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Using an Integrating Factor The expression \(u(x)\) is an integrating factor for \(y^{\prime}+P(x) y=Q(x)\) . Which of the following is equal to \(u^{\prime}(x) ?\) Verify your answer. $$ \begin{array}{ll}{\text { (a) } P(x) u(x)} & {\text { (b) } P^{\prime}(x) u(x)} \\ {\text { (c) } Q(x) u(x)} & {\text { (d) } Q^{\prime}(x) u(x)}\end{array} $$

Short Answer

Expert verified
The derivative of the integrating factor \(u^{\prime}(x) = P(x) u(x)\), so the correct answer is (a) \(P(x) u(x)\).

Step by step solution

01

Understanding Integrating Factors

First, understand that an integrating factor is a function that is used to solve linear differential equations of the form \(y^{\prime}+P(x) y=Q(x)\). The integrating factor, in this case, is \(u(x)\), and the differential equation has a coefficient \(P(x)\) and a function \(Q(x)\).
02

Understand the Derivative of the Integrating Factor

The derivative of the integrating factor \(u(x)\), or \(u^{\prime}(x)\), is defined as \(u^{\prime}(x) = u(x)P(x)\). This is based on the property of integrating factors.
03

Matching the Answer

Now you match the definition of \(u^{\prime}(x)\) from step 2, which is \(u(x)P(x)\) with the options given in the exercise, and find the answer is (a) \(P(x) u(x)\).

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Key Concepts

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

Linear Differential Equations
Linear differential equations are a fundamental part of mathematics, used to model many real-world phenomena. They typically have the form:
  • First-order: \( y^{\prime} + P(x)y = Q(x) \)
  • Higher-order: Involving higher derivatives, but can often be simplified into first-order forms.
In the equation format above, \( y^{\prime} \) represents the derivative of \( y \) with respect to the variable \( x \), \( P(x) \) is a coefficient function, and \( Q(x) \) is a driving function or source term. Applying an integrating factor is a standard method for solving such equations.
By choosing an appropriate integrating factor, you can transform a complex differential equation into one that is simpler to solve, often reducing it to a standard integration problem.
Derivative of a Function
The derivative of a function often measures how a function's output values change as its input changes. In calculus, it provides critical insights into the behavior of functions, especially when analyzing curves and slopes. For example, if you have a function \( f(x) \), its derivative \( f^{\prime}(x) \) tells you the rate of change of \( f \) with respect to \( x \).
Derivatives can be used to find:
  • Slopes of tangents to curves at given points.
  • Instantaneous rates of change, like velocity and acceleration in physics.
  • Critical points where functions reach maximum or minimum values.
When solving linear differential equations, understanding the derivative helps because the process often involves differentiating unknown and known functions. For integrating factors, knowing the derivative formula is key for determining the correct factor to simplify the equation.
Function of a Single Variable
Functions of a single variable are crucial because they relate each value from a domain (input) to a value from a range (output). Consider a function \( f(x) \). Here, \( x \) is the independent variable, and the function produces an output based on \( x \) alone.
These functions provide a foundational framework for understanding broader mathematical concepts, serving as building blocks for more complex multi-variable functions.
In the context of differential equations:
  • The functions \( P(x) \) and \( Q(x) \) are examples of functions of a single variable, where the behavior of these functions can directly influence the outcome of the equation.
  • Simplifying and manipulating these functions of a single variable are often necessary to solve or approximate solutions to equations.
Having a solid grasp of single-variable functions is essential for navigating topics such as derivatives and linear differential equations, as they frequently interact in solving mathematical problems.

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Most popular questions from this chapter

Mixture In Exercises \(35-38\) , consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. A 200 -gallon tank is full of a solution containing 25 pounds of concentrate. Starting at time \(t=0\) , distilled water is admitted to the tank at a rate of 10 gallons per minute, and the well-stirred solution is withdrawn at the same rate. (a) Find the amount of concentrate \(Q\) in the solution as a function of \(t .\) (b) Find the time at which the amount of concentrate in the tank reaches 15 pounds. (c) Find the quantity of the concentrate in the solution as \(t \rightarrow \infty\) .

Solving a Logistic Differential Equation In Exercises \(55-58\) , find the logistic equation that passes through the given point. $$ \frac{d y}{d t}=\frac{3 y}{20}-\frac{y^{2}}{1600}, \quad(0,15) $$

Solving a Bernoulli Differential Equation In Exercises \(59-66,\) solve the Bernoulli differential equation. The Berchoulli equation is a well-known nonlinear equation of the form $$ y^{\prime}+P(x) y=Q(x) y^{n} $$ that can be reduced to a linear form by a substitution. The general solution of a Bernoulli equation is $$ y^{1-n} e^{\int(1-n) P(x) d x}=\int(1-n) Q(x) e^{f(1-n) P(x) d x} d x+C $$ $$ y^{\prime}+\left(\frac{1}{x}\right) y=x y^{2} $$

Finding Orthogonal Trajectories In Exercises \(41-46,\) find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. $$ x^{2}+y^{2}=C $$

Finding a General Solution Using Separation of Variables In Exercises \(1-14,\) find the general solution of the differential equation. $$ x y^{\prime}=y $$

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