Question

The equation of the curve satisfying the equation $\left(xy-x^{2}\right) \frac{dy}{dx} = y^{2}$ and passing through the point (-1, 1) is

• A. y = (log y -1)x
• B. y = (log y +1)x
• C. x = (log x-1) y
• D. x = (log x+1) y

Answer 1

Answer: (A)

We have, $\left(xy-x^{2}\right) \frac{dy}{dx} = y^{2}$
$\Rightarrow \quad y^{2}\frac{dx}{dy} = xy -x^{2} \Rightarrow \frac{1}{x^{2}} \frac{dx}{dy} -\frac{1}{x}. \frac{1}{y} = \frac{-1}{y^{2}}$
Putting $\frac{1}{x} = u$ so that $\frac{-1}{x^{2}} \frac{dx}{dy} = \frac{du}{dy}$
We obtain $\frac{du}{dy}+\frac{u}{y} = \frac{1}{y^{2}}$. Which is linear.
$I.F. = e^{\int\frac{1}{y}dy} = e^{log\,y} = y$
Hence the solution is $uy = \int \frac{1}{y^{2}}.ydy+C$
or$\quad \frac{y}{x} = log \,y + C \quad$ or $\quad y = x\left(log \,y +C\right)$
This passes through the point $\left(-1,\, 1\right)$,
$\therefore \quad1 = -1\, \left(log \,1 + C\right) \,i.e \,C = -1$
Thus, the equation of the curve is $y = x\left(log \,y - 1\right).$