Question
Download Solution PDFWhat are the value of α and β that make \(dF(x, y)=\left(\frac{1}{x^2+2}+\frac{\alpha}{y}\right)dx\space + (xy^\beta+1)dy\) an exact differential equation?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Consider an exact differential equation:
M dx + N dy = 0
The condition of an exact differential equation:
\({\partial M\over \partial y}={\partial N\over \partial x}\)
Calculation:
Given, \(dF(x, y)=\left(\frac{1}{x^2+2}+\frac{\alpha}{y}\right)dx\space + (xy^\beta+1)dy\)
\(M=\frac{1}{x^2+2}+\frac{\alpha}{y}\) and \(N=(xy^\beta+1)\)
\(-\alpha y^{-2}=y^\beta\)
α = -1, β = -2
Last updated on Jun 23, 2025
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