Question

Solution of the differential equation
$\{\frac{1}{x}-\frac{y^2}{(x-y{)}^2}\}dx+\{\frac{x^2}{(x-y{)}^2}-\frac{1}{y}\}dy=0$ is
(where $c$ is arbitrary constant).

• A. $\ln \left|\frac{x}{y}\right|+\frac{xy}{(x-y)}=c$
• B. $\ln |xy|+\frac{xy}{(x-y)}=c$
• C. $\frac{xy}{(x-y)}=c{e}^{x/y}$
• D. $\frac{xy}{(x-y)}=c{e}^{xy}$

Answer 1

The correct option is A $\ln \left|\frac{x}{y}\right|+\frac{xy}{(x-y)}=c$

It's not linear separable form but we can see that it can be written in homogeneous form.

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

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

Put $\frac{y}{x}=v\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=\frac{-(\frac{1}{x}-\frac{v^2}{(1-v{)}^2})}{\frac{1}{(1-v{)}^2}-\frac{1}{vx}}$

Shift $v$ to $RHS$ and perform the $LCM$ and then do subtraction

$=x\frac{dv}{dx}=\frac{\frac{v^2}{(1-v{)}^2}-\frac{v}{(1-v{)}^2}}{\frac{1}{(1-v{)}^2}-\frac{1}{vx}}$

Solving above by taking $L.C.M$ and then dividing by $x$ we get,

$\frac{dv}{dx}=\frac{v^2(v-1)}{vx-(1-v{)}^2}$

$\frac{dx}{dv}=\frac{vx-(1-v{)}^2}{v^2(v-1)}$

$\frac{dx}{dv}-\frac{x}{v(v-1)}=\frac{1-v}{v^2}$

It is $First-orderlineardifferentialequation.$

We find $u=e^{\int -\frac{1}{v(v-1)}dv}$

$u=e^{\int (\frac{1}{v}-\frac{1}{v-1})dv}$

$u=e^{ln\left(\frac{v}{v-1}\right)}=\frac{v}{v-1}$

Solution of differential equation is $ux=\int u\left(\frac{(1-v)}{v^2}\right)dx$

Put value of $u$,

$\frac{v}{v-1}x=-\int \frac{1}{v}dv$

$\frac{v}{v-1}x=-ln|v|+c$

put $v=\frac{y}{x}$,

$\frac{xy}{x-y}+ln|\frac{x}{y}|=c$

Hence,$\left(A\right)$