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. $tan1$
• C. $tan1+1$
• D. $tan1-1$

Answer 1

Answer: (D)

Substitute $x+y=z$
$\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$
So, the given equation becomes
$\frac{dz}{dx}-1=\frac{a^2}{z^2}$
$\Rightarrow \frac{dz}{dx}=\frac{a^2+z^2}{z^2}$
$\Rightarrow \frac{z^2}{a^2+z^2}dz=dx$
$\Rightarrow x+c=z-ata{n}^{-1}\left(\frac{z}{a}\right)$
$\Rightarrow ata{n}^{-1}\left(\frac{x+y}{a}\right)=y-c$
$x=0$,
$y=0$
$\Rightarrow c=0$
$\Rightarrow \frac{y}{a}=ta{n}^{-1}\left(\frac{x+y}{a}\right)$
$\therefore y=a$
$\Rightarrow 1=ta{n}^{-1}\left(\frac{x}{a}+1\right)$
$\Rightarrow \frac{x}{a}=tan1-1$.