Question

The solution of differential equation $(2x-y+1)dx+(2y-x+1)dy=0$ is
(where $c$ is integration constant)

• A. ${(y-1)}^2-(y+1)(x+1)+{(x-1)}^2=c^2$
• B. ${(y+1)}^2-(y+1)(x+1)+{(x+1)}^2=c^2$
• C. ${(y+1)}^2-(y-1)(x-1)+2{(x+1)}^2=c^2$
• D. $3{(y+1)}^2-(y+1)(x+1)+2{(x+1)}^2=c^2$

Answer 1

The correct option is B ${(y+1)}^2-(y+1)(x+1)+{(x+1)}^2=c^2$

$(2x-y+1)dx=-(2y-x+1)dy$
$\Rightarrow \frac{dy}{dx}=\frac{-(2x-y+1)}{(2y-x+1)}$
Let $y=Y+k$ and $x=X+h.$ Then
$\frac{dy}{dx}=\frac{dY}{dX}=\frac{-(2X-Y+2h-k+1)}{(2Y-X+2k-h+1)}$
Now, $h$ and $k$ can be chosen such that
$2h-k+1=0\cdots \left(i\right)$
and $2k-h+1=0\cdots \left(ii\right)$
By solving these equations, $h=-1,k=-1.$
$\Rightarrow \frac{dY}{dX}=\frac{-(2X-Y)}{(2Y-X)}$

Now, it's a homogenous equation, so put $Y=vX$
$\Rightarrow \frac{dY}{dX}=v+X\frac{dv}{dX}$
$\Rightarrow v+X\frac{dv}{dX}=\frac{-(2-v)}{(2v-1)}\Rightarrow X\frac{dv}{dX}=-(\frac{2-v}{2v-1}+v)$
$\Rightarrow X\frac{dv}{dX}=-\left(\frac{2-v+2v^2-v}{2v-1}\right)\Rightarrow \int \frac{(2v-1)}{(2v^2-2v+2)}dv=-\int \frac{dX}{X}$
$\Rightarrow \frac{1}{2}\int \frac{(2v-1)}{(v^2-v+1)}dv=-\int \frac{dX}{X}\Rightarrow \frac{1}{2}\ln |v^2-v+1|=-\ln |X|+\ln |c|$
$\Rightarrow \ln \left|\sqrt{(v^2-v+1)}X\right|=\ln |c|$
$\Rightarrow \left|X\sqrt{(v^2-v+1)}\right|=|c|\Rightarrow \sqrt{(Y^2-XY+X^2)}=|c|$
Now substituting $Y=y-k$ and $X=x-h$, we get
${(y+1)}^2-(y+1)(x+1)+{(x+1)}^2=c^2$