Question

The curve $y\left(x\right)$ satisfying the differential equation $y-x\frac{dy}{dx}=a(y^2+\frac{dy}{dx})$ passes through $(1,1)$, then the possible of $a$ is
(assuming the constant of integration to be zero)

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

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

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