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Chapter 8: Problem 12

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, because a constant solution implies \( q(x) = C p(x) \).

Step by step solution

01

Understand Constant Solution Condition

Assume the solution \( y(x) = C \), where \( C \) is a constant. Substituting this into the differential equation \( \frac{d y}{d x} + p(x) y = q(x) \), we have \( \frac{d(C)}{d x} + p(x) C = q(x) \). Since \( \frac{d(C)}{d x} = 0 \), the equation simplifies to \( p(x) C = q(x) \).
02

Analyze Simplified Equation

The equation \( p(x) C = q(x) \) indicates that \( q(x) \) must equal \( C p(x) \). This implies that \( q(x) \) is a constant multiple of \( p(x) \), with the constant being \( C \).
03

Conclusion

From the derived condition \( q(x) = C p(x) \), we conclude that for the differential equation to have a constant solution, \( q(x) \) indeed must 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.

First-Order Linear Differential Equation
A first-order linear differential equation is a special type of differential equation that involves a first derivative. It has the standard form: \[ \frac{dy}{dx} + p(x) y = q(x) \] Where:
  • \( \frac{dy}{dx} \) is the first derivative of \( y \) with respect to \( x \).
  • \( p(x) \) is a function that multiplies the unknown function \( y \).
  • \( q(x) \) is the function on the right-hand side of the equation.

The goal when solving these equations is to find the unknown function \( y(x) \). They are fundamental in mathematics because they often describe real-world processes, such as cooling rates or populations changes.
To solve these equations, you often integrate, apply particular methods like an integrating factor or even transform the equations if necessary. Understanding the form and components of the equation is crucial for those solutions.
Constant Solution
A constant solution to a differential equation means the solution is a constant value; it does not change with \( x \). Imagine it as a horizontal line, like \( y(x) = C \), where \( C \) is a constant.
In the context of first-order linear differential equations, let's consider: \[ \frac{dy}{dx} + p(x) y = q(x) \] Substitute \( y(x) = C \) into the equation and see what happens:
  • The derivative \( \frac{dC}{dx} \) becomes zero, as \( C \) does not change.
  • We then simplify to: \( p(x) C = q(x) \).

For the equation to hold true, \( q(x) \) must be equal to \( p(x) \times C \), showing \( q(x) \) must be a constant multiple of \( p(x) \). This result is crucial, linking the nature of \( q(x) \) to the possibility of having a constant solution.
Function Substitution
Function substitution is a technique where you replace a function in the equation with something else to simplify the problem.
When dealing with differential equations, substituting with a simpler form, like a constant, allows us to analyze certain solutions. Let's demonstrate this with the first-order differential equation \[ \frac{dy}{dx} + p(x) y = q(x) \]
Assume the solution \( y(x) = C \). Substituting, you get:
  • Because \( \frac{dC}{dx} = 0 \), simplifying gives \( p(x) C = q(x) \).
  • This means \( q(x) \) is \( Cp(x) \) proving that this constant solution is valid when \( q(x) \) is a constant multiple of \( p(x) \).

Function substitution can thus quickly reveal insights, especially in showing dependencies of different parts of a differential equation, making the solving process more straightforward.

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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 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 \(1.33 \%\) per year the world population would double every 52 years. This result agrees with the Rule of \(70,\) since \(70 / 1.33 \approx 52.6 .\) 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.

Suppose that the growth of a population \(y=y(t)\) is given by the logistic equation $$ y=\frac{60}{5+7 e^{-t}} $$ (a) What is the population at time \(t=0\) ? (b) What is the carrying capacity \(L ?\) (c) What is the constant \(k ?\) (d) When does the population reach half of the carrying capacity? (e) Find an initial-value problem whose solution is \(y(t)\)

Suppose that a particle moving along the \(x\) -axis encounters a resisting force that results in an acceleration of \(a=d v / d t=-0.02 \sqrt{v}\). Given that \(x=0 \mathrm{cm}\) and \(v=9\) \(\mathrm{cm} / \mathrm{s}\) at time \(t=0,\) find the velocity \(v\) and position \(x\) as a function of \(t\) for \(t \geq 0 .\)

Find a solution to the initial-value problem. $$ y^{\prime}+4 x=2, y(0)=3 $$

Polonium-2 10 is a radioactive element with a half-life of 140 days. Assume that 10 milligrams of the element are placed in a lead container and that \(y(t)\) is the number of milligrams present t days later. (a) Find an initial-value problem whose solution is \(y(t) .\) (b) Find a formula for \(y(t) .\) (c) How many milligrams will be present after 10 weeks? (c) How many milligrams will be present after 10 weeks? (d) How long will it take for \(70 \%\) of the original sample to decay?
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