Question

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

• A. $(x+ a)(1+a y)=-4 a^{2} y$
• B. $(x+ a)(1-a y)=4 a^{2} y$
• C. $(x+ a)(1-a y)=-4 a^{2} y$
• D. None of these

Answer 1

The Correct Option is C

We have $y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)$
$\Rightarrow y d x-x d y=a y^{2} d x+ a d y$
$\Rightarrow y(1-a y) d x=(x +a) d y$
$\Rightarrow \frac{d x}{x+a}-\frac{d y}{y(1-a y)}=0$
Integrating, we get
$\log (x +a)-\log y+\log (1-a y)=\log C$
or $\log \frac{(a+ x)(1-a y)}{y}=\log C$
i.e.$(x +a)(1-a y)=C y$
Since the curve passes through $\left(a,-\frac{1}{a}\right)$
$\therefore 2 a \times(1+1)=-\frac{C}{a}$
i.e $C=-4 a^{2}$
So, $(x+ a)(1-a y)=-4 a^{2} y$