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