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Chapter 1: Problem 56

Consider the differential equation \(d y / d x=5-y\). (a) Either by inspection, or by the method suggested in Problems 33-36, find a constant solution of the DE. (b) Using only the differential equation, find intervals on the \(y\)-axis on which a nonconstant solution \(y=\phi(x)\) is increasing. Find intervals on the \(y\)-axis on which \(y=\phi(x)\) is decreasing.

Short Answer

Expert verified
(a) The constant solution is \(y = 5\). (b) Increasing for \(y < 5\); decreasing for \(y > 5\).

Step by step solution

01

Identify Constant Solution

To find a constant solution for the differential equation \(\frac{dy}{dx} = 5 - y\), set \(\frac{dy}{dx} = 0\), as constant functions have zero derivative. Solving \(5 - y = 0\), we get \(y = 5\). Hence, a constant solution is \(y = 5\).
02

Determine Condition for Increasing Function

For a function \(y = \phi(x)\) to be increasing, its derivative must be positive: \(\frac{dy}{dx} > 0\). Substituting the differential equation, \(5 - y > 0\), we get \(y < 5\). Thus, the solution is increasing for \(y < 5\).
03

Determine Condition for Decreasing Function

For a function \(y = \phi(x)\) to be decreasing, its derivative must be negative: \(\frac{dy}{dx} < 0\). Substituting the differential equation, \(5 - y < 0\), we get \(y > 5\). Hence, the solution is decreasing for \(y > 5\).

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Key Concepts

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

Constant Solutions
The concept of constant solutions in differential equations refers to solutions where the dependent variable does not change with respect to the independent variable. In our given differential equation \( \frac{dy}{dx} = 5 - y \), a constant solution occurs when the derivative is zero because constant solutions mean that there is no change. Thus, we set \( \frac{dy}{dx} = 0 \), leading us to solve \( 5 - y = 0 \). Upon solving this equation, we find that \( y = 5 \). This is the value of the dependent variable \( y \) that would keep it constant, regardless of how \( x \) changes. Hence, \( y = 5 \) is a constant solution, providing a flat line with a slope of zero when graphed.
Increasing and Decreasing Functions
Understanding when a function is increasing or decreasing is essential for interpreting solutions to differential equations. For a solution \( y = \phi(x) \) to be increasing, its slope, determined by the derivative \( \frac{dy}{dx} \), must be positive.
  • For the given equation \( \frac{dy}{dx} = 5 - y \), the derivative is positive when \( 5 - y > 0 \). Solving this inequality gives us \( y < 5 \). This implies the function will rise, or is increasing, anytime \( y \) is less than 5.
On the other hand, for a function to be decreasing, the derivative must be negative.
  • Solving \( \frac{dy}{dx} < 0 \) for \( \frac{dy}{dx} = 5 - y \) gives \( y > 5 \). This tells us the function will fall, or is decreasing, when \( y \) is greater than 5.
Through these conditions, we can predict the behavior of the different solutions depending on the range of \( y \). This analysis is foundational for sketching solution curves and understanding dynamic systems.
Solution Intervals
Solution intervals in differential equations denote ranges over which the solutions exhibit specific behaviors, such as being constant, increasing, or decreasing. For the differential equation \( \frac{dy}{dx} = 5 - y \), we can describe intervals on the \( y \)-axis based on the derivative's sign.
  • The interval \( y = 5 \) is the point where the function neither increases nor decreases but remains constant鈥攖his corresponds to our constant solution.
  • The interval \( y < 5 \) is where the solutions are increasing. Any initial condition starting in this range will cause the function to rise towards \( y = 5 \).
  • Conversely, \( y > 5 \) represents the interval for decreasing solutions, where the initial value will result in the function decreasing towards \( y = 5 \).
These intervals not only help visualize the function's behavior for different initial conditions but also provide insight into the equilibrium stability at \( y = 5 \), which acts as an asymptote in this context.

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Most popular questions from this chapter

(a) Verify that \(y=-1 /(x+c)\) is a one-parameter family of solutions of the differential equation \(y^{\prime}=y^{2}\). (b) Since \(f(x, y)=y^{2}\) and \(\partial f l \partial y=2 y\) are continuous everywhere, the region \(R\) in Theorem \(1.2 .1\) can be taken to be the entire \(x y\) -plane. Find a solution from the family in part (a) that satisfies \(y(0)=1\). Find a solution from the family in part (a) that satisfies \(y(0)=-1\). Determine the largest interval \(I\) of definition for the solution of each initial-value problem.

Find a function \(y=f(x)\) whose second derivative is \(y^{\prime \prime}=\) \(12 x-2\) at each point \((x, y)\) on its graph and \(y=-x+5\) is tangent to the graph at the point corresponding to \(x=1\).

Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ 2 y^{\prime}+y=0 ; \quad y=e^{-x / 2} $$

Consider the differential equation \(d y / d x=y(a-b y)\), where \(a\) and \(b\) are positive constants. (a) Either by inspection, or by the method suggested in Problems 33-36, find two constant solutions of the DE. (b) Using only the differential equation, find intervals on the \(y\)-axis on which a nonconstant solution \(y=\phi(x)\) is increasing. On which \(y=\phi(x)\) is decreasing. (c) Using only the differential equation, explain why \(y=a / 2 b\) is the \(y\)-coordinate of a point of inflection of the graph of a nonconstant solution \(y=\phi(x)\). (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the \(x y\)-plane into three regions. In each region, sketch the graph of a nonconstant solution \(y=\phi(x)\) whose shape is suggested by the results in parts (b) and (c).

Use the concept that \(y=c,-\infty
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