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Magazine Chapter 8: Problem 7
Find the general solution of each of the following differential equations. \(\left(1+e^{x}\right) y^{\prime}+2 e^{x} y=\left(1+e^{x}\right) e^{x}\)
Short Answer
Expert verified
The general solution is \( y = \frac{(1+e^x)}{3} + \frac{C}{(1+e^x)^2} \).
Step by step solution
01
Identify the type of differential equation
The given differential equation is \(\left(1+e^{x}\right) y^{\prime}+2 e^{x} y=\left(1+ e^{x}\right) e^{x}\). This is a first-order linear differential equation in the form: \( a(x)y' + b(x)y = c(x) \), where \( a(x) = 1 + e^x \), \( b(x) = 2e^x \), and \( c(x) = (1 + e^x) e^x \).
02
Rewrite the differential equation in standard form
Standard form for a first-order linear differential equation is: \( y' + P(x)y = Q(x) \). Divide both sides by \(1 + e^x \):\[ y' + \frac{2e^x}{1+e^x} y = \frac{(1+e^x) e^x}{1+e^x} \]. Simplify to get: \[ y' + \frac{2e^x}{1+e^x} y = e^x \].
03
Identify integrating factor
The integrating factor \( \mu(x) \) is determined by \( \mu(x) = e^{\int P(x) dx} \). Here, \( P(x) = \frac{2e^x}{1+e^x} \). Find the integral of \(P(x) \): \[ \int \frac{2e^x}{1+e^x} dx \]. Let \( u = 1+e^x \) then \( du = e^x dx \). The integral becomes \[ \int \frac{2du}{u} = 2 \ln|u| = 2 \ln|1+e^x| \]. Thus, \( \mu(x) = e^{2 \ln|1+e^x|} = (1+e^x)^2 \).
04
Multiply through by the integrating factor
Multiply the entire differential equation by the integrating factor \( (1+e^x)^2 \) to get: \[ (1+e^x)^2 y' + \frac{2e^x (1+e^x)}{1+e^x} y = (1+e^x)^2 e^x \]. This simplifies to: \[ (1+e^x)^2 y' + 2e^x (1+e^x) y = (1+e^x)^2 e^x \].
05
Recognize and simplify left-hand side
Notice that the left-hand side of the equation is a derivative of the product of the integrating factor and y: \[ \frac{d}{dx}\left[ (1+e^x)^2 y \right] = (1+e^x)^2 e^x \].
06
Integrate both sides
Integrate both sides with respect to x: \[ \int \frac{d}{dx}\left[ (1+e^x)^2 y \right] dx = \int (1+e^x)^2 e^x dx \]. The left-hand side becomes \[ (1+e^x)^2 y \]. For the right-hand side, let \( u = 1+e^x \), then \( du = e^x dx \): \[ \int u^2 du = \frac{u^3}{3} = \frac{(1+e^x)^3}{3} + C \].
07
Solve for y
Divide both sides by \( (1+e^x)^2 \) to solve for \( y \): \[ y = \frac{(1+e^x)^3/3 + C}{(1+e^x)^2} = \frac{(1+e^x)}{3} + \frac{C}{(1+e^x)^2} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
An integrating factor is a function used to multiply through a differential equation to make it easier to solve. This technique is particularly useful for first-order linear differential equations.
For a differential equation of the form:
\( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by:
\( \mu(x) = e^{\int P(x) dx} \).
This integrating factor, when multiplied with the entire equation, helps to transform the left-hand side into the derivative of a product of the integrating factor and \( y \).
In our example, \( P(x) = \frac{2e^x}{1 + e^x} \). Integrating \( P(x) \), we get:
\( \int \frac{2e^x}{1 + e^x} dx = 2 \ln|1 + e^x| \. \)
Thus, the integrating factor:
\( \mu(x) = e^{2 ln|1 + e^x|} = (1 + e^x)^2 \).
For a differential equation of the form:
\( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by:
\( \mu(x) = e^{\int P(x) dx} \).
This integrating factor, when multiplied with the entire equation, helps to transform the left-hand side into the derivative of a product of the integrating factor and \( y \).
In our example, \( P(x) = \frac{2e^x}{1 + e^x} \). Integrating \( P(x) \), we get:
\( \int \frac{2e^x}{1 + e^x} dx = 2 \ln|1 + e^x| \. \)
Thus, the integrating factor:
\( \mu(x) = e^{2 ln|1 + e^x|} = (1 + e^x)^2 \).
Differential Equations
Differential equations involve derivatives and are used to model various real-world phenomena like population growth, heat conduction, and motion. The equation in our exercise is a first-order linear differential equation of the form:
\( (1 + e^x)y' + 2e^x y = (1 + e^x)e^x \. \)
Here:
The goal is to find a function \( y(x) \) that satisfies this equation. By following systematic solution techniques, we can uncover this function.
\( (1 + e^x)y' + 2e^x y = (1 + e^x)e^x \. \)
Here:
- \( y' \) represents the first derivative of \( y \)
- \( \left(1+e^x\right) \) and \( 2e^x \) are functions of \( x \) that multiply \( y' \) and \( y \) respectively
- \( (1 + e^x) e^x \) is the function on the right side of the equation
The goal is to find a function \( y(x) \) that satisfies this equation. By following systematic solution techniques, we can uncover this function.
Solution Techniques
To find the solution of a first-order linear differential equation, we employ specific techniques. The main steps are:
In our specific example, after calculating the integrating factor:
\( (1 + e^x)^2 \, \) we multiply it throughout the differential equation, transforming it into:
\( \frac{d}{dx}[(1 + e^x)^2 y] = (1 + e^x)^2 e^x \. \)
Upon integration and subsequent simplification, we ultimately derive the general solution:
\( y = \frac{1 + e^x}{3} + \frac{C}{(1 + e^x)^2} \. \)
This systematic approach helps to find the solution efficiently and correctly.
- Rewrite the equation in standard linear form \( y' + P(x)y = Q(x) \)
- Identify and calculate the integrating factor \( \mu(x) \)
- Multiply the entire equation by the integrating factor
- Recognize and simplify the resulting expression
- Integrate both sides to find the solution
In our specific example, after calculating the integrating factor:
\( (1 + e^x)^2 \, \) we multiply it throughout the differential equation, transforming it into:
\( \frac{d}{dx}[(1 + e^x)^2 y] = (1 + e^x)^2 e^x \. \)
Upon integration and subsequent simplification, we ultimately derive the general solution:
\( y = \frac{1 + e^x}{3} + \frac{C}{(1 + e^x)^2} \. \)
This systematic approach helps to find the solution efficiently and correctly.
