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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
9장, 문제 9.PE.19

In Exercises 1–22, solve the differential equation.


dy + x(2y - e^(x-x²))dx = 0

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1
Rewrite the given differential equation in the form \(M(x,y)\,dx + N(x,y)\,dy = 0\). Here, the equation is \(dy + x(2y - e^{x - x^{2}})\,dx = 0\), which can be rearranged as \(x(2y - e^{x - x^{2}})\,dx + dy = 0\).
Identify the functions \(M(x,y)\) and \(N(x,y)\) from the equation: \(M(x,y) = x(2y - e^{x - x^{2}})\) and \(N(x,y) = 1\).
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 \(\frac{\partial N}{\partial x}\).
If the equation is not exact, look for an integrating factor that depends on either \(x\) or \(y\) to make it exact. Determine the integrating factor by using the formula involving \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\).
Once the equation is exact (either originally or after multiplying by the integrating factor), find the 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 include a function of \(y\), then differentiate with respect to \(y\) and equate to \(N\) to find that function. Finally, write the implicit solution \(\Psi(x,y) = C\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
07:39
Classifying Differential Equations

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.
추천 영상:
07:39
Classifying Differential Equations

Integrating Factor

If a differential equation is not exact, an integrating factor is a function used to multiply the equation to make it exact. Finding the correct integrating factor often depends on the variables involved and simplifies solving the equation.
추천 영상:
06:18
Integration by Parts for Definite Integrals