Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
The solution of $\frac{dy}{x^2 + y^2} = \left(\frac{1}{x^2 + y^2} - 1\right) dx$ is:
$y = x\cot(c - x)$
$\cos^{-1}\left(\frac{y}{x}\right) = -x + c$
$y = x\tan(c - x)$
$\frac{y^2}{x^2} = \tan(c - x)$
Step-by-Step Solution
Key Concept: The differential form $\frac{xdy - ydx}{x^2+y^2}$ is the exact derivative of $\tan^{-1}(y/x)$.
Rewrite the equation as $\frac{xdy - ydx}{x^2 + y^2} = -dx$ and recognize the left side as $d\left(\tan^{-1}\frac{y}{x}\right)$. Integrating both sides gives $\tan^{-1}\frac{y}{x} = -x + c$, which simplifies to $\frac{y}{x} = \tan(c-x)$.
Correct Answer: 3