Problem 22 - Calculus Single Variable

Question

In each of Exercises \(17-28,\) solve the given initial value problem. $$ \frac{d y}{d x}+2 x y=\exp \left(-x^{2}\right) \quad y(0)=5 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{x + 5}{\exp(x^2)} \).

Step 1: 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 \( Q(x) = \exp(-x^2) \).

Step 2: 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 \( \mu(x) = \exp(x^2) \).

Step 3: 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 \( \frac{d}{dx} (\exp(x^2) y) \).

Step 4: 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 \( \exp(x^2)y = x + C \).

Step 5: 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 \( C = 5 \).

Step 6: 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 \( y = \frac{x + 5}{\exp(x^2)} \).

Key Concepts

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.

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.

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.

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.