Problem 72
Question
Solve the initial value problems. $$\frac{d y}{d x}=10-x, \quad y(0)=-1$$
Step-by-Step Solution
Verified Answer
The solution is \( y = 10x - \frac{x^2}{2} - 1 \).
1Step 1: Integrating the Differential Equation
The given differential equation is \( \frac{dy}{dx} = 10 - x \). To solve for \( y \), we integrate both sides with respect to \( x \): \[ y = \int (10 - x) \, dx \]},
2Step 2: Solving the Integral
Calculate the integral \( \int (10 - x) \, dx \), which can be solved by integrating each term separately:\[ y = \int 10 \, dx - \int x \, dx = 10x - \frac{x^2}{2} + C \]Where \( C \) is the constant of integration.
3Step 3: Applying the Initial Condition
Use the initial condition \( y(0) = -1 \) to solve for \( C \). Substitute \( x = 0 \) and \( y = -1 \) into the equation:\[ -1 = 10(0) - \frac{0^2}{2} + C \]\[ -1 = C \] Thus, \( C = -1 \).
4Step 4: Writing the Particular Solution
Now substitute \( C = -1 \) back into the integrated equation to get the particular solution:\[ y = 10x - \frac{x^2}{2} - 1 \]
Key Concepts
Differential EquationsIntegrationParticular SolutionConstant of Integration
Differential Equations
A differential equation is a mathematical equation involving an unknown function and its derivatives. It expresses how the function changes along with its variables. In this context, the given equation \( \frac{dy}{dx} = 10 - x \) describes how the function \( y \) changes with respect to \( x \).
Differential equations can represent various physical phenomena like motion, heat, and population dynamics.
These equations are classified into different types, such as ordinary (ODEs) and partial differential equations (PDEs), based on the number of variables and their derivatives. This particular problem involves an ordinary differential equation.
Differential equations can represent various physical phenomena like motion, heat, and population dynamics.
These equations are classified into different types, such as ordinary (ODEs) and partial differential equations (PDEs), based on the number of variables and their derivatives. This particular problem involves an ordinary differential equation.
- An ordinary differential equation (ODE) contains one independent variable and its derivatives.
- Solutions to differential equations can help us predict phenomena and understand underlying relationships.
Integration
Integration is a fundamental operation in calculus that is essentially the reverse process of differentiation. It allows us to find the original function given its derivative. In the solution, integration is used to find \( y \) from \( \frac{dy}{dx} = 10 - x \).
The integral of a function \( f(x) \) with respect to \( x \) is denoted by \( \int f(x) \, dx \). Performing integration on \( 10 - x \) involves breaking it into separate terms and integrating each:
The integral of a function \( f(x) \) with respect to \( x \) is denoted by \( \int f(x) \, dx \). Performing integration on \( 10 - x \) involves breaking it into separate terms and integrating each:
- \( \int 10 \, dx = 10x \)
- \( \int x \, dx = \frac{x^2}{2} \)
Particular Solution
A particular solution is a specific function that satisfies the differential equation along with any given initial conditions. Applying the initial value, such as \( y(0) = -1 \), transforms a general solution into a particular one.
In this exercise, after integration, we find the general solution, \( y = 10x - \frac{x^2}{2} + C \). The initial condition helps in finding the value of \( C \), making the solution particular. Here:
In this exercise, after integration, we find the general solution, \( y = 10x - \frac{x^2}{2} + C \). The initial condition helps in finding the value of \( C \), making the solution particular. Here:
- Substitute \( x = 0 \) and \( y = -1 \) into the equation.
- Solve the resulting equation for \( C \).
- Finally, the particular solution is \( y = 10x - \frac{x^2}{2} - 1 \).
Constant of Integration
The constant of integration \( C \) arises during indefinite integration, reflecting the family of curves that can all be potential solutions to a differential equation. However, a unique solution requires information beyond the equation itself, such as initial conditions.
In our problem, after integrating the right-hand side of \( \frac{dy}{dx} = 10 - x \), we find the general solution containing \( C \): \( y = 10x - \frac{x^2}{2} + C \). Given \( y(0) = -1 \), we solve for \( C \) to tailor the general solution to a specific scenario.
In our problem, after integrating the right-hand side of \( \frac{dy}{dx} = 10 - x \), we find the general solution containing \( C \): \( y = 10x - \frac{x^2}{2} + C \). Given \( y(0) = -1 \), we solve for \( C \) to tailor the general solution to a specific scenario.
- The integration constant allows for the flexibility of multiple solutions around a given derivative.
- It ensures calculus operations remain reversible, reconnecting derivative processes back to the original functions.
Other exercises in this chapter
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