Problem 68
Question
Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=10-x, \quad y(0)=-1 $$
Step-by-Step Solution
Verified Answer
The solution is \( y(x) = 10x - \frac{x^2}{2} - 1 \).
1Step 1: Understand the Problem
We are given a first-order ordinary differential equation \( \frac{dy}{dx} = 10 - x \) with the initial condition \( y(0) = -1 \). Our goal is to find the function \( y(x) \) that satisfies both the differential equation and the initial condition.
2Step 2: Integrate the Differential Equation
To solve the differential equation, we integrate both sides with respect to \( x \). The equation becomes: \[\int \frac{dy}{dx} \, dx = \int (10 - x) \, dx.\] The left-hand side simplifies to \( y \) (the antiderivative of \( \frac{dy}{dx} = 1 \)), and the right-hand side is the antiderivative of \( 10 - x \), which is \( 10x - \frac{x^2}{2} + C \), where \( C \) is the constant of integration.
3Step 3: Include the Constant of Integration
After integration, we find that \[y = 10x - \frac{x^2}{2} + C.\] Here, \( C \) is an arbitrary constant that we need to find using the initial condition.
4Step 4: Apply the Initial Condition
We use the initial condition \( y(0) = -1 \) to find the value of \( C \). Substituting \( x = 0 \) and \( y = -1 \) into the integrated equation, we get:\[-1 = 10(0) - \frac{0^2}{2} + C.\]Simplifying, we find:\[C = -1.\]
5Step 5: Write the Final Solution
Substitute \( C = -1 \) back into the expression for \( y(x) \). The final solution is:\[y = 10x - \frac{x^2}{2} - 1.\]This equation satisfies both the differential equation and the initial condition.
Key Concepts
Ordinary Differential EquationIntegrationConstant of IntegrationInitial Condition
Ordinary Differential Equation
An ordinary differential equation (ODE) is an equation involving a function and its derivatives. In simplest terms, it's a mathematical expression that relates an unknown function, often representing a physical process, to its rate of change. For instance, if you think about how fast a car is accelerating, the speed is the first derivative, while the position is the function you're trying to find.
In the given problem, the ODE is \( \frac{dy}{dx} = 10 - x \). This tells us that the rate at which \( y \) changes with respect to \( x \) is \( 10 - x \). Our task is to determine the function \( y(x) \) from this equation. By solving it, we can find \( y \) as a function of \( x \) that describes the entire relationship between the two variables.
In the given problem, the ODE is \( \frac{dy}{dx} = 10 - x \). This tells us that the rate at which \( y \) changes with respect to \( x \) is \( 10 - x \). Our task is to determine the function \( y(x) \) from this equation. By solving it, we can find \( y \) as a function of \( x \) that describes the entire relationship between the two variables.
Integration
Integration is the process of finding a function from its derivative, essentially reversing differentiation. When you integrate a function, you're looking for what you started with before you differentiated it. In other words, integration helps us to work backwards to find the original function.
In our example, to solve \( \frac{dy}{dx} = 10 - x \), we integrate both sides of the equation with respect to \( x \). The integral of \( \frac{dy}{dx} \) simplifies to \( y \), while integrating \( 10 - x \) gives us \( 10x - \frac{x^2}{2} + C \), where \( C \) is known as the constant of integration.
In our example, to solve \( \frac{dy}{dx} = 10 - x \), we integrate both sides of the equation with respect to \( x \). The integral of \( \frac{dy}{dx} \) simplifies to \( y \), while integrating \( 10 - x \) gives us \( 10x - \frac{x^2}{2} + C \), where \( C \) is known as the constant of integration.
Constant of Integration
The constant of integration, represented by \( C \), is an essential part of solving differential equations. When we integrate a function, we are finding a family of solutions. Each element of this family is a shifted version of another, distinguished only by \( C \).
Because the derivative of a constant is zero, when we differentiate any element of this family, we lose the constant. That's why we need to introduce \( C \) when integrating—it accounts for all possible vertical shifts of the graph of the function. In this problem, once we've integrated to find \( y = 10x - \frac{x^2}{2} + C \), the constant \( C \) enables us to satisfy any specific initial condition provided.
Because the derivative of a constant is zero, when we differentiate any element of this family, we lose the constant. That's why we need to introduce \( C \) when integrating—it accounts for all possible vertical shifts of the graph of the function. In this problem, once we've integrated to find \( y = 10x - \frac{x^2}{2} + C \), the constant \( C \) enables us to satisfy any specific initial condition provided.
Initial Condition
An initial condition is a value that allows us to find the specific solution from the family of solutions given by an indefinite integral. In many real-world problems, initial conditions are parameters we know or measure, like the initial position or temperature at \( t = 0 \).
In order to find the particular value of the constant \( C \) in the equation \( y = 10x - \frac{x^2}{2} + C \), we use the initial condition \( y(0) = -1 \). By substituting these values into the equation, we find that \( C = -1 \). This creates the unique solution \( y = 10x - \frac{x^2}{2} - 1 \), which satisfies both the differential equation and initial condition.
In order to find the particular value of the constant \( C \) in the equation \( y = 10x - \frac{x^2}{2} + C \), we use the initial condition \( y(0) = -1 \). By substituting these values into the equation, we find that \( C = -1 \). This creates the unique solution \( y = 10x - \frac{x^2}{2} - 1 \), which satisfies both the differential equation and initial condition.
Other exercises in this chapter
Problem 67
Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=2 x-7, \quad y(2)=0 $$
View solution Problem 67
Odd functions If an odd function \(g(x)\) has a local minimum value at \(x=c,\) can anything be said about the value of \(g\) at \(x=-c ?\) Give reasons for you
View solution Problem 68
We know how to find the extreme values of a continuous function \(f(x)\) by investigating its values at critical points and endpoints. But what if there are no
View solution Problem 68
Sketch a smooth connected curve \(y=f(x)\) with $$ \begin{array}{ll}{f(-2)=8,} & {f^{\prime}(2)=f^{\prime}(-2)=0} \\ {f(0)=4,} & {f^{\prime}(x) 0 \text { for }|
View solution