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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.4.11
Chapter 7, Problem 7.4.11

Solve the differential equation in Exercises 9–22.
11. (dy/dx) = e^(x-y)

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Rewrite the given differential equation \( \frac{dy}{dx} = e^{x - y} \) to separate the variables \( y \) and \( x \). Start by expressing \( e^{x - y} \) as \( e^x \cdot e^{-y} \), so the equation becomes \( \frac{dy}{dx} = e^x e^{-y} \).
Separate the variables by multiplying both sides by \( e^y \) and multiplying both sides by \( dx \), giving \( e^y dy = e^x dx \). This puts all \( y \)-terms on one side and all \( x \)-terms on the other.
Integrate both sides: integrate \( e^y \) with respect to \( y \) on the left, and integrate \( e^x \) with respect to \( x \) on the right. This yields \( \int e^y dy = \int e^x dx \).
After integrating, write the general solution including the constant of integration \( C \). The integrals of \( e^y \) and \( e^x \) are both \( e^y \) and \( e^x \) respectively, so the equation becomes \( e^y = e^x + C \).
If needed, solve explicitly for \( y \) by taking the natural logarithm of both sides, resulting in \( y = \ln(e^x + C) \), which represents the implicit or explicit general solution to the differential equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the solution.
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Integration of Exponential Functions

Integrating exponential functions involves recognizing the form e^(u) and applying substitution if necessary. The integral of e^(u) with respect to u is e^(u) plus a constant, which is essential when solving differential equations involving exponentials.
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Integrals of General Exponential Functions

Implicit and Explicit Solutions

Solutions to differential equations can be implicit, where y is defined implicitly by an equation involving both x and y, or explicit, where y is expressed directly as a function of x. Understanding both forms helps interpret and verify solutions.
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Finding The Implicit Derivative
Related Practice
Textbook Question

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.

126. eʸ = y^(ln x)

Textbook Question

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.

25. y = ln(ln(x))

1
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Textbook Question

Use l’Hôpital’s rule to find the limits in Exercises 7–52.


7. lim (x → 2) (x - 2) / (x² - 4)

Textbook Question

In Exercises 59–86, find the derivative of y with respect to the given independent variable.

59. y = 2^x

Textbook Question

133. Find the absolute maximum value of

f(x) = x^2 * ln(1/x)

and say where it is assumed.

Textbook Question

Solve the differential equation in Exercises 9–22.

15. √x (dy/dx) = e^(y+√x), x > 0