Question

Solution of $(x+y{)}^2\frac{dy}{dx}=a^2$ ('$a$' being a constant) is

• A. $\frac{(x+y)}{a}=\tan \frac{y+c}{a},c$ is an arbitrary constant
• B. $xy=a\tan cx,c$ is an arbitrary constant
• C. $\frac{x}{a}=\tan \frac{y}{c},c$ is an arbitrary constant
• D. $xy=\tan (x+c),c$ is an arbitrary constant

Answer 1

The correct option is A $\frac{(x+y)}{a}=\tan \frac{y+c}{a},c$ is an arbitrary constant

$(x+y{)}^2\frac{dy}{dx}=a^2$
Let $x+y=z$
$\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$

$(x+y{)}^2\frac{dy}{dx}=a^2\Rightarrow z^2(\frac{dz}{dx}-1)=a^2\Rightarrow \frac{dz}{dx}=\frac{a^2+z^2}{z^2}\Rightarrow dx=\frac{z^2}{z^2+a^2}dz\Rightarrow dx=dz-\frac{a^2}{z^2+a^2}dz$

Integrating both the sides
$x=z-a{\tan }^{-1}\frac{z}{a}+c\Rightarrow \mathrm{x/}=\mathrm{x/}+y-a{\tan }^{-1}\frac{x+y}{a}+c\Rightarrow \frac{y+c}{a}={\tan }^{-1}\frac{x+y}{a}\Rightarrow \tan \frac{y+c}{a}=\frac{(x+y)}{a}$