A) \[2y={{\log }_{e}}x+\frac{1}{{{\log }_{e}}x}\]
B) \[y={{\log }_{e}}x+\frac{2}{{{\log }_{e}}x}\]
C) \[y{{\log }_{e}}x={{\log }_{e}}x+1\]
D) \[y={{\log }_{e}}x+e\]
Correct Answer: A
Solution :
[a]:Given,\[\frac{dy}{dx}+\frac{y}{x{{\log }_{e}}x}=\frac{1}{x}\] Here,\[I.F.={{e}^{\int_{{}}^{{}}{\frac{dx}{x{{\log }_{e}}x}}}}={{e}^{\log (lo{{g}_{e}}x)}}={{\log }_{e}}x\] The solution of differential equation will be given by \[y{{\log }_{e}}x=\int_{{}}^{{}}{{{\log }_{e}}x\frac{1}{x}dx}\] \[\Rightarrow \]\[y{{\log }_{e}}x=\frac{{{\left( {{\log }_{e}}x \right)}^{2}}}{2}+c\] ...(i) Given, y = 1 when x = e Therefore from (1) we get \[1\times 1=\frac{1}{2}+c\Rightarrow c=\frac{1}{2}\] Hence,\[y{{\log }_{e}}x=\frac{{{(lo{{g}_{e}}x)}^{2}}}{2}+\frac{1}{2}\] \[\Rightarrow \]\[2y={{\log }_{e}}x+\frac{1}{{{\log }_{e}}x}\]
You need to login to perform this action.
You will be redirected in
3 sec
