Question

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

• A. $y=\left(\log y-1\right)x$
• B. $y=\left(\log y+1\right)x$
• C. $y=\left(\log x-1\right)y$
• D. $y=\left(\log x+1\right)y$

Answer 1

The correct option is A

$y=\left(\log y-1\right)x$

Explanation for the correct option :

Finding equation of the curve :

Given, $\left(xy-x^2\right)\frac{dy}{dx}=y^2$

$$\begin{gathered} \Rightarrow y^2\frac{dx}{dy}=\left(xy-y^2\right) \\ \Rightarrow \frac{1}{x^2}\frac{dx}{dy}-\frac{1}{x}\times \frac{1}{y}=-\frac{1}{y^2} \end{gathered}$$

Putting $\frac{1}{x}=u$ so that $\frac{-1}{x^2}\times \frac{dx}{dy}=\frac{du}{dy}$

$\frac{du}{dy}+\frac{u}{y}=\frac{1}{y^2}$

Integrating factor $=e^{\int \frac{1}{y}dy}=e^{\log y}=y$

$$\begin{gathered} \frac{y}{x}=\log y+C \\ \Rightarrow y=x\left(\log y+C\right)\dots \left(1\right) \end{gathered}$$

$$\begin{gathered} 1=-1\left(1+\log C\right) \\ \Rightarrow C=1 \end{gathered}$$

Put the value of $C$ in equation $\left(1\right)$.

Thus, the equation of curve is $y=\left(\log y-1\right)x$.

Hence, the correct answer is Option(A).