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Solve the following differential equation : $ x\: \log\: x\: \frac{dy}{dx}+y=\large\frac{2}{x}$$\log\:x $

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Dividing both sides by $x \log x$, we can rewrite the given equation as $\large\frac{dy}{dx}$$ + \large\frac{y}{\log x}$ $ = \large\frac{2}{x^2}$
The integrating factor is $= e^{\int \frac{dx}{x \log x}} $$= e^{\log (\log x)}$$ = \log x$
Multiplying the equation by $\ \log x$ we get: $\log x \large\frac{dy}{dx}$$ + \large\frac{y}{x}$$ = \large\frac{2}{x^2}$$ \log x$
Integrating, we get: $y \log x = \large \int \frac{2}{x^2} $$ \log x dx + c $
$\Rightarrow y \log x = 2 \large \int $$ \log x \times x^{-2} dx + c$
Using integration by parts and solving, we get: $y \log x = \large\frac{-2}{x}$$ (1+x) + c$
answered Mar 20, 2014 by balaji.thirumalai

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