Question

The equation to the curve which is such that portion of the axis of $x$ cut off between the origin and the tangent at any point is proportional to the ordinate of that point is:

• A. $x=y(C-K\log y)$
• B. $\log x=K{y}^2+C$
• C. $x^2=y(C-K\log y)$
• D. none of these

Answer 1

Answer: (A)

$x=y(C-K\log y)$

Let the curve be $y=f\left(x\right)$. The equation of tangent at any point $(x,y)$ is given by $Y-y=f{}^{\prime }\left(x\right)(X-x)$.
So the portion of the axis of $x$ which is cut off between the origin and the tangent at any point is obtained by putting $Y=0$.
Therefore, $x-\frac{y}{f{}^{\prime }\left(x\right)}=Ky\Rightarrow x-y\frac{dx}{dy}=Ky\Rightarrow \frac{dx}{dy}-\frac{x}{y}=-K$
which is linear equation in $x$, so its integrating factor is $e^{-\int \left(1/y\right)dy}=y^{-1}$.
Therefore, multiplying by $y^{-1}$. we have
$\frac{d}{dy}\left(x{y}^{-1}\right)=K{y}^{-1}\Rightarrow x{y}^{-1}=-K\log y+C$
$x=y(C-K\log y)$
where $C$ is arbitrary constant