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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 9, Problem 9.PE.14

In Exercises 1–22, solve the differential equation.
e⁻ˣ dy + (e⁻ˣ y - 4x) dx = 0

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Rewrite the given differential equation in the form \(M(x,y)\,dx + N(x,y)\,dy = 0\). Here, identify \(M(x,y)\) and \(N(x,y)\) from the equation \(e^{-x} dy + (e^{-x} y - 4x) dx = 0\). Rearranging, we get \((e^{-x} y - 4x) dx + e^{-x} dy = 0\), so \(M(x,y) = e^{-x} y - 4x\) and \(N(x,y) = e^{-x}\).
Check if the differential equation is exact by verifying if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Compute \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) separately.
If the equation is not exact, find an integrating factor. Since \(M\) and \(N\) involve \(x\) and \(y\), try to find an integrating factor that depends on \(x\) or \(y\) alone by using the formulae for integrating factors for non-exact equations.
Once the integrating factor is found, multiply the entire differential equation by it to make the equation exact. Then, solve the exact differential equation by finding a potential function \(\Psi(x,y)\) such that \(\frac{\partial \Psi}{\partial x} = M\) and \(\frac{\partial \Psi}{\partial y} = N\).
Integrate \(M\) with respect to \(x\) and \(N\) with respect to \(y\) to find \(\Psi(x,y)\), including the function of integration as needed. Finally, set \(\Psi(x,y) = C\) (a constant) to express the implicit 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.

Differential Equations

A differential equation relates a function with its derivatives. Solving it means finding the function that satisfies this relationship. In this problem, the equation involves dy and dx terms, indicating a first-order differential equation.
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Exact Differential Equations

An exact differential equation can be written as the total differential of some function equal to zero. To check exactness, verify if the partial derivatives of the involved functions satisfy a specific equality. If exact, the solution is found by integrating these functions.
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Integrating Factor

When a differential equation is not exact, an integrating factor is a function used to multiply the equation to make it exact. It often depends on x or y alone and simplifies solving the equation by enabling the use of exact equation methods.
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