Problem 72
Question
Solve the initial value problems in Exercises \(71-90\) . $$\frac{d y}{d x}=10-x, \quad y(0)=-1$$
Step-by-Step Solution
Verified Answer
The solution is \( y(x) = 10x - \frac{x^2}{2} - 1 \).
1Step 1: Identify the Given Differential Equation
We are given the 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 \( \frac{dy}{dx} = 10 - x \), we integrate both sides with respect to \( x \): \[ y(x) = \int (10 - x) \, dx = \int 10 \, dx - \int x \, dx \]. This gives us \( y(x) = 10x - \frac{x^2}{2} + C \), where \( C \) is the constant of integration.
3Step 3: Apply the Initial Condition
We use the initial condition \( y(0) = -1 \) to find the constant \( C \). Substituting \( x = 0 \) and \( y = -1 \) into the equation, we get: \( -1 = 10(0) - \frac{(0)^2}{2} + C \), which simplifies to \( -1 = C \). Therefore, \( C = -1 \).
4Step 4: Write the Particular Solution
Substitute \( C = -1 \) back into the expression for \( y(x) \) to find the particular solution: \( y(x) = 10x - \frac{x^2}{2} - 1 \). This function satisfies both the differential equation and the initial condition.
Key Concepts
First Order Differential EquationIntegrationConstant of IntegrationInitial Condition
First Order Differential Equation
To begin with, a **first order differential equation** is one of the simplest types of differential equations. It involves derivatives of the function we are trying to find, but only its first derivative. In simpler terms, if you have a function like \( y(x) \), the differential equation involves only \( \frac{dy}{dx} \). This specific equation: \( \frac{dy}{dx} = 10 - x \), is a classic example. It tells us how the function \( y(x) \) changes with respect to \( x \).
Understanding this change is crucial because it lets you predict the future behavior or state of the function based on its current state. Solving such equations gives us a formula for \( y(x) \) rather than just its rate of change.
Understanding this change is crucial because it lets you predict the future behavior or state of the function based on its current state. Solving such equations gives us a formula for \( y(x) \) rather than just its rate of change.
Integration
Solving the first order differential equation involves a mathematical process called **integration**. Integration is like the reverse of taking a derivative; instead of finding the rate of change, we try to reconstruct the original function.
In our exercise, we started with \( \frac{dy}{dx} = 10 - x \), which requires us to integrate both sides with respect to \( x \). This leads to \( y(x) = \int (10 - x) \, dx \). The integration splits into simpler integrals: \( \int 10 \, dx \) which results in \( 10x \), and \( \int x \, dx \) which gives \( \frac{x^2}{2} \). Thus, we arrive at \( y(x) = 10x - \frac{x^2}{2} + C \), where \( C \) is an unknown constant.
In our exercise, we started with \( \frac{dy}{dx} = 10 - x \), which requires us to integrate both sides with respect to \( x \). This leads to \( y(x) = \int (10 - x) \, dx \). The integration splits into simpler integrals: \( \int 10 \, dx \) which results in \( 10x \), and \( \int x \, dx \) which gives \( \frac{x^2}{2} \). Thus, we arrive at \( y(x) = 10x - \frac{x^2}{2} + C \), where \( C \) is an unknown constant.
Constant of Integration
When integrating, we often encounter something called a **constant of integration**, represented as \( C \). This is because integration can result in a family of functions. If we only know the rate from earlier parts, there could be many functions that fit.
The constant \( C \) accounts for all those possible vertical shifts in the function graph. Unless we have more information, we don't know exactly where the graph is on the vertical axis. That's why finding \( C \) is essential in creating a particular solution from a general one.
The constant \( C \) accounts for all those possible vertical shifts in the function graph. Unless we have more information, we don't know exactly where the graph is on the vertical axis. That's why finding \( C \) is essential in creating a particular solution from a general one.
Initial Condition
An **initial condition** helps us determine the specific constant of integration. It works like a clue, fixing one point on the function's graph, allowing us to pinpoint the exact form of the function.
In our step-by-step solution, the initial condition is \( y(0) = -1 \). By substituting \( x = 0 \) and \( y = -1 \) into our equation \( y(x) = 10x - \frac{x^2}{2} + C \), we solve for \( C \):
In our step-by-step solution, the initial condition is \( y(0) = -1 \). By substituting \( x = 0 \) and \( y = -1 \) into our equation \( y(x) = 10x - \frac{x^2}{2} + C \), we solve for \( C \):
- \( -1 = 10 \times 0 - \frac{(0)^2}{2} + C \), simplifies to \( -1 = C \).
Other exercises in this chapter
Problem 71
Let \(f\) be differentiable at every value of \(x\) and suppose that \(f(1)=1,\) that \(f^{\prime}0\) on \((1, \infty).\) \begin{equation}\begin{array}{l}{\text
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Determine the values of constants \(a, b, c,\) and \(d\) so that \(f(x)=a x^{3}+b x^{2}+c x+d\) has a local maximum at the point \((0,0)\) and a local minimum a
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Graph the functions in Exercises \(71-74 .\) Then find the extreme values of the function on the interval and say where they occur. $$ g(x)=|x-1|-|x-5|, \quad-2
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