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, $\left(xy-x^2\right)\frac{dy}{dx}=y^2$
$\Rightarrow y^2\frac{dx}{dy}=xy-x^2\Rightarrow \frac{1}{x^2}\frac{dx}{dy}-\frac{1}{x}.\frac{1}{y}=\frac{-1}{y^2}$
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.
$I.F.=e^{\int \frac{1}{y}dy}=e^{logy}=y$
Hence the solution is $uy=\int \frac{1}{y^2}.ydy+C$
or$\frac{y}{x}=logy+C$ or $y=x\left(logy+C\right)$
This passes through the point $\left(-1,1\right)$,
$\therefore 1=-1\left(log1+C\right)i.eC=-1$
Thus, the equation of the curve is $y=x\left(logy-1\right).$