(25 \mathrm{mg}\) per hour.

An aquarium pool has volume \(2 \cdot 10^{6}\) liters. The pool initially contains pure fresh water. At \(t=0\) minutes, water containing 10 grams/liter of salt is poured into the pool at a rate of 60 liters/minute. The salt water instantly mixes with the fresh water, and the excess mixture is drained out of the pool at the same rate (60 liters/minute). (a) Write a differential equation for \(S(t),\) the mass of salt in the pool at time \(t.\) (b) Solve the differential equation to find \(S(t).\) (c) What happens to \(S(t)\) as \(t \rightarrow \infty ?\)

Each of the differential equations (i)-(iv) represents the position of a 1 gram mass oscillating on the end of a damped spring. Pick the differential equation representing the system which answers the question. (i) \(\quad s^{\prime \prime}+s^{\prime}+4 s=0\) (ii) \(s^{\prime \prime}+2 s^{\prime}+5 s=0\) (iii) \(s^{\prime \prime}+3 s^{\prime}+3 s=0\) (iv) \(\quad s^{\prime \prime}+0.5 s^{\prime}+2 s=0\) Which spring has the largest coefficient of damping?
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