Question

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

• A. ${\tan }^{-1}(x+y)=x+c$
• B. ${\cot }^{-1}(x+y)=c$
• C. ${\tan }^{-1}(x+y)=0$
• D. ${\cot }^{-1}(x+y)=x+c$

Answer 1

Answer: (A)

$x+y=z\Rightarrow \frac{dz}{dx}=1+z^2$
$\int \frac{1}{1+z^2}dz=\int 1dx$
${\mathrm{Tan}}^{-1}(x+y)=x+c$