Question

Solve:

$(x-y)dy=(x+y+1)dx$

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

Answer 1

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

Given differential eqn can eb written as
$\frac{dy}{dx}=\frac{x+y+1}{x-y}$ .... $\left(1\right)$
which is not a homogeneous differential equation.
Here, $\frac{a_1}{a_2}=1;\frac{b_1}{b_2}=-1$
$\Rightarrow \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
So, let $x=X+h,y=Y+k$
Now, $\frac{dy}{dx}=\frac{dy}{dY}\frac{dY}{dX}\frac{dX}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{dY}{dX}$ $\because \frac{dy}{dY}=\frac{dx}{dX}=1$
So eqn $\left(1\right)$ becomes
$\frac{dY}{dX}=\frac{X+h+Y+k+1}{X+h-Y-k}$
$\Rightarrow \frac{dY}{dX}=\frac{(X+Y)+(h+k+1)}{(X-Y)+(h-k)}$ ...$\left(2\right)$
Let h and k be such that
$h+k+1=0$ .....$\left(3\right)$
$h-k=0$ .....$\left(4\right)$
Solving $\left(3\right)$ and $\left(4\right)$, we get
$h=k=-\frac{1}{2}$
So, eqn $\left(2\right)$ becomes
$\Rightarrow \frac{dY}{dX}=\frac{X+Y}{X-Y}$ .....$\left(4\right)$
which is a homogeneous differential eqn.
Put $Y=VX$
$\frac{dY}{dX}=V+X\frac{dV}{dX}$
So, eqn $\left(4\right)$ becomes
$\Rightarrow V+X\frac{dV}{dX}=\frac{X(1+V)}{X(1-V)}$
$\Rightarrow X\frac{dV}{dX}=\frac{V^2+1}{1-V}$
$\Rightarrow \frac{1-V}{V^2+1}dV=\frac{dX}{X}$
Integrating both sides,
$\Rightarrow \int (\frac{1}{V^2+1}-\frac{V}{V^2+1})dV=\log X+\log k$
Put $V^2+1=t$ in the second integral
$\Rightarrow VdV=\frac{1}{2}dt$
$\Rightarrow {\tan }^{-1}V-\frac{1}{2}\log (V^2+1)=\log X+\log c$
$\Rightarrow {\tan }^{-1}V=\log \sqrt{1+V^2}+\log X+\log c$
$\Rightarrow {\tan }^{-1}V=\log \left(cX\sqrt{1+V^2}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{Y}{X}\right)=\log \left(c\sqrt{X^2+Y^2}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{y-k}{x-h}\right)=\log \left(c\sqrt{{(x-h)}^2+{(y-k)}^2}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{y+\frac{1}{2}}{x+\frac{1}{2}}\right)=\log \left(c\sqrt{{(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{2y+1}{2x+1}\right)=\log \left(c\sqrt{x^2+y^2+x+y+\frac{1}{2}}\right)$