Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
Solution of the equation $\frac{xdx + ydy}{xdy - ydx} = \sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ is:
$\sqrt{x^2-y^2} = a\tan\left\{\tan^{-1}\left(\frac{y}{x}\right)+c\right\}$
$\sqrt{x^2+y^2} = a\sin\left\{\tan^{-1}\left(\frac{y}{x}\right)+c\right\}$
$\sqrt{x^2+y^2} = a\tan\left\{\tan^{-1}\left(\frac{y}{x}\right)+c\right\}$
$\sqrt{x^2-y^2} = a\cos\left\{\tan^{-1}\left(\frac{y}{x}\right)+c\right\}$
Step-by-Step Solution
Key Concept: Convert to polar coordinates using x = r cos θ and y = r sin θ to transform the differential equation into separable form. Recognize that xdx + ydy = rdr and xdy - ydx = r²dθ, which simplifies the equation to dr/dθ = r√(a² - r²)/r² = √(a² - r²)/r.
Substitute $x = r\cos heta$ and $y = r\sin heta$ where $x^2 + y^2 = r^2$ and $ heta = an^{-1}(y/x)$. Differentiate the first relation to get $xdx + ydy = rdr$, and differentiate the second to get $xdy - ydx = x^2\sec^2 heta d heta$. From the original differential equation, extract the coefficients and substitute these parametric relations. This yields a separable equation in $r$ and $ heta$. Integrate to obtain the solution in the form $r^2 = c\sec^2 heta$ or equivalently $x^2 + y^2 = ce^{2 an^{-1}(y/x)}$.
Correct Answer: 2