Mathematics | VITEEE 2017 | Discussion Page

The solution of the differential equation $\log_{}{}x\frac{dy}{dx}+\frac{y}{x} = \sin2x is$

$y\log_{}{}|x|= C - 1/2 \cos x$

$y\log_{}{}|x|= C + 1/2 \cos 2x$

$y\log_{}{}|x|= C - 1/2 \cos 2x$

$xy\log_{}{}|x|= C - 1/2 \cos 2x$

Option C

$\frac{dy}{dx}+\frac{y}{x\log_{}{}x} =\frac{\sin2x}{\log_{}{}x}$

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

.'. I.F. = $e^{\int_{}^{}\frac{1}{t}dt}$ = $e^{\log_{}{}t}$ = t = log |x|

solution is given by

y(I.F.) = $\int_{}^{}Q(I.F.)dx + C$

y log |x| = $\int_{}^{}\frac{sin 2x}{log |x|}(log|x|)dx +C = -\frac{Cos 2x}{2}+C$

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