Q. If $y = f(x)$ is the equation of the curve an its differential equation is given by $\frac{dy}{dx} = \frac{x + 2}{y + 3},$ then the equation of the curve, if it passes through (2, 2), is
A) $x^{2} - y^{2} + 4x - 6y + 4 = 0$
B) $x^{2} - y^{2} + 4x + 6y = 0$
C) $x^{2} - y^{2} - 4x - 6y = 0$
D) $x^{2} - y^{2} - 4x - 6y - 4 = 0$
Explanation
(a)
The given equation is
$(y + 3)dy = (x + 2)dx$
$\Rightarrow \frac{y^{2}}{2} + 3y = \frac{x^{2}}{2} + 2x + c$
Since, it passes through (2, 2).
$\therefore 2 + 6 = 2 + 4 + c\ \Rightarrow c = 2$
$\therefore\frac{y^{2}}{2} + 3y = \frac{x^{2}}{2} + 2x + 2$
$\Rightarrow y^{2} + 6y = x^{2} + 4x + 4$
$\Rightarrow x^{2} + 4x - y^{2} - 6y + 4 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.