Question

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

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

Answer 1

Answer: (A)

$y+\frac{\pi }{4}={\tan }^{-1}(x+y)$

Put $x+y=t\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-1$
$\Rightarrow t^2\frac{dt}{dx}=1+t^2\Rightarrow \int dx=\int \frac{t^2}{1+t^2}dt$
$\Rightarrow x=\int (1-\frac{1}{1+t^2})dt$

$\Rightarrow x=t-{\tan }^{-1}t+c$

$\Rightarrow x=(x+y)-{\tan }^{-1}(x+y)+c$

$\Rightarrow y={\tan }^{-1}(x+y)+C$

$(C=-c)$

Now $y\left(1\right)=0\Rightarrow 0={\tan }^{-1}1+C\Rightarrow C=-\frac{\pi }{4}$
$y={\tan }^{-1}(x+y)-\frac{\pi }{4}$ or
$y+\frac{\pi }{4}={\tan }^{-1}(x+y)$

Hence, option A.