Chapter 31: Problem 16
Solve the given differential equations. $$e^{x+y}(d x+d y)+4 x d x=0$$
Short Answer
Expert verified
The solution is \( y = \ln(2x - 2 + Ce^{x}) - x \).
Step by step solution
01
Simplify the Differential Equation
Start with the given equation: \[ e^{x+y}(dx + dy) + 4x \, dx = 0. \]Distribute the \( e^{x+y} \) term:\[ e^{x+y} dx + e^{x+y} dy + 4x \, dx = 0. \]Combine the terms involving \( dx \):\[ (e^{x+y} + 4x) \, dx + e^{x+y} \, dy = 0. \]
02
Isolate the dy Term
Rearrange the equation to isolate the \( dy \) term:\[ e^{x+y} \, dy = -(e^{x+y} + 4x) \, dx \]Or equivalently,\[ rac{dy}{dx} = -1 - \frac{4x}{e^{x+y}}. \]
03
Use an Integrating Factor
Recognize that we can't separate the variables directly, so consider a substitution or an integrating factor. Here we can set \( v = e^{x+y} \), so \( rac{dv}{dx} = rac{d}{dx}(e^{x+y}) = e^{x+y}(1 + \frac{dy}{dx}) \).Using the differentiated form and substituting back, rearrange to bring the equation:\[ rac{dv}{dx} = v + 4x \].
04
Solve for v Using the Integrating Factor Method
Observe that this is a linear differential equation in terms of \( v \). The standard form is \[ rac{dv}{dx} - v = 4x. \]The integrating factor is \( e^{-x} \), multiply through:\[ e^{-x} \frac{dv}{dx} - e^{-x}v = 4xe^{-x}. \]Integrate the left side and solve:\[ rac{d}{dx}(ve^{-x}) = 4xe^{-x} \]Integrate the right side:\[ ve^{-x} = rac{4}{2}xe^{-x} - rac{4}{2}rac{1}{e^{x}} + C. \]Simplify to find \( v \):\[ v = 2x - 2 + Ce^{x}. \]
05
Re-substitute to Find Final Solution
Recall that \( v = e^{x+y} \).So, \[ e^{x+y} = 2x - 2 + Ce^{x}. \]Taking the natural logarithm to solve for \( y \):\[ y = ext{ln}(2x - 2 + Ce^{x}) - x. \]
06
Conclusion
The solution for the differential equation is:\[ y = ext{ln}(2x - 2 + Ce^{x}) - x, \]where \( C \) is a constant determined by initial conditions if provided.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
The integrating factor is a crucial concept when solving linear differential equations. It functions as a strategic multiplier that simplifies the process of finding a solution. The technique is particularly useful when dealing with non-separable linear differential equations. When dealing with an equation in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is derived from the function \( e^{\int P(x) \, dx} \). This essentially converts the left side of the equation into the derivative of a product, making the equation more tractable.
- Begin by identifying the standard form of the equation.
- Calculate the integrating factor \( \mu(x) = e^{\int P(x) \ dx} \).
- Multiply every term in the differential equation by this integrating factor.
- The equation is then transformed into a simpler form that is easier to solve by integration.
Linear Differential Equations
Linear differential equations play a prominent role in the study of differential equations. These are equations that involve derivatives of a function and the function itself, appearing linearly. In formal terms, a first-order linear differential equation can be written as \( \frac{dy}{dx} + P(x)y = Q(x) \). The significance of these equations lies in their wide applicability across sciences and engineering.
- They are called 'linear' because the dependent variable \( y \) and its derivative appear to the first power and there are no products of \( y \) and \( \frac{dy}{dx} \).
- These equations can be homogeneous (\( Q(x) = 0 \)) or non-homogeneous (\( Q(x) eq 0 \)).
- The solution approach often leverages methods such as separation of variables or integrating factors.
Substitution Method
The substitution method is a powerful technique in solving differential equations, especially when direct methods like separation of variables fail or become cumbersome. This involves substituting a part of the differential equation with a new variable, simplifying its structure, and allowing for easier manipulation.
- Start by identifying a substitution that simplifies the equation. Often, this relates to terms or expressions that repeat or look algebraically complex.
- Introduce a new variable to replace these expressions.
- Rewrite the differential equation with this substitution, aiming for a form that is easier to integrate.
- Find a solution in terms of the new variable, then substitute back to express the solution in terms of the original variables.