Question

Solution of the differential equation $(x-y{)}^2\frac{dy}{dx}=a^2$ is

• A. $y=\frac{a}{2}\log \left|\frac{x-y-a}{x-y+a}\right|+c$
• B. $x=\frac{a}{2}\log \left|\frac{x-y+a}{x-y-a}\right|+c$
• C. $y^2=a\log \left|\frac{x-y+a}{x-y-a}\right|+c$
• D. none of these

Answer 1

The correct option is A $y=\frac{a}{2}\log \left|\frac{x-y-a}{x-y+a}\right|+c$

Given, Differential equation as $(x-y)\frac{dy}{dx}=a^2---->A$
Let $x-y=t$
On differentiating both sides
$\frac{d}{dx}(x-y{)}^2=\frac{dt}{dx}$
$(1-\frac{dy}{dx})=\frac{dt}{dx}$
$\frac{dy}{dx}=1-\frac{dt}{dx}---->B$
Substituting B in A
$\Rightarrow t^2\ast (1-\frac{dt}{dx})=a^2$
$\Rightarrow t^2\ast -t^2\frac{dt}{dx})=a^2$
$\Rightarrow \frac{dt}{dx}=\frac{(t^2-a^2)}{t^2}$
$\Rightarrow \frac{t^2}{(t^2-a^2)}dt=dx$
On integrating both sides
$\Rightarrow \int \frac{t^2}{(t^2-a^2)}dt=\int dx$
$\Rightarrow \int (1+\frac{a^2}{(t^2-a^2)}dt=\int dx$
$\Rightarrow (t+a^2\ast \frac{1}{2a}ln|\frac{(t-a)}{(t+a)}|)=x+C$
But $t=(x-y)$
$\Rightarrow \left(\right(x-y)+\frac{a}{2}ln|\frac{(x-y-a)}{(x-y+a)}|)=x+C$
$\Rightarrow y=\frac{a}{2}ln|\frac{(x-y-a)}{(x-y+a)}|+C$