A) \[y=\log x+c\]
B) \[y=\log {{x}^{2}}+c\]
C) \[y\log x={{(\log x)}^{2}}+c\]
D) \[y=x\log x+c\]
Correct Answer: C
Solution :
\[x\log x\frac{dy}{dx}+y=2\log x\] Þ \[\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x}\] This is linear differential equation in y. \[\therefore \] I.F. \[={{e}^{\int_{x}^{{}}{\frac{1}{\log x}dx}}}={{e}^{{{\log }_{e}}{{\log }_{e}}x}}=\log x\] Þ \[y.\](I.F.)\[=\int_{{}}^{{}}{Q\ .\ (I.F.)\,dx}\] Þ \[y\log x=\int_{{}}^{{}}{\frac{2}{x}}.\log xdx\] Þ \[y\log x={{(\log x)}^{2}}+c\].
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