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Chapter 9: Problem 7

Write an equivalent first-order differential equatio and initial condition for \(y\). $$y=-1+\int_{1}^{x}(t-y(t)) d t.$$

Short Answer

Expert verified
Differential equation: \( \frac{dy}{dx} = x - y \); Initial condition: \( y(1) = -1 \).

Step by step solution

01

Understand the Integral Equation

The given equation is \( y = -1 + \int_{1}^{x} (t - y(t)) \, dt \), an integral equation expressing \( y \) in terms of an integral of \( t - y(t) \). We need to transform this into a first-order differential equation.
02

Differentiate Both Sides with Respect to x

To eliminate the integral, differentiate both sides of the equation with respect to \( x \). Using the Fundamental Theorem of Calculus, we have: \( \frac{d}{dx} \left(y \right) = \frac{d}{dx} \left( -1 + \int_{1}^{x} (t - y(t)) \, dt \right) \). This simplifies to \( \frac{dy}{dx} = x - y(x) \).
03

Identify Initial Condition from the Equation

Evaluate the given equation at \( x = 1 \). Substituting \( x = 1 \) into the integral, we have \( y(1) = -1 \). This provides us with the initial condition: \( y(1) = -1 \).
04

Write the First-Order Differential Equation and Initial Condition

The first-order differential equation derived is \( \frac{dy}{dx} = x - y \) and the initial condition is \( y(1) = -1 \). These encapsulate the same relationship as the original integral equation.

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Key Concepts

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

Integral Equations
An integral equation involves a function that is defined in terms of an integral. In the context of the given problem, we have the equation:\[ y = -1 + \int_{1}^{x} (t - y(t)) \, dt. \]Here, the unknown function \( y(x) \) is expressed as part of an integral calculation, where the integration is performed with respect to \( t \) over the domain \([1, x]\).
  • In this setup, the variable of integration, \( t \), is different from \( x \), which is the variable with respect to which the integration limits change.
  • Integral equations like this one can often be re-expressed as differential equations, making them easier to solve using techniques specific to differential equations.
By differentiating the entire equation, we can simplify it into a first-order differential equation, which often makes the relationship clearer.
Initial Conditions
An initial condition is a value that specifies the state of a function at a particular point. In differential equations, initial conditions are crucial for finding a unique solution.
  • In the problem, the initial condition is given at \( x = 1 \).
  • It's stated as \( y(1) = -1 \), meaning when \( x \) is 1, the value of \( y \) should also be -1.
These initial conditions help lock down the constant of integration that typically appears when solving these equations. Without this information, we could end up with a family of solutions rather than a single, specific one that fits our integral equation.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation and integration. It essentially states that differentiation and integration are inverse processes.
  • To move from an integral equation to a differential equation, we use this theorem.
  • In our worked solution, differentiating both sides of the integral equation gave us the first-order differential equation \( \frac{dy}{dx} = x - y(x) \).
This transformation is possible because the Fundamental Theorem of Calculus tells us that if a function is continuous over an interval, the integral over that interval can be differentiated to recover the original function's rate of change. Thus, this theorem is key to converting the problem from the integral form into a differential one, facilitating key solution strategies for equational analysis.

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Most popular questions from this chapter

Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$y^{\prime}=y e^{x}, \quad y(0)=2, \quad d x=0.5$$

Use a CAS to explore graphically each of the differential equations.Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y\) -window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b].\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error ( \(y\) (exact) \(-y\) (Euler)) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error.$$\begin{array}{l}y^{\prime}=(\sin x)(\sin y), \quad y(0)=2 ; \quad-6 \leq x \leq 6, \quad-6 \leq y \leq 6 \\\b=3 \pi / 2\end{array}.$$

Obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. $$\begin{aligned}. \(y^{\prime}=(\sin x)(\sin y), \quad y(0)=2 ; \quad-6 \leq x \leq 6, \quad-6 \leq y \leq 6\)

A tank initially contains 400 L of brine in which \(20 \mathrm{kg} / \mathrm{L}\) of salt are dissolved. A brine containing \(0.2 \mathrm{kg} / \mathrm{L}\) of salt runs into the tank at the rate of \(20 \mathrm{L} /\) min. The mixture is kept uniform by stirring and flows out of the tank at the rate of \(16 \mathrm{L} / \mathrm{min}\). a. At what rate (kilograms per minute) does salt enter the tank at time \(t ?\) b. What is the volume of brine in the tank at time \(t ?\) c. At what rate (kilograms per minute) does salt leave the tank at time \(t ?\) d. Write down and solve the initial value problem describing the mixing process. e. Find the concentration of salt in the tank 25 min after the process starts.

Write an equivalent first-order differential equatio and initial condition for \(y\). $$y=\int_{1}^{x} \frac{1}{t} d t.$$
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