Find the general solution of the differential equation $\frac{dy}{dx} = \frac{x+1}{2-y}, (y \neq 2)$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $x^2 + y^2 + 2x - 4y + C = 0$ ##
We have
$\frac{dy}{dx} = \frac{x+1}{2-y} \quad \dots(1)$
Separating the variables in equation (1), we get
$(2-y) \, dy = (x+1) \, dx \quad \dots(2)$
Integrating both sides of equation (2), we get
$\int (2-y) \, dy = \int (x+1) \, dx$
$\text{or } 2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1$
$\text{or } x^2 + y^2 + 2x - 4y + 2C_1 = 0$
$\text{or } x^2 + y^2 + 2x - 4y + C = 0, \text{ where } C = 2C_1$
which is the general solution of equation (1).