Solve the following differential equation : $x\: \log\: x\: \frac{dy}{dx}+y=\large\frac{2}{x}$$\log\:x 1 Answer Dividing both sides by x \log x, we can rewrite the given equation as \large\frac{dy}{dx}$$ + \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$