Question

The solution of the differential equation $\frac{dy}{dx}=1+x+y+xy$ is

• A. $log(1+y)=x+\frac{x^2}{2}+c$
• B. $(1+y{)}^2=x+\frac{x^2}{2}+c$
• C. $log(1+y)=log(1+x)+c$
• D. None of these

Answer 1

The correct option is A

$log(1+y)=x+\frac{x^2}{2}+c$

$\frac{dy}{dx}=1+x+y+xy\Rightarrow \frac{dy}{dx}=(1+x)(1+y)\Rightarrow \frac{dy}{1+y}=(1+x)dx$
On integrating, we get
$log(1+y)=\frac{x^2}{2}+x+c.$