Question

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

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

Answer 1

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

Given differential eqn
$\frac{dy}{dx}=\frac{y^2-x}{2y(x+1)}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{2(x+1)}-\frac{x}{2y(x+1)}$
$\Rightarrow \frac{dy}{dx}-\frac{y}{2(x+1)}=-\frac{x}{2y(x+1)}$ ....(1)
which is a Bernoulli's equation.
Multiplying eqn (1) by $y$, we get
$y\frac{dy}{dx}-\frac{y^2}{2(x+1)}=-\frac{x}{2(x+1)}$ .....(2)
Put $y^2=z$
$2y\frac{dy}{dx}=\frac{dz}{dx}$
So, eqn (2) becomes
$\frac{1}{2}\frac{dz}{dx}-\frac{z}{2(x+1)}=-\frac{x}{2(x+1)}$
$\Rightarrow \frac{dz}{dx}-\frac{z}{(x+1)}=-\frac{x}{(x+1)}$
which is a linear differential equation with z as dependent variable.
Here, $P=-\frac{1}{x+1};Q=-\frac{x}{(x+1)}$
Integrating factor $I.F.=e^{\int Pdx}$
$=e^{\int -\frac{1}{x+1}dx}$
$=e^{-\log (x+1)}$
$I.F.=\frac{1}{x+1}$
Solution of differential equation is given by
$z\frac{1}{x+1}=\int \frac{1}{x+1}\times \frac{-x}{x+1}dx$
$z\frac{1}{x+1}=-\int \frac{x+1-1}{(x+1{)}^2}dx$
$z\frac{1}{x+1}=-\log (x+1)-\frac{1}{(x+1)}+\log c$
$\Rightarrow z=(x+1)\log \frac{c}{(x+1)}-1$
$\Rightarrow y^2=(x+1)\log \frac{c}{(x+1)}-1$