Question

The general solution of the differential equation x2 + y2 – 2xy $\frac {dy}{dx}$ = 0 is (where C is a constant of integration.)

• A. 2(x2 – y2) + x = C
• B. x2 + y2 = Cy
• C. x2 – y2 = Cx
• D. x2 + y2 = Cx

Answer 1

Answer: (C)

x2 – y2 = Cx

Answer 2

Given x2 + y2 – 2xy $\frac {dy}{dx}$ = 0
$\frac {dy}{dx}$ = $\frac {x^2 +y^2}{2xy}$
$\frac {dy}{dx}$ = $\frac {x}{2y} + \frac {y}{2x}$
Let $\frac {y}{x}$ = v
$\frac {dy}{dx}$= v +x $\frac {dv}{dx}$
v + x$\frac {dv}{dx}$ = $\frac {v}{2}$ + $\frac 12$v
x$\frac {dv}{dx}$= -v + $\frac {v}{2}$+ $\frac 12$ v
x$\frac {dv}{dx}$ = -$\frac {v}{2}$+ $\frac 12$v
∫ $\frac {2v}{1-v^2}$ dv = ∫ $\frac {1}x{}$ dx
ln $\frac {1}{1-v^2}$ = ln Cx
Where C is a arbitrary constant.
Now put v = $\frac {y}{x}$
ln $\frac {1}{1-(\frac {y}{x})^2}$= ln Cx
$\frac {x^2}{x^2 - y^2}$ = Cx
x2 - y2 = Cx is also a constant.
Therefore, the correct answer is (C) x2 – y2 = Cx