Question

If $\left(x+y\right)^{2} \,\frac{dy}{dx}=a^{2}$, $y=0$ when $x = 0$, then $y = a$ if $\frac{x}{a}=$

• A. $1$
• B. $tan\, 1$
• C. $tan\, 1 + 1$
• D. $tan\, 1 - 1$

Answer 1

Answer: (D)

Substitute $x+y=z$
$ \Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$
So, the given equation becomes
$\frac{dz}{dx}-1=\frac{a^{2}}{z^{2}}$
$\Rightarrow \frac{dz}{dx}=\frac{a^{2}+z^{2}}{z^{2}}$
$\Rightarrow \frac{z^{2}}{a^{2}+z^{2}} dz=dx$
$\Rightarrow x+c=z-a\,tan^{-1}\left(\frac{z}{a}\right)$
$\Rightarrow a\,tan^{-1}\left(\frac{x+y}{a}\right)=y-c$
$x=0$,
$y=0$
$\Rightarrow c=0$
$\Rightarrow \frac{y}{a}=tan^{-1}\left(\frac{x+y}{a}\right)$
$\therefore y=a$
$\Rightarrow 1=tan^{-1}\left(\frac{x}{a}+1\right)$
$\Rightarrow \frac{x}{a}=tan\,1-1$.