JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y = y(x)\) be the solution of the differential equation, \(x\frac{{dy}}{{dx}} + y = x\,{\log _e}\,x,\,\left( {x > 1} \right)\) If \(2y(2) = log_e\, 4 -1\), then \(y(e)\) is equal to
- A \(-\frac{e}{2}\)
- B \( - \frac{{{e^2}}}{2}\)
- C \(\frac{e}{4}\)
- D \( \frac{{{e^2}}}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{e}{4}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\frac{1}{x} y=\log x\) \(\mathrm{IF}_{.}=\mathrm{e}^{\int \frac{\mathrm{dx}}{\mathrm{x}}}=\mathrm{x}\) \(y x=\int x \ln x d x\) \(x y=\ln x \frac{x^{2}}{2}-\int \frac{x}{2} d x\) \(x y=\ln \frac{x^{2}}{2}-\frac{x^{2}}{4}+C\) Putting \(x=2\)…
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