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Chapter 7: Problem 14

In Exercises \(9-18\), solve the differential equation. $$ y^{\prime}=e^{x-y} $$

Short Answer

Expert verified
The general solution to the given differential equation \( y'=e^{x-y} \) is \( y(x)=\ln(e^x + C) \).

Step by step solution

01

Separating Variables

We will rewrite the given equation so that all the terms involving x are on one side and all the terms involving y are on the other side. To achieve this, we will multiply both sides by \(e^y\) and divide by \(e^x\). $$ e^y y'=e^x \\ e^y dy = e^x dx $$
02

Integration

Now, we will integrate both sides of the equation with respect to their respective variables (y and x): $$ \int e^y dy = \int e^x dx $$ $$ e^y = e^x + C $$
03

Rearranging the equation for y

Now, we will take the natural logarithm of both sides to solve for y: $$ y = \ln (e^x +C) $$ Thus, the general solution to the given differential equation is: $$ y(x)= \ln(e^x + C) $$

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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 powerful technique often used to solve ordinary differential equations (ODEs). This approach is particularly useful when the ODE is expressed in a form where all terms involving one variable can be arranged on one side of the equation, and all terms involving the other variable can be placed on the opposite side.

In the given example, the differential equation is: \( y' = e^{x-y} \).To separate the variables:
  • We multiply both sides by \( e^y \) to get \( e^y y' = e^x \).
  • Next, we express the expression as \( e^y dy = e^x dx \), successfully separating y-terms and x-terms.
By separating the variables, we set the stage for the next step: integration. This technique helps us treat the problem in parts and simplify solving complex relationships between variables by isolating them.
Integration
Integration is the process of finding a function from its derivative. Once the differential equation's variables are separated, we integrate each side with respect to its specific variable.

In our example, the separated equation is: \( e^y dy = e^x dx \). Integrating both sides, we have:
  • For the y-side: \( \int e^y dy = e^y + C_1 \)
  • For the x-side: \( \int e^x dx = e^x + C_2 \)
Since both sides will result in their own constants, we can simplify it as: \( e^y = e^x + C \), where \( C \) is a general constant that encompasses \( C_1 - C_2 \).
Integration here transforms our separated variables from their differential form into an expression containing an arbitrary constant, an essential component of the general solutions to differential equations.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. Natural logarithms are commonly used in differential equations when solving exponential forms.

In the final step of solving the differential equation, we have:\( e^y = e^x + C \).To solve for \( y \), we employ the natural logarithm:
  • Taking the natural logarithm of both sides gives us \( y = \ln(e^x + C) \).
By applying \( \ln \), we linearize the exponentials, making it easier to isolate \( y \). Understanding and using the natural logarithm in these contexts helps simplify expressions involving exponential growth or decay. It is a critical tool in transitioning from exponential forms to linear equations.

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Most popular questions from this chapter

Learning Curves The American Court Reporter Institute finds that the average student taking elementary machine shorthand will progress at a rate given by $$ \frac{d Q}{d t}=k(80-Q) $$ in a 20 -week course, where \(k\) is a positive constant and \(Q(t)\) measures the number of words of dictation a student can take per minute after \(t\) weeks in the course. If the average student can take 50 words of dictation per minute after 10 weeks in the course, how many words per minute can the average student take after completing the course?

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