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True-False Determine whether the statement is true or false. Explain your answer. If the first-order linear differential equation $$ \frac{d y}{d x}+p(x) y=q(x) $$ has a solution that is a constant function, then \(q(x)\) is a constant multiple of \(p(x)\).

Short Answer

Expert verified
True; \(q(x) = C \cdot p(x)\) for constant solution \(y=C\).

Step by step solution

01

Understand Constant Solution

A constant function solution for the differential equation is of the form \( y = C \), where \( C \) is a constant. This means that the derivative \( \frac{dy}{dx} = 0 \).
02

Substitute Constant Solution

Substitute \( y = C \) and \( \frac{dy}{dx} = 0 \) into the equation. This gives us:\[ 0 + p(x)C = q(x) \]which simplifies to:\[ p(x)C = q(x) \]
03

Analyze Relationship

The equation \( p(x)C = q(x) \) shows that \( q(x) \) must be equal to \( p(x) \) times the constant \( C \). This means \( q(x) \) is a constant multiple of \( p(x) \).
04

Conclusion

Based on the simplification, the statement is true. If the differential equation has a constant solution \( y = C \), \( q(x) \) must indeed be a constant multiple of \( p(x) \).

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Key Concepts

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

Constant Function Solution
When we talk about a constant function solution in the context of a differential equation, we refer to a scenario where the solution is not changing, regardless of the input value. In mathematical terms, this means the solution is simply a constant value, expressed as \( y = C \), where \( C \) is a fixed number. This translates to a derivative of zero, \( \frac{dy}{dx} = 0 \), because a constant does not change.In the context of the first-order linear differential equation \( \frac{dy}{dx} + p(x)y = q(x) \), substituting the constant solution gives:
  • \( \frac{dy}{dx} = 0 \)
  • \( y = C \)
This results in a condition \( p(x)C = q(x) \). So, a key characteristic of a constant function solution is that the right-hand side of the differential equation must be effectively neutral or balanced, aligning \( q(x) \) with a constant multiple of \( p(x) \). This setup confirms that a specific relation between \( p(x) \) and \( q(x) \) is necessary to maintain the solution as constant for all \( x \).
Relationship Between Functions
Understanding the relationship between functions in a first-order linear differential equation is crucial. When a differential equation like \( \frac{dy}{dx} + p(x)y = q(x) \) has a constant function as a solution, specific conditions must hold. After substituting the constant function solution \( y = C \), we obtain \( p(x)C = q(x) \).This relationship tells us a lot:
  • It indicates that \( q(x) \) must be a scaled version of \( p(x) \) with \( C \) as the scaling factor.
  • The factor \( C \) is constant, implying that \( q(x) \) must change proportionally to \( p(x) \) across all values of \( x \).
  • Ultimately, if this proportional relationship did not exist, \( y = C \) could not remain a valid solution, as it would not satisfy the equation for every \( x \) in its domain.
Thus, the intricate balance between \( p(x) \) and \( q(x) \) is central to maintaining a constant solution, highlighting how functions interact and depend on each other's form.
Differential Equation Analysis
Analyzing differential equations involves breaking down the equation into understandable parts. For a first-order linear differential equation, analysis begins by examining each component and understanding how they interact.Given the form \( \frac{dy}{dx} + p(x)y = q(x) \), the differential component \( \frac{dy}{dx} \) reflects how the function changes. For a constant solution, this component becomes zero, simplifying the equation to focus on \( p(x) \) and \( q(x) \). Key points to consider:
  • The function \( p(x) \) acts as a multiplier to \( y \), indicating its role in influencing changes in \( y \).
  • When \( y \) is constant, \( p(x) \) alone must fulfill the balance equation with \( q(x) \).
  • If \( q(x) \) mirrors the changes in \( p(x) \), i.e., it is proportional, it becomes apparent why \( p(x)C = q(x) \) must hold true.
Analytical steps thus ensure that all parts of the differential equation support a balanced and solvable scenario, whether the solution is constant or variable. This thorough exploration guarantees we comprehend each element’s contribution towards maintaining equilibrium in the solution.

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Most popular questions from this chapter

(a) There is a trick, called the Rule of 70, that can be used to get a quick estimate of the doubling time or half-life of an exponential model. According to this rule, the doubling time or half-life is roughly 70 divided by the percentage growth or decay rate. For example, we showed in Example 5 that with a continued growth rate of \(1.10 \%\) per year the world population would double every 63 years. This result agrees with the Rule of 70 , since \(70 / 1.10 \approx 63.6\). Explain why this rule works. (b) Use the Rule of 70 to estimate the doubling time of a population that grows exponentially at a rate of \(1 \%\) per year. (c) Use the Rule of 70 to estimate the half-life of a population that decreases exponentially at a rate of \(3.5 \%\) per hour. (d) Use the Rule of 70 to estimate the growth rate that would be required for a population growing exponentially to double every 10 years.

Consider the initial-value problem \(y^{\prime}=y, y(0)=1\), and let \(y_{n}\) denote the approximation of \(y(1)\) using Euler's Method with \(n\) steps. (a) What would you conjecture is the exact value of \(\lim _{n \rightarrow+\infty} y_{n} ?\) Explain your reasoning. (b) Find an explicit formula for \(y_{n}\) and use it to verify your conjecture in part (a).

Suppose that an initial population of 10,000 bacteria grows exponentially at a rate of \(2 \%\) per hour and that \(y=y(t)\) is the number of bacteria present \(t\) hours later. (a) Find an initial-value problem whose solution is \(y(t)\). (b) Find a formula for \(y(t)\). (c) How long does it take for the initial population of bacteria to double? (d) How long does it take for the population of bacteria to reach \(45,000 ?\)

Find a solution to the initial-value problem. \(y^{\prime}-y^{2} \equiv 0, y(1)=2 \quad[\) Hint: Assume the solution has an inverse function \(x=x(y)\). Find, and solve, a differential equation that involves \(x^{t}(y)\).]

A bullet of mass \(m\), fired straight up with an initial velocity of \(v_{0}\), is slowed by the force of gravity and a drag force of air resistance \(k v^{2}\), where \(k\) is a positive constant. As the bullet moves upward, its velocity \(v\) satisfies the equation $$ m \frac{d v}{d t}=-\left(k v^{2}+m g\right) $$ where \(g\) is the constant acceleration due to gravity. (a) Show that if \(x=x(t)\) is the height of the bullet above the barrel opening at time \(t\), then $$ m v \frac{d v}{d x}=-\left(k v^{2}+m g\right) $$ (b) Express \(x\) in terms of \(v\) given that \(x=0\) when \(v=v_{0}\). (c) Assuming that $$ \begin{aligned} &v_{0}=988 \mathrm{~m} / \mathrm{s}, \quad g=9.8 \mathrm{~m} / \mathrm{s}^{2} \\\ &m=3.56 \times 10^{-3} \mathrm{~kg}, \quad k=7.3 \times 10^{-6} \mathrm{~kg} / \mathrm{m} \end{aligned} $$ use the result in part (b) to find out how high the bullet rises. [Hint: Find the velocity of the bullet at its highest point.

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