WBJEE · Maths · Differential Equations
The solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}\) under the condition \(y=1\) when \(x=e\) is
- A \(2 y=\log _{e} x+\frac{1}{\log _{e} x}\)
- B \(y=\log _{e} x+\frac{2}{\log _{e} x}\)
- C \(y \log _{e} x=\log _{2} x+1\)
- D \(y=\log _{e} x+e\)
Answer & Solution
Correct Answer
(A) \(2 y=\log _{e} x+\frac{1}{\log _{e} x}\)
Step-by-step Solution
Detailed explanation
Given, differential equation is \(\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}\) It is a linear equation of the form \(\frac{d y}{d x}+P y=Q\) where \(\quad P=\frac{1}{x \log _{e} x}\) and \(Q=\frac{1}{x}\) \(\therefore\) Integrating factor, \(\mathrm{IF}=e\)…
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