Question

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

• A. $1-y^2=\left|cx\right|$
• B. $1+y^2=\left|cx\right|$
• C. $1-x^2=\left|cy\right|$
• D. $1+x^2=\left|cy\right|$

Answer 1

Answer: (B)

$1+y^2=\left|cx\right|$

$\Rightarrow 2xy\frac{dy}{dx}-(1+y^2)=0$

$\Rightarrow \frac{2yx}{x^2}\frac{dy}{dx}-\frac{(1+y^2)}{x^2}=0$

$\Rightarrow \frac{2y}{x}\frac{dy}{dx}+(1+y^2)\frac{d}{dx}\left(1/x\right)=0$

This is in the form of $u\frac{dv}{dx}+v\frac{du}{dx}=0$

$\Rightarrow \frac{d}{dx}\frac{1+y^2}{x}=0$

$\Rightarrow \frac{(1+y^2)}{\left|x\right|}=\left|C\right|$

$\Rightarrow 1+y^2=\left|Cx\right|$ [$\because$ | | is because as $a+y^2$ is +ve $C_x$ should always be ${+}^{ve}$]