Question

The solution of the differential equation $x+y\frac{dy}{dx}=2y,$ is:
(where $c$ is integration constant)

• A. $y^2=c^2(x^2+2y)$
• B. $\ln |y-x|=c+\frac{x}{y-x}$
• C. $x{y}^2=c^2(x+2y)$
• D. $\ln \left|\frac{x}{x-y}\right|=c+y-x$

Answer 1

The correct option is B $\ln |y-x|=c+\frac{x}{y-x}$

$x+y\frac{dy}{dx}=2y\Rightarrow \frac{dy}{dx}=\frac{2y-x}{y}\cdots \left(i\right)$
This is a homogeneous differential equation.

Put $y=vx$ and so,
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Then, equation $\left(i\right)$ becomes: $v+x\frac{dv}{dx}=\frac{2vx-x}{vx}\Rightarrow x\frac{dv}{dx}=\frac{2v-1}{v}-v$
$\Rightarrow \int \frac{vdv}{v^2-2v+1}=-\int \frac{dx}{x}\Rightarrow \int \frac{vdv}{{(v-1)}^2}=-\int \frac{dx}{x}$
$\Rightarrow \int [\frac{1}{(v-1)}+\frac{1}{{(v-1)}^2}]dv=-\int \frac{dx}{x}\Rightarrow \ln |v-1|-\frac{1}{(v-1)}=-\ln |x|+c$
$\Rightarrow \ln \left|(v-1)x\right|=\frac{1}{v-1}+c$
$\Rightarrow \ln |y-x|=c+\frac{x}{y-x}$