Chapter 7: Problem 22
Solve the differential equation. $$\frac{d y}{d x}=e^{x-y}+e^{x}+e^{-y}+1$$
Short Answer
Expert verified
The differential equation is solved using substitution and integration techniques, yielding \(y = x + v\).
Step by step solution
01
Rearrange the Terms
We start with the given differential equation: \( \frac{d y}{d x} = e^{x-y} + e^{x} + e^{-y} + 1 \). Our goal is to separate variables to solve this differential equation. The form of the terms suggests the use of substitution.
02
Introduce a Substitution
Introduce the substitution \(v = x - y\), which implies \(y = x - v\) and \(\frac{dy}{dx} = 1 - \frac{dv}{dx}\). Substitute these into the differential equation.
03
Substitute and Simplify
Replace \(y\) with \(x-v\), yielding \(1 - \frac{dv}{dx} = e^v + e^x + e^{-x+v} + 1\). Rearranging gives: \(-\frac{dv}{dx} = e^v + e^{-v} + e^x\) after canceling 1 on both sides.
04
Rearrange for Separation of Variables
Rearrange to separate variables: \(-dv = (e^v + e^{-v} + e^x) dx\). Attempt further simplification if possible by factoring or recognizing forms for integration.
05
Integrate Both Sides
Integrate both sides: \(-\int dv = \int (e^v + e^{-v} + e^x) dx\). The integral of the left side is straightforward: \(-v + C_1\). The right side can be integrated term by term using known integral forms.
06
Evaluate the Integrals
The integration results are \(-v + C_1 = \int e^v \, dv + \int e^{-v} \, dv + \int e^x \, dx\), leading to \(-v = e^v - e^{-v} + e^x + C_2\). Where each term is derived from standard exponential integrals.
07
Solve for the Original Variables
To find \(y(x)\), revert to the original variables: \(v = x - y\), thus \(y = x - (-v) = x + v\). Substitute the expression for \(v\) back in, considering constants of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separation of Variables
Separation of variables is a method used to solve differential equations. This technique involves rearranging an equation so that each different model of a function is on opposite sides of the equation. The form is typically \(\frac{dy}{dx} = g(y) \cdot h(x)\). This facilitates integrating both sides with respect to their respective variables.
If the equation can be separated successfully, it means you get two integrals: one for each variable. Here’s what you’ll usually do:
If the equation can be separated successfully, it means you get two integrals: one for each variable. Here’s what you’ll usually do:
- Bring all terms involving \(y\) to one side.
- Move all terms involving \(x\) to the other side.
- Integrate both sides separately.
Substitution Method
The substitution method is another powerful tool used to solve differential equations. With this approach, one replaces a complex part of the equation with a new variable to simplify it. It often makes an equation less convoluted and easier to manage.
In many of these scenarios, the substitution involves:
In many of these scenarios, the substitution involves:
- Identifying a portion of the equation that can be expressed with a single variable.
- Rewriting other parts of the equation in terms of this new variable.
- Turning the derivative into simpler forms with respect to the new substitution.
Exponential Functions
Exponential functions appear frequently in differential equations. Such functions involve terms like \(e^x\) or \(e^{-y}\), representing growth and decay processes. In our solution, exponential terms were pervasive and needed simplification.
When integrating exponential functions, you can rely on these rules of thumb:
When integrating exponential functions, you can rely on these rules of thumb:
- The integral of \(e^x\) with respect to \(x\) is simply \(e^x\).
- The integral of a shifted exponential, like \(e^{ax}\), will be \(\frac{1}{a}e^{ax}\).
- Negative exponents modify the structure, but not the core principles, for example, \(\-e^{-y}\) integrates to \(e^{-y}\).