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JEE Mains · Maths · STD 12 - 9. differential equations

જો \(y = y(x)\) એ વિકલ સમીકરણ \(x\frac{{dy}}{{dx}} + y = x\,{\log _e}\,x,\,\left( {x > 1} \right)\)  નો ઉકેલ છે અને \(2y(2) = log_e\, 4 -1\) હોય તો  \(y(e)\) મેળવો.

  1. A \(-\frac{e}{2}\)
  2. B \( - \frac{{{e^2}}}{2}\)
  3. C \(\frac{e}{4}\)
  4. D \(  \frac{{{e^2}}}{4}\)
Verified Solution

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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