Question

General solution of $(x+y{)}^2\frac{dy}{dx}=a^2,a\ne 0$ is ($c$ is arbitrary constant)

• A. $\frac{x}{a}=\tan \frac{y}{a}+c$
• B. $\tan xy=c$
• C. $\tan (x+y)=c$
• D. $\tan \frac{y+c}{a}=\frac{x+y}{a}$

Answer 1

The correct option is D $\tan \frac{y+c}{a}=\frac{x+y}{a}$

$\frac{dy}{dx}=\frac{a^2}{(x+y{)}^2}$
Put $x+y=t\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$
$\Rightarrow \frac{dt}{dx}=1+\frac{a^2}{t^2}\Rightarrow \frac{t^2}{t^2+a^2}dt=dx$
$\Rightarrow dt-\frac{a^{2}dt}{t^2+a^2}=dx$
Integrating both sides
$\Rightarrow t-a{\tan }^{-1}\frac{t}{a}+c=x$
$\Rightarrow y+c=a{\tan }^{-1}\frac{x+y}{a}$
$\Rightarrow \tan \left(\frac{y+c}{a}\right)=\frac{x+y}{a}$