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Magazine Chapter 8: Problem 2
Solve the differential equation by the method of integrating factors. \(\frac{d y}{d x}+2 x y=x\)
Short Answer
Expert verified
The solution is \( y = \frac{1}{2} + Ce^{-x^2} \).
Step by step solution
01
Identify the Standard Form
The given first-order linear differential equation is \( \frac{dy}{dx} + 2xy = x \). The standard form for applying the method of integrating factors is \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, we have \( P(x) = 2x \) and \( Q(x) = x \).
02
Determine the Integrating Factor
The integrating factor \( \mu(x) \) is found using \( \mu(x) = e^{\int P(x) \, dx} \). In this case, \( P(x) = 2x \), so we compute the integral: \( \int 2x \, dx = x^2 \). Therefore, the integrating factor is \( \mu(x) = e^{x^2} \).
03
Multiply Through by the Integrating Factor
Multiply every term in the differential equation by the integrating factor \( e^{x^2} \): \( e^{x^2} \frac{dy}{dx} + 2x e^{x^2} y = x e^{x^2} \).
04
Rewrite the Left Side as a Derivative
Notice that the left side of the equation \( e^{x^2} \frac{dy}{dx} + 2x e^{x^2} y \) can be written as a single derivative: \( \frac{d}{dx}(e^{x^2}y) \). Thus, the equation becomes \( \frac{d}{dx}(e^{x^2}y) = x e^{x^2} \).
05
Integrate Both Sides
Integrate both sides with respect to \( x \): \( e^{x^2}y = \int x e^{x^2} \, dx \). To evaluate the integral on the right, use substitution: let \( u = x^2 \), so \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2}du \). The integral becomes \( \frac{1}{2} \int e^u \, du = \frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C \).
06
Solve for \( y \)
From \( e^{x^2}y = \frac{1}{2}e^{x^2} + C \), solve for \( y \): Divide both sides by \( e^{x^2} \): \( y = \frac{1}{2} + Ce^{-x^2} \). This is the general solution to the differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Differential Equation
Generally, a linear differential equation of the first order can be expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). This is an ordinary differential equation where both the dependent variable \( y \) and its derivative \( \frac{dy}{dx} \) appear linearly.
There are a few key points to keep in mind with these equations:
There are a few key points to keep in mind with these equations:
- The equation is "linear" because it can be rearranged to resemble a straight line in terms of \( y \) and its derivatives, except for the functions \( P(x) \) and \( Q(x) \).
- The term \( P(x)y \) signifies that \( y \) is multiplied by a function of \( x \), affecting the overall behavior of the solution.
- The term \( Q(x) \) is independent of \( y \) and serves as a "forcing function" that influences the equation's output directly.
Integrating Factor Calculation
The method of integrating factors is a powerful technique to solve first-order linear differential equations. Before anything, you need the equation in its standard form \( \frac{dy}{dx} + P(x)y = Q(x) \). Next, compute the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
Let鈥檚 discuss the step-by-step approach:
Let鈥檚 discuss the step-by-step approach:
- Identify \( P(x) \). It's the function with which \( y \) is multiplied. In our example, \( P(x) = 2x \).
- Compute the integral \( \int P(x) \, dx \). Here, you integrate \( 2x \), resulting in \( x^2 \).
- Compute the integrating factor as \( \mu(x) = e^{x^2} \).
First-Order Differential Equation
A first-order differential equation involves only the first derivative of the function and no higher derivatives. In our example, we are given an equation \( \frac{dy}{dx} + 2xy = x \), where \( \frac{dy}{dx} \) is the first derivative of \( y \) with respect to \( x \).
Here are some important points regarding first-order differential equations:
Here are some important points regarding first-order differential equations:
- These equations generally describe how the rate of change of a variable \( y \) depends on \( x \) and possibly on \( y \) itself.
- Since only the first derivative is involved, they are usually simpler to solve than higher-order equations, where second or third derivatives would be included.
- Solutions to first-order equations often represent the family of curves describing the relationship between the variables.
