Question

Let $y=f\left(x\right)$ satisfy the differential equation $x^2\frac{dy}{dx}=y^2{e}^{1/x}(x\ne 0)$ and $\underset{x\to 0^{-}}{lim}f\left(x\right)=1.$ Then which of the following is correct?

• A. Range of $f\left(x\right)$ is $(0,1)-\left\{\frac{1}{2}\right\}$.
• B. $f\left(x\right)$ is bounded.
• C. $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
• D. $\underset{0}{\overset{e}{\int }}f\left(x\right)dx>\underset{0}{\overset{1}{\int }}f\left(x\right)dx$

Answer 1

Answer: (A), (B), (D)

$\underset{0}{\overset{e}{\int }}f\left(x\right)dx>\underset{0}{\overset{1}{\int }}f\left(x\right)dx$

$x^2\frac{dy}{dx}=y^2{e}^{1/x}$
$\Rightarrow \frac{dy}{y^2}=\frac{e^{1/x}}{x^2}dx$
Integrating both sides, we get
$-\frac{1}{y}=\int \frac{e^{1/x}}{x^2}dx+C$
Let $\frac{1}{x}=t$
$\Rightarrow -\frac{1}{x^2}dx=dt$
$\therefore -\frac{1}{y}=-\int e^{t}dt+C$
$\Rightarrow \frac{1}{y}=e^{1/x}-C$
$\Rightarrow y=\frac{1}{e^{1/x}-C}$
Since $\underset{x\to 0^{-}}{lim}f\left(x\right)=1,$
$\therefore \underset{x\to 0^{-}}{lim}\frac{1}{e^{1/x}-C}=1$
$\Rightarrow \underset{h\to 0}{lim}\frac{1}{e^{-1/h}-C}=1$
$\Rightarrow \frac{1}{0-C}=1\Rightarrow C=-1$
$\therefore y=\frac{1}{e^{1/x}+1}$

$\frac{dy}{dx}=-\frac{e^{1/x}}{{(1+e^{1/x})}^2}\cdot (-\frac{1}{x^2})$
$\Rightarrow \frac{dy}{dx}=\frac{e^{1/x}}{x^2{(1+e^{1/x})}^2}$
$\therefore \frac{dy}{dx}>0\forall x\in R-\left\{0\right\}$
$\underset{x\to \pm \infty }{lim}\frac{1}{e^{1/x}+1}=\frac{1}{2}$
$\underset{x\to 0^{-}}{lim}f\left(x\right)=1$
and $\underset{x\to 0^{+}}{lim}\frac{1}{e^{1/x}+1}=\underset{h\to 0}{lim}\frac{1}{e^{1/h}+1}=0$
$\therefore$ Range of $f\left(x\right)$ is $(0,1)-\left\{\frac{1}{2}\right\}$

As $f\left(x\right)>0,$
so, $\underset{0}{\overset{e}{\int }}f\left(x\right)dx>\underset{0}{\overset{1}{\int }}f\left(x\right)dx$