Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
The solution of $\frac{xdx + ydy}{xdy - ydx} = \frac{a^2 - x^2 - y^2}{x^2 + y^2}$ is:
$\sqrt{x^2 + y^2} = a\left\{\sin\left(\tan^{-1}\frac{y}{x} + C\right)\right\}$
$\sqrt{x^2 + y^2} = a\cos\left(\tan^{-1}\frac{y}{x} + C\right)$
$\sqrt{x^2 + y^2} = a\{\tan(\sin^{-1}y/x) + C\}$
None of these
Step-by-Step Solution
Key Concept: Polar coordinates naturally simplify equations involving $\sqrt{x^2+y^2}$ and $\tan^{-1}(y/x)$ terms.
Use polar substitution $x = r\cos\theta$, $y = r\sin\theta$ to transform the equation into $\frac{rdr}{r^2d\theta}\sqrt{a^2-r^2}/r^2 = \frac{dr}{d\theta}\sqrt{a^2-r^2}$. This simplifies to $\frac{dr}{d\theta} = \sqrt{a^2-r^2}$, which integrates to $C + \sin^{-1}r/a = \tan^{-1}(y/x)$ or $\sqrt{x^2+y^2} = a\sin(\tan^{-1}(y/x) + \text{const})$.
Correct Answer: 1