Question

The solution of the differential equation $(x+y{)}^2\frac{dy}{dx}=1$, satisfying the condition $y\left(1\right)=0$ is:

• A. $y+\frac{\pi }{4}=ta{n}^{-1}(x+y)$
• B. $y-\frac{\pi }{4}=ta{n}^{-1}(x+y)$
• C. $y={\tan }^{-1}x$
• D. $y={\tan }^{-1}\left(lnx\right)+1$

Answer 1

The correct option is A $y+\frac{\pi }{4}=ta{n}^{-1}(x+y)$

$(x+y{)}^2\frac{dy}{dx}=1$
$x+y=t⟹\frac{dy}{dx}=\frac{dt}{dx}-1$
$⟹t^2\frac{dt}{dx}=1+t^2$
$⟹\int \frac{t^2}{t^2+1}dt=\int dx$
$⟹x=1-\tan (-1)t+c$
$⟹x=1-\tan (-1)(x+y)+c$
$y=0$ when $x=1$
$c=\frac{\pi }{4}$
$x=1-\frac{\pi }{4}+\tan -1(x+y)$