Chapter 3: Problem 4
Finding a Particular Solution Find the particular solution to the differential equation \(y^{\prime}=2 x\) passing through the point \((2,7)\).
Short Answer
Expert verified
The particular solution is \(y = x^2 + 3\).
Step by step solution
01
Find the General Solution
We start by integrating the differential equation. The given equation is \(y' = 2x\). To find the general solution, we need to integrate both sides with respect to \(x\). \[ y = \int 2x \, dx \]The integral of \(2x\) is \(x^2 + C\), where \(C\) is the constant of integration. Thus, the general solution is:\[ y = x^2 + C \]
02
Use Initial Condition to Find Particular Solution
We are given that the solution passes through the point \((2,7)\). This means when \(x = 2\), \(y = 7\). We substitute these values into the general solution \(y = x^2 + C\):\[ 7 = 2^2 + C \]Simplify the equation:\[ 7 = 4 + C \]Solve for \(C\) by subtracting \(4\) from both sides:\[ C = 3 \]
03
Write the Particular Solution
Now that we have determined \(C = 3\), we can substitute it back into the general solution to get the particular solution:\[ y = x^2 + 3 \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Particular Solutions in Differential Equations
A particular solution of a differential equation provides a specific function that satisfies the equation for given conditions. A differential equation can often have many solutions, but by applying specific conditions - like passing through a certain point - we can pinpoint a unique solution.
For our exercise, the differential equation is represented by the derivative of a function, namely \( y' = 2x \). By solving the equation, we find a general form, \( y = x^2 + C \). This represents a family of curves varying with the constant \( C \). The specific condition, often termed the initial condition, indicates that the curve must pass through given coordinates
In summary, particular solutions are essential when we want a precise function from many solutions of differential equations, fitting specific conditions.
For our exercise, the differential equation is represented by the derivative of a function, namely \( y' = 2x \). By solving the equation, we find a general form, \( y = x^2 + C \). This represents a family of curves varying with the constant \( C \). The specific condition, often termed the initial condition, indicates that the curve must pass through given coordinates
- (x,y), like (2,7) in this exercise.
In summary, particular solutions are essential when we want a precise function from many solutions of differential equations, fitting specific conditions.
Integration as a Tool in Solving Differential Equations
Integration is the process used to solve differential equations and find the function whose derivative matches the given equation. To integrate is to essentially "reverse" the process of differentiation. It accumulates the rate of change represented by the differential equation over a particular range.
In our problem, we started with the equation \( y' = 2x \). By integrating \( 2x \) with respect to \( x \), we found the antiderivative, leading to the equation \( y = x^2 + C \). The constant \( C \) is crucial as it represents any constant shift vertically in the graph of the function.
In our problem, we started with the equation \( y' = 2x \). By integrating \( 2x \) with respect to \( x \), we found the antiderivative, leading to the equation \( y = x^2 + C \). The constant \( C \) is crucial as it represents any constant shift vertically in the graph of the function.
- Integration can sometimes be straightforward when dealing with simple functions like polynomials.
- However, more complex functions may require special techniques or transformations.
The Importance of Initial Conditions
Initial conditions are specific values assigned to a system, allowing us to find a particular solution from a general solution of a differential equation. They are like pointers telling us precisely which curve among a family of curves matches the requirements of a real-world scenario or a mathematical problem.
In the current example, the initial condition given is the point \((2,7)\). This implies that when \( x = 2 \), the function \( y \) should equal 7. By substituting these values into the general solution \( y = x^2 + C \), we solved for \( C \) (i.e., \( C = 3 \)). This calculation moved us from the general equation to our particular solution \( y = x^2 + 3 \).
In the current example, the initial condition given is the point \((2,7)\). This implies that when \( x = 2 \), the function \( y \) should equal 7. By substituting these values into the general solution \( y = x^2 + C \), we solved for \( C \) (i.e., \( C = 3 \)). This calculation moved us from the general equation to our particular solution \( y = x^2 + 3 \).
- Initial conditions help customize and refine solutions to suit specific needs or scenarios.
- They ensure the solution model fits precisely into the context defined by the initial values.