Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
If the solution of the differential equation $\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$ be $f(x,y) + \sqrt{1+f(x,y)} = c\left(\sqrt{\frac{x+y}{\sqrt{f(x,y)}}}\right)$, then $f(x,y)$ is:
$x^2 + y^2$
$1 + x^2 - y^2$
$x^2 - y^2$
$\frac{x^2-y^2}{x^2+y^2}$
Step-by-Step Solution
Key Concept: Parametric substitution with polar coordinates transforms the differential equation into a separable form.
Substitute $x = r\sec heta$ and $y = r an heta$ into the given differential equation. From $x^2 - y^2 = r^2$ and $\sin heta = y/x$, differentiate to get $xdx - ydy = rdr$. Using the second relation $xdy - ydx = x^2\cos heta d heta = r^2\sec^2 heta d heta$, substitute both into the original equation to obtain $ rac{dr}{r^2\sec heta d heta} = \sec heta d heta$. Integrating both sides yields $\ln(r + \sqrt{1+r^2}) = \ln(\sec heta + an heta) + \ln c$, which simplifies to $(r + \sqrt{1+r^2}) = c(\sec heta + an heta)$.
Correct Answer: 3