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Magazine Chapter 7: Problem 22
In each of Exercises \(17-28,\) solve the given initial value problem. $$ \frac{d y}{d x}+2 x y=\exp \left(-x^{2}\right) \quad y(0)=5 $$
Short Answer
Expert verified
The solution is \( y(x) = \frac{x + 5}{\exp(x^2)} \).
Step by step solution
01
Identify the Equation Type
This problem presents a first-order linear differential equation in the standard form \( \frac{dy}{dx} + P(x) y = Q(x) \). Here, \( P(x) = 2x \) and \( Q(x) = \exp(-x^2) \).
02
Find the Integrating Factor
To solve this differential equation, we first need to find the integrating factor, \( \mu(x) \), which is calculated as \( \exp\left(\int P(x) \, dx\right) = \exp\left(\int 2x \, dx\right) \). Integrating, we get \( \int 2x \, dx = x^2 \), thus \( \mu(x) = \exp(x^2) \).
03
Use the Integrating Factor
Multiply the entire differential equation by \( \mu(x) = \exp(x^2) \), yielding \( \exp(x^2) \frac{dy}{dx} + 2x\exp(x^2)y = \exp(0) \). The left side is the derivative of \( \exp(x^2)y \) with respect to \( x \), thus it simplifies to \( \frac{d}{dx} (\exp(x^2) y) \).
04
Integrate Both Sides
Integrate both sides with respect to \( x \). The left side simplifies to \( \exp(x^2)y \) and the right side integrates to \( x + C \), so we have \( \exp(x^2)y = x + C \).
05
Apply the Initial Condition
Use the initial condition \( y(0) = 5 \) to solve for \( C \). Substitute \( x = 0 \) and \( y = 5 \) into \( \exp(x^2)y = x + C \): \( \exp(0) \cdot 5 = 0 + C \) so \( C = 5 \).
06
Solve for y(x)
Substitute \( C = 5 \) back into the equation \( \exp(x^2) y = x + C \). We have \( \exp(x^2) y = x + 5 \), and thus \( y = \frac{x + 5}{\exp(x^2)} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equations
A first-order linear differential equation is an equation that relates a function, its derivative, and a variable. In this exercise, the differential equation given is in the form \( \frac{dy}{dx} + P(x) y = Q(x) \), which is a typical structure.
For our specific equation, \( P(x) = 2x \), and \( Q(x) = \exp(-x^2) \), where \( dy/dx \) is the derivative of \( y \) with respect to \( x \).
The task is to solve for \( y \) as a function of \( x \), taking into account the initial condition \( y(0) = 5 \).
This type of equation nicely lends itself to the integrating factor method, making it solvable via straightforward steps.
For our specific equation, \( P(x) = 2x \), and \( Q(x) = \exp(-x^2) \), where \( dy/dx \) is the derivative of \( y \) with respect to \( x \).
The task is to solve for \( y \) as a function of \( x \), taking into account the initial condition \( y(0) = 5 \).
This type of equation nicely lends itself to the integrating factor method, making it solvable via straightforward steps.
Integrating Factor Method
The integrating factor method is crucial for solving first-order linear differential equations. It simplifies the equation so that we can easily integrate both sides.
To find the integrating factor \( \mu(x) \), we calculate \( \exp(\int P(x) \, dx) \).
For our equation, \( P(x) = 2x \), so we integrate: \( \int 2x \, dx = x^2 \).
This results in the integrating factor \( \mu(x) = \exp(x^2) \). Applying the integrating factor to the differential equation transforms it, making one side the derivative of a product.
This method systematically guides us straight to the solution, so using it effectively is important.
To find the integrating factor \( \mu(x) \), we calculate \( \exp(\int P(x) \, dx) \).
For our equation, \( P(x) = 2x \), so we integrate: \( \int 2x \, dx = x^2 \).
This results in the integrating factor \( \mu(x) = \exp(x^2) \). Applying the integrating factor to the differential equation transforms it, making one side the derivative of a product.
This method systematically guides us straight to the solution, so using it effectively is important.
Solving Differential Equations
Once we have applied the integrating factor, we can multiply the entire differential equation by \( \mu(x) \), which is \( \exp(x^2) \) in our case.
The equation then appears as \( \exp(x^2) \frac{dy}{dx} + 2x\exp(x^2)y = \exp(0) \).
This can be rewritten as a derivative: \( \frac{d}{dx} (\exp(x^2) y) \). This allows us to integrate both sides easily.Upon integrating, we simplify the equation to \( \exp(x^2) y = x + C \), where \( C \) is a constant of integration.
This process of finding \( y \) requires careful algebraic manipulation along with understanding the properties of exponential functions.
The equation then appears as \( \exp(x^2) \frac{dy}{dx} + 2x\exp(x^2)y = \exp(0) \).
This can be rewritten as a derivative: \( \frac{d}{dx} (\exp(x^2) y) \). This allows us to integrate both sides easily.Upon integrating, we simplify the equation to \( \exp(x^2) y = x + C \), where \( C \) is a constant of integration.
This process of finding \( y \) requires careful algebraic manipulation along with understanding the properties of exponential functions.
Applying Initial Conditions
Initial conditions are given values that solve for unknown constants in differential equations.
Here, the initial condition is \( y(0) = 5 \). They are pivotal in finding specific solutions from general ones.
Substituting \( x = 0 \) and \( y = 5 \) into our integrated equation \( \exp(x^2)y = x + C \) yields values \( \exp(0) \cdot 5 = 0 + C \). This simplifies to \( C = 5 \).By plugging \( C = 5 \) back, we finalize our solution as \( y = \frac{x + 5}{\exp(x^2)} \).
Initial conditions ensure that our solution \( y(x) \) satisfies the original problem setup, thus verifying correctness.
Here, the initial condition is \( y(0) = 5 \). They are pivotal in finding specific solutions from general ones.
Substituting \( x = 0 \) and \( y = 5 \) into our integrated equation \( \exp(x^2)y = x + C \) yields values \( \exp(0) \cdot 5 = 0 + C \). This simplifies to \( C = 5 \).By plugging \( C = 5 \) back, we finalize our solution as \( y = \frac{x + 5}{\exp(x^2)} \).
Initial conditions ensure that our solution \( y(x) \) satisfies the original problem setup, thus verifying correctness.
