Question

The particular solution of the differential equation $x\frac{dy}{dx}=y\ln \left(\frac{y}{x}\right),x\ne 0$ is $y=f\left(x\right)$. If $f\left(1\right)=e^2$, then

• A. $f\left(x\right)=x{e}^{1+x^2}$
• B. $f\left(x\right)=x{e}^{1+x}$
• C. there are more than one tangent parallel to $x-$axis to the curve $y=f\left(x\right)$
• D. there is exactly one tangent parallel to $x-$axis to the curve $y=f\left(x\right)$

Answer 1

Answer: (B), (D)

The correct options are
B $f\left(x\right)=x{e}^{1+x}$
D there is exactly one tangent parallel to $x-$axis to the curve $y=f\left(x\right)$

$x\frac{dy}{dx}=y\ln \left(\frac{y}{x}\right)\cdots \left(1\right)$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}\ln \left(\frac{y}{x}\right)$
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\therefore v+\frac{xdv}{dx}=v\ln v\Rightarrow x\frac{dv}{dx}=v(\ln v-1)\Rightarrow \int \frac{dv}{v(\ln v-1)}=\int \frac{dx}{x}$
$\Rightarrow \ln |\ln \left(v\right)-1|=\ln |x|+\ln |C|\Rightarrow \ln \left(v\right)-1=Cx\Rightarrow \ln \frac{y}{x}=1+Cx$
Now, $f\left(1\right)=e^2\Rightarrow C=1$
$\therefore$ The particular solution is $y=x{e}^{1+x}\cdots \left(2\right)$

If tangent is parallel to $x-$axis, then
$\frac{dy}{dx}=0$,
$\Rightarrow e^{1+x}+x{e}^{1+x}=0\Rightarrow x=-1(\because e^{1+x}\ne 0)$

Putting $x=-1$ in equation $\left(2\right)$,
$y=-1$
So, only one tangent is parallel to $x-$axis to the curve $y=f\left(x\right)$