Question

The solution of the differential equation $\left(y -x \frac{dy}{dx}\right) = a \left(y^{2} + \frac{dy}{dx}\right) $ is

• A. $y = k (1 - ay) (x + a)$
• B. $y = k (1 + ay) (x - a)$
• C. $y = k (1 + ay) (x + a)$
• D. $y = k (1 - ay) (x - a)$

Answer 1

Answer: (A)

Given, $ \left(y-x \frac{dy}{dx}\right)=a\left(y^{2}+\frac{dy}{dx}\right)$
$\Rightarrow y-x \frac{dy}{dx}=ay^{2}+a \frac{dy}{dx}$
$\Rightarrow \left(x + a\right)\frac{dy}{dx}=y-ay^{2}$
$\Rightarrow \left(x + a\right)\frac{dy}{dx}=y\left(1-ay\right)$
$\Rightarrow \frac{dy}{y\left(1-ay\right)}=\left(\frac{1}{x + a}\right)dx$
On integrating
$\int \frac{dy}{y\left(1-ay\right)}=\int\frac{dx}{x + a}$
$\Rightarrow \int \left(\frac{1}{y}+\frac{a}{1-ay}\right) dy=\int\frac{dx}{x + a}$
$\Rightarrow \log\left|y\right|-\log\left|1-ay\right|=\log\left| x + a\right|+\log\,k$
$\Rightarrow \log \left|\frac{y}{1-ay}\right|=\log\left| x + a\right|k$
$\Rightarrow \frac{y}{1-ay}=k\left(x + a\right)$
$\Rightarrow y=k\left(1-ay\right)\left(x + a\right)$