Question

The solution of the differential equation $x\log x\frac{dy}{dx} + y = 2\log x$ is

• A. $y = \log x + c$
• B. $y = \log x^{2} + c$
• C. $y\log x = (\log x)^{2} + c$
• D. $y = x\log x + c$

Answer 1

Answer: (C)

$y\log x = (\log x)^{2} + c$

Answer 2

$x\log x\frac{dy}{dx} + y = 2\log x \Rightarrow \frac{dy}{dx} + \frac{1}{x\log x}y = \frac{2}{x}$

The is linear differential equation in y.

$\therefore$ I.F. $= e^{\int_{}^{}{\frac{1}{x\log x}dx}} = e^{\log_{e}{\log_{e}x}} = \log x$

$\Rightarrow y(I.F.) = \int_{}^{}Q(I.F.)dx \Rightarrow y\log x = \int_{}^{}\frac{2}{x}.\log{}xdx$

$\Rightarrow$y log x = (log x)² + c