Question

If the curve $y=y\left(x\right)$ satisfies the differential equation $y-x\frac{dy}{dx}=a(y^2+\frac{dy}{dx})$ and always passes through a fixed point $(1,1),$ then the total number of possible values of $a$ is
(Assume the constant of integration to be zero)

Answer 1

$y-x\frac{dy}{dx}=a(y^2+\frac{dy}{dx})\Rightarrow y-a{y}^2=(x+a)\frac{dy}{dx}\Rightarrow \int \frac{dy}{y(1-ay)}=\int \frac{dx}{x+a}\Rightarrow \int \frac{1}{y}dy+\frac{a}{1-ay}dy=\int \frac{dx}{x+a}\Rightarrow \ln |y|-\ln |1-ay|=\ln |x+a|$

Putting $(1,1)$,
$0-\ln |1-a|=\ln |1+a|\Rightarrow \frac{1}{|1-a|}=|1+a|\Rightarrow |(1+a)(1-a)|=1\Rightarrow 1-a^2=\pm 1\Rightarrow a^2=0,2\Rightarrow a=0,\pm \sqrt{2}$

Hence, the total number of values of $a$ is $3$