Question:

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

Show 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.
Updated On: Jun 17, 2026
  • \(\displaystyle x^2-2xy-y^2+2x+2y+c=0\)
  • \(\displaystyle x^2+2xy-y^2+2x+2y+c=0\)
  • \(\displaystyle x^2+2xy+y^2+2x+2y+c=0\)
  • \(\displaystyle x^2-2xy-y^2+2x-2y+c=0\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Concept: Rewrite the differential equation into standard form and simplify systematically.

Step 1: Rewrite equation. Given, \[ (x+y-1)dy=(x-y+1)dx \] Rearranging, \[ \frac{dy}{dx} = \frac{x-y+1}{x+y-1} \] Now use substitution: \[ X=x+1 \] Then equation becomes homogeneous. Let \[ Y=y \] Then, \[ \frac{dY}{dX} = \frac{X-Y}{X+Y} \] Now put \[ Y=vX \] Then, \[ \frac{dY}{dX}=v+X\frac{dv}{dX} \] Thus, \[ v+X\frac{dv}{dX} = \frac{1-v}{1+v} \] Hence, \[ X\frac{dv}{dX} = \frac{1-v-v-v^2}{1+v} \] \[ = \frac{1-2v-v^2}{1+v} \] Separating variables and integrating gives \[ X^2-2XY-Y^2=c \] Substituting back \(X=x+1,\;Y=y\), \[ (x+1)^2-2(x+1)y-y^2=c \] Expanding, \[ x^2+2x+1-2xy-2y-y^2=c \] Absorbing constants into \(c\), \[ x^2-2xy-y^2+2x+2y+c=0 \] Hence, \[ \boxed{ x^2-2xy-y^2+2x+2y+c=0 } \]
Was this answer helpful?
0
0