Question

The solution of the differential equation $x\frac{dy}{dx}=\frac{y}{1+\log x}$ is :

• A. $y=\log x+C$
• B. $y=\frac{C}{1+\log x}$
• C. $y=C(x+\log x)$
• D. $y=x+\log \left(Cx\right)$
• E. $y=C(1+\log x)$

Answer 1

Answer: (E)

$y=C(1+\log x)$

$x\frac{dy}{dx}=\frac{y}{1+\log x}\frac{dy}{y}=\frac{dx}{x(1+\log x)}$

Put $t=\log x$

$\frac{dt}{dx}=\frac{1}{x}\Rightarrow dt=\frac{dx}{x}\Rightarrow \frac{dy}{y}=\frac{dt}{1+t}\int \frac{dy}{y}=\int \frac{dt}{1+t}\log y=\log (1+t)+\log C\log y=\log C(1+t)\Rightarrow y=C(1+t)$

Substituting $t$

$\Rightarrow y=C(1+\log x)$

So option $E$ is correct.