Question

The general solution of the equation
$\frac{dy}{dx}=\frac{y^2-x}{2y(x+1)}$ is :

• A. $y^2=(1+x)\log \frac{c}{1+x}-1$
• B. $y^2=(1+x)\log (1+x)-c$
• C. $y^2=(1+x)\log \frac{c}{1-x}-1$
• D. $y^2=(1-x)\log \frac{c}{1+x}-1$

Answer 1

The correct option is A $y^2=(1+x)\log \frac{c}{1+x}-1$

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

Putting $y^2=t$
$\Rightarrow 2y\frac{dy}{dx}=\frac{dt}{dx}$

$\frac{dt}{dx}-\frac{t}{x+1}=\frac{-x}{1+x}$

$\text{I.F.}=\exp \left(\int \frac{-1}{x+1}dx\right)=\frac{1}{x+1}\Rightarrow \frac{t}{x+1}=\int (\frac{1}{x+1}\times \frac{-x}{1+x})dx\Rightarrow \frac{t}{x+1}=\int \frac{1}{x+1}\times (\frac{1}{1+x}-1)dx\Rightarrow \frac{t}{x+1}=\int (\frac{1}{(1+x{)}^2}-\frac{1}{x+1})dx\Rightarrow \frac{t}{x+1}=\frac{-1}{1+x}-\log (1+x)+\log c$, where $c$ is a constant.
$\Rightarrow y^2=(1+x)\log \frac{c}{1+x}-1$