Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 49
Solve the initial value. $$\frac{d^{2} y}{d x^{2}}=2 e^{-x}, \quad y(0)=1 \quad \text { and } \quad y^{\prime}(0)=0$$
Short Answer
Expert verified
The solution is \(y(x) = 2e^{-x} + 2x - 1\).
Step by step solution
01
Identify the differential equation and initial conditions
The given problem is a second-order differential equation \(\frac{d^2 y}{dx^2} = 2e^{-x}\) with the initial conditions \(y(0) = 1\) and \(y'(0) = 0\). Our goal is to find the function \(y(x)\) that satisfies both the differential equation and the initial conditions.
02
Integrate to find the first derivative
Integrate the second-order differential equation to obtain the first derivative:\[\int \frac{d^2 y}{dx^2} \, dx = \int 2e^{-x} \, dx\]This results in:\[\frac{dy}{dx} = -2e^{-x} + C_1\]where \(C_1\) is an integration constant.
03
Apply the initial condition to find C1
Use the initial condition \(y'(0) = 0\) to solve for \(C_1\):\[-2e^{0} + C_1 = 0\]This simplifies to:\[C_1 = 2\]Thus, the expression for the first derivative becomes:\[\frac{dy}{dx} = -2e^{-x} + 2\]
04
Integrate to find y(x)
Now integrate the first derivative equation to find \(y(x)\):\[\int \frac{dy}{dx} \, dx = \int (-2e^{-x} + 2) \, dx\]This results in:\[y = 2e^{-x} + 2x + C_2\]where \(C_2\) is another integration constant.
05
Apply the second initial condition to find C2
Use the initial condition \(y(0) = 1\) to determine \(C_2\):\[2e^{0} + 2 \times 0 + C_2 = 1\]This simplifies to:\[2 + C_2 = 1\]Thus, \(C_2 = -1\).
06
Write the final solution
Substitute \(C_2 = -1\) into the expression for \(y(x)\):\[y(x) = 2e^{-x} + 2x - 1\]This is the function that satisfies both the differential equation and the initial conditions.
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical expressions that relate a function with its derivatives. They describe how a certain quantity changes with respect to another. In the given exercise, the equation \( \frac{d^2 y}{dx^2} = 2e^{-x} \) is a second-order differential equation. This means it involves the second derivative of \( y \) with respect to \( x \).
Differential equations are vital in various fields such as physics, engineering, and economics because they model real-world phenomena. Solving them means finding the unknown function \( y(x) \) that satisfies the equation. In this case, both the second derivative and initial conditions are used to find that function.
Differential equations are vital in various fields such as physics, engineering, and economics because they model real-world phenomena. Solving them means finding the unknown function \( y(x) \) that satisfies the equation. In this case, both the second derivative and initial conditions are used to find that function.
Initial Value Problems
An initial value problem involves solving a differential equation with specific initial conditions provided. Initial conditions are values known at a particular point and are used to find specific solutions, not general ones. In the problem, the initial conditions are \( y(0) = 1 \) and \( y'(0) = 0 \).
These conditions mean that at \( x = 0 \), the function \( y \) equals 1, and its first derivative (slope) is 0. These constraints are crucial because they allow us to find the constants in the integrals when solving differential equations.
These conditions mean that at \( x = 0 \), the function \( y \) equals 1, and its first derivative (slope) is 0. These constraints are crucial because they allow us to find the constants in the integrals when solving differential equations.
Integration
Integration is the mathematical process of finding the antiderivative or the integral of a function. It is the reverse operation of differentiation. To solve the given differential equation, we integrate twice: first to find the first derivative \( \frac{dy}{dx} \), and then to find \( y(x) \) itself.
When we integrated \( 2e^{-x} \), we arrived at \( -2e^{-x} + C_1 \), where \( C_1 \) is the constant of integration. After applying initial conditions, another integration was done to find \( y(x) \), resulting in \( 2e^{-x} + 2x + C_2 \). Each integration step brings in a new constant, which is determined using the initial conditions.
When we integrated \( 2e^{-x} \), we arrived at \( -2e^{-x} + C_1 \), where \( C_1 \) is the constant of integration. After applying initial conditions, another integration was done to find \( y(x) \), resulting in \( 2e^{-x} + 2x + C_2 \). Each integration step brings in a new constant, which is determined using the initial conditions.
Second Order Derivatives
A second-order derivative \( \frac{d^2 y}{dx^2} \) refers to the derivative of a derivative鈥攈ow the slope of a function changes. In the problem, \( \frac{d^2 y}{dx^2} = 2e^{-x} \) highlights that the curve has a specific rate of change in curvature.
Understanding second-order derivatives helps in assessing the concavity of a function and the behavior of dynamic systems, such as how an object moves under the influence of forces. Solving for \( y(x) \) involved integrating this second derivative twice, because the goal is to get back to the original function. These two integrations, constrained by the initial conditions, yield a unique function that meets the criteria of the problem.
Understanding second-order derivatives helps in assessing the concavity of a function and the behavior of dynamic systems, such as how an object moves under the influence of forces. Solving for \( y(x) \) involved integrating this second derivative twice, because the goal is to get back to the original function. These two integrations, constrained by the initial conditions, yield a unique function that meets the criteria of the problem.
