Question

The solution of the differential equation $\frac{dy}{dx}+\frac{y}{x{\log }_{e}x}=\frac{1}{x}$ under the condition $y=1$ when $x=e$ is

• 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. $y={\log }_{e}x+e$

Answer 1

The correct option is A $2y={\log }_{e}x+\frac{1}{{\log }_{e}x}$

$\frac{dy}{dx}+\frac{y}{x\log e^x}=\frac{1}{x}\frac{dy}{dx}=\frac{1}{x\log e^x}\left(y\right)=\frac{1}{x}I.F=e^{\int \frac{1}{x\log e^x}dx}=e^{{\log }_e\left({\log }_{e}x\right)}={\log }_{e}xG.S=y\cdot I.F=\int Q(I.F)dxy{\log }_{e}x=\int \frac{1}{x}{\log }_{e}xdyy{\log }_{e}x=\frac{({\log }_{e}x{)}^2}{2}+Cy=1,x=e\Rightarrow 1\left(1\right)=\frac{(1{)}^2}{2}+C\Rightarrow C=\frac{1}{2}$

Solution: $y{\log }_{e}x=\frac{({\log }_{e}x{)}^2}{2}+\frac{1}{2}\Rightarrow 2y={\log }_{e}x+\frac{1}{{\log }_{e}x}$