Question

The solution of $y\frac{dy}{dx}=1+y^2$ is:

• A. $2x=log\left[c(1+y^2)\right]$
• B. $x=c{y}^2$
• C. $c(1+y^2)=x$
• D. $2y=log\left\{c(1+x^2)\right\}$

Answer 1

The correct option is A $2x=log\left[c(1+y^2)\right]$

$y\frac{dy}{dx}=1+y^2$
$\Rightarrow \frac{2ydy}{1+y^2}=2dx$
Let$1+y^2=t\Rightarrow 2ydy=dt$
$\Rightarrow \frac{dt}{t}=2dx$
$\Rightarrow logt=2x+c$
$\Rightarrow log(1+y^2)=2x+c$