Question:
Lety $=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation, $x \frac{d y}{d x}+y=x \log _{e} x,(x>1)$. If $2 \mathrm{y}(2)=\log _{\mathrm{e}} 4-1$, then $\mathrm{y}(\mathrm{e})$ is equal to :-
Correct Option: , 2
Solution:
$\frac{d y}{d x}=\frac{y}{x}=\ell n x$
$e^{\int \frac{1}{x} d x}=x$
$x y=\int x \ell n x+C$
$\ell n x \frac{x^{2}}{2}-\int \frac{1}{x} \cdot \frac{x^{2}}{2}$
$x y=\frac{x}{2} \ell n x-\frac{x^{2}}{4}+C$, for $2 y(2)=2 \ell n 2-1$
$\Rightarrow C=0$
$y=\frac{x}{2} \ln x-\frac{x}{4}$
$y(e)=\frac{e}{4}$
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