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Magazine Chapter 1: Problem 3
Solve the given differential equation. $$e^{x+y} d y-d x=0$$
Short Answer
Expert verified
The short answer to the given differential equation \(e^{x+y} dy - dx = 0\) is: \( y(x) = e^{-y} x + C + 1 \), where C is the constant of integration.
Step by step solution
01
Rewrite the differential equation in the standard form
We want to rewrite the given equation as a first-order linear differential equation of the form:
\( \frac{dy}{dx} + P(x) y = Q(x) \)
The given equation is:
\( e^{x+y} dy - dx = 0 \)
To rewrite the equation in the standard form, divide both sides by \(e^{x+y} dx\):
\[ \frac{dy}{dx} - e^{-y} = 0 \]
Now, we have the equation in the form of \( \frac{dy}{dx} + P(x) y = Q(x) \), with \(P(x) = 0\) and \(Q(x) = e^{-y}\).
02
Find the integrating factor
An integrating factor (\(IF\)) is a function, which when multiplied by our linear equation, turns it into an exact differential equation. The formula for finding the integrating factor for a first-order linear differential equation is:
\( IF = e^{\int P(x) dx} \)
In our case, \(P(x) = 0\), and the integral of 0 with respect to x is simply 0.
So, the integrating factor is \( IF = e^0 = 1 \).
03
Multiply the equation by the integrating factor
Now, multiply the equation, \( \frac{dy}{dx} - e^{-y} = 0 \), by the integrating factor, 1.
\[ 1 \cdot \left(\frac{dy}{dx}-e^{-y}\right)= 0\left(1\right) \]
The equation remains unchanged:
\[ \frac{dy}{dx} - e^{-y} = 0 \]
04
Solve the differential equation
To solve the first-order linear differential equation, we can apply the separation of variables method. Move the functions involving y to one side and the functions involving x to the other side:
\[ dy + e^{-y} dx = 0 \]
Now, integrate both sides with respect to their respective variables:
\[ \int_{1}^y dy + \int_{0}^x e^{-y} dx = \int_0^0 0 \]
The left side consists of two integrals, so we have to calculate them separately:
\[ \int_{1}^y dy = y - 1 \]
and
\[ \int_{0}^x e^{-y} dx = -e^{-y} x + C \],
where C is the constant of integration.
Now, we can combine the results of both integrals:
\[ y - 1 - e^{-y} x = C \]
05
Solve for y
To present the solution in terms of the function y(x), add 1 to both sides of the equation:
\[y(x) = e^{-y} x + C + 1\]
And there we have our final solution for the given differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
Understanding first-order linear differential equations is fundamental in the study of differential equations. These are equations that involve the derivatives of a function and can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \) alone. The key to solving them is to first identify \( P(x) \) and \( Q(x) \) in the given equation.
In the exercise, we're given the equation \( e^{x+y} dy - dx = 0 \) which we can manipulate to fit the standard form mentioned above. By dividing both sides by \( e^{x+y} \) and rearranging, we obtain \( \frac{dy}{dx} - e^{-y} = 0 \) with \( P(x) = 0 \) and \( Q(x) = e^{-y} \). Notice that since \( P(x) \) is zero, it greatly simplifies the process of finding an integrating factor, which leads us to our next section.
In the exercise, we're given the equation \( e^{x+y} dy - dx = 0 \) which we can manipulate to fit the standard form mentioned above. By dividing both sides by \( e^{x+y} \) and rearranging, we obtain \( \frac{dy}{dx} - e^{-y} = 0 \) with \( P(x) = 0 \) and \( Q(x) = e^{-y} \). Notice that since \( P(x) \) is zero, it greatly simplifies the process of finding an integrating factor, which leads us to our next section.
Integrating Factor
The concept of an integrating factor is a neat trick that allows us to turn a non-exact differential equation into an exact one, by simply multiplying every term by a strategically chosen function. To find this special factor for a first-order linear differential equation, we typically use the formula \( IF = e^{\int P(x) dx} \).
In the exercise, \( P(x) \) is zero, and the integral of zero is zero. This means our integrating factor \( IF \) simplifies to \( e^0 \) or 1. Multiplying the equation by 1 doesn't change the equation, which is a unique and fortunate circumstance; otherwise, we'd need to multiply every term by the \( IF \) to proceed. An integrating factor usually makes it easier to combine the separate terms into a single derivative, streamlining the path to solving the equation.
In the exercise, \( P(x) \) is zero, and the integral of zero is zero. This means our integrating factor \( IF \) simplifies to \( e^0 \) or 1. Multiplying the equation by 1 doesn't change the equation, which is a unique and fortunate circumstance; otherwise, we'd need to multiply every term by the \( IF \) to proceed. An integrating factor usually makes it easier to combine the separate terms into a single derivative, streamlining the path to solving the equation.
Separation of Variables
The separation of variables method is a powerful technique for solving differential equations where variables can be separated on different sides of the equation. To achieve this separation, we rearrange the equation so that all terms involving \( y \) are on one side, and those involving \( x \) are on the other. We then integrate both sides, treating \( y \) and \( x \) as independent variables.
For the given exercise, we organize the terms to get \( dy + e^{-y} dx = 0 \). After separating the variables, we integrate each side with respect to its variable, leading to \( \int dy = y \) and \( \int e^{-y} dx = -e^{-y} x + C \), where \( C \) is the constant of integration. The final step is putting together these integrals to find \( y \) as a function of \( x \) which is the solution of the differential equation.
For the given exercise, we organize the terms to get \( dy + e^{-y} dx = 0 \). After separating the variables, we integrate each side with respect to its variable, leading to \( \int dy = y \) and \( \int e^{-y} dx = -e^{-y} x + C \), where \( C \) is the constant of integration. The final step is putting together these integrals to find \( y \) as a function of \( x \) which is the solution of the differential equation.
