Question

The solution of the differential equation $\frac{dy}{dx} = (x +y)^2$ is

• A. $ \frac{1}{x+y} = c$
• B. $ \sin^{-1} (x + y) =x +c$
• C. $ \tan^{-1} (x +y) = c$
• D. $ \tan^{-1} (x +y) = x +c$

Answer 1

Answer: (D)

Let $x + y = t \: \Rightarrow \: 1 + \frac{dy}{dx} = \frac{dt}{dx} $
$\frac{dt}{dx}-1=t^{2} \Rightarrow \frac{dt}{dx} =t^{2} +1 \Rightarrow \int\frac{dt}{t^{2}+1} = \int dx $
$\Rightarrow \tan^{-1}t =x+c \Rightarrow \tan^{-1}\left(x+y\right)=x+c$