Question

Find the integral curve of the differential equation, $x(1-x\ln y).\frac{dy}{dx}+y=0$ which passes through $(1,\frac{1}{e})$, then $(2+\ln y)x$ is:

• A. $1$
• B. $2$
• C. $0$
• D. None of these

Answer 1

The correct option is A $1$

Given differential equation is $x(1-x\ln y).\frac{dy}{dx}+y=0$

$\Rightarrow \frac{dy}{dx}(x-x^2\ln y)+y=0\Rightarrow x-x^2\ln y+\frac{dx}{dy}y=0\Rightarrow y\frac{dx}{dy}+x=x^2\ln y$

Putting $xy=t\Rightarrow y\frac{dx}{dy}+x=\frac{dt}{dy}$

We get $\frac{dt}{dy}=\frac{t^2}{y^2}\ln y\Rightarrow \int \frac{dt}{t^2}=\int \frac{\ln y}{y^2}dy$ ...(1)

Let $I=\int \frac{\ln y}{y^2}dy$

Putting $u=\ln y\Rightarrow du=\frac{dy}{y}$, we get

$I=\int \frac{u}{e^u}du=\int u{e}^{-u}du=\frac{u{e}^{-u}}{-1}\int e^{-u}du=-u{e}^{-u}-e^{-u}$

Therefore from (1)

$-\frac{1}{t}=-e^{-\ln y}(\ln y+1)=-\frac{1}{y}(\ln y+1)$

$\Rightarrow -\frac{1}{t}=-\frac{1}{y}(\ln y+1)+c\Rightarrow \frac{1}{xy}=-\frac{1}{y}(\ln y+1)+c$

$\Rightarrow \frac{1}{x}=(\ln y+1)+c$ ...(2)

As this curve passes through $(1,\frac{1}{e})$

$1=(\ln e^{-1}+1+c)\Rightarrow c=1$

Therefore (2) becomes

$\frac{1}{x}=(\ln y+1)+1\Rightarrow (2+\ln y)x=1$