JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(x \log _e x \frac{d y}{d x}+y=x^2 \log _e x,(x > 1)\). If \(y (2)=2\), then \(y ( e )\) is equal to
- A \(\frac{4+ e ^2}{4}\)
- B \(\frac{1+ e ^2}{4}\)
- C \(\frac{2+ e ^2}{2}\)
- D \(\frac{1+ e ^2}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{4+ e ^2}{4}\)
Step-by-step Solution
Detailed explanation
\(x \log _e x \frac{d y}{d x}+y=x^2 \log _e x,(x > 1)\) \(\Rightarrow \frac{d y}{d x}+\frac{y}{x \ln x}=x\) \(\text { Linear differential equation }\) \(\text { I.F. }=e^{\int \frac{1}{x \ln x} d x}=|\ln x|\) \(y|\ln x|=\int x|\ln x| d x\)…
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