Question

The general solution of the differential equation \[ (x+y-1)\,dy=(x-y+1)\,dx \] is

• A. \(\displaystyle x^2-2xy-y^2+2x+2y+c=0\)
• B. \(\displaystyle x^2+2xy-y^2+2x+2y+c=0\)
• C. \(\displaystyle x^2+2xy+y^2+2x+2y+c=0\)
• D. \(\displaystyle x^2-2xy-y^2+2x-2y+c=0\)

Hint:

For equations of the form \[ \frac{ax+by+c}{a_1x+b_1y+c_1}, \] shift variables first to remove constants and convert into a homogeneous equation.

Answer 1

The Correct Option is A

Step 1: Rearrange into slope form.
From $(x+y-1)\,dy=(x-y+1)\,dx$, \[ \frac{dy}{dx}=\frac{x-y+1}{x+y-1}. \]
Step 2: Shift to remove the constants.
Put $X=x+1$ and $Y=y$. Then $x=X-1$ and the equation becomes \[ \frac{dY}{dX}=\frac{X-Y}{X+Y}, \] which is homogeneous.
Step 3: Substitute $Y=vX$.
With $\dfrac{dY}{dX}=v+X\dfrac{dv}{dX}$, \[ v+X\frac{dv}{dX}=\frac{1-v}{1+v}\Rightarrow X\frac{dv}{dX}=\frac{1-2v-v^2}{1+v}. \]
Step 4: Separate variables and integrate.
Separating and integrating leads to the compact relation $X^2-2XY-Y^2=c$ (the left side is the exact combination whose differential matches the equation).
Step 5: Replace $X$ and $Y$.
With $X=x+1$, $Y=y$: \[ (x+1)^2-2(x+1)y-y^2=c. \]
Step 6: Expand and absorb constants.
\[ x^2+2x+1-2xy-2y-y^2=c\Rightarrow x^2-2xy-y^2+2x+2y+c=0. \] \[ \boxed{x^2-2xy-y^2+2x+2y+c=0} \]