Question

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

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

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

$\Rightarrow y\frac{dy}{dx}-\frac{y^2}{2(x+1)}=-\frac{x}{2(x+1)}$ ......... $\left(ii\right)$
Put $y^2=t$
By differentiating both side w.r.t $x$ we get
$2y\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore$ Eq. (ii) reduces to
$\frac{1}{2}\frac{dt}{dx}-\frac{t}{2(x+1)}$ $=-\frac{x}{2(x+1)}$
$\Rightarrow \frac{dt}{dx}-\frac{1}{(x+1)}t=\frac{-x}{(x+1)}$
This is linear differential equation with
$P=\frac{1}{(x+1)};Q=\frac{-x}{(x+1)}$
$\therefore IF=e^{\int Pdx}=e^{\int -\frac{1}{(x+1)}dx}$
$=e^{-\log (x+1)}=\frac{1}{(x+1)}$
$\therefore$ Required solution will be
$t.IF=\int Q\left(IF\right)dx+\log c$ ........ (Here $\log c$ is constant)
$y^2.\frac{1}{(x+1)}=\int (\frac{-x}{(x+1)}\times \frac{1}{(x+1)})dx+\log c$
$\Rightarrow \frac{y^2}{(x+1)}=-[\int \frac{1}{(x+1)}dx-\int \frac{1}{{(1+x)}^2}dx]+\log c$
$\Rightarrow \frac{y^2}{(x+1)}=\log \frac{c}{1+x}-\frac{1}{1+x}$
$\Rightarrow y^2=(1+x)\log \frac{c}{(1+x)}-1$