Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
The solution of $\left(\frac{dy}{dx}\right)^2 + 2y\cot x\frac{dy}{dx} = y^2$ is:
$y - \frac{c}{1+\cos x} = 0$
$y = \frac{c}{1-\cos x}$
$x = 2\sin^{-1}\left(\sqrt{\frac{c}{2y}}\right)$
$x = 2\cos^{-1}\left(\sqrt{\frac{c}{2y}}\right)$
Step-by-Step Solution
Key Concept: Separate variables and integrate the trigonometric differential equation, simplifying using half-angle and trigonometric identities.
Solving $\frac{dy}{dx} = y(-\cot x \pm \cos ecx)$ by separating variables gives $\frac{dy}{y} = (-\cot x + \cos ecx)dx$. Integrating yields $\ln y = -\ln\sin x + \log\tan\frac{x}{2} + \log c$, which simplifies to $y = \frac{c\tan\frac{x}{2}}{\sin x} = \frac{c}{1+\cos x}$. For the second form, solving the negative case gives $y = \frac{c}{1-\cos x}$ and $x = 2\sin^{-1}\sqrt{\frac{C}{2y}}$.
Correct Answer: 1,2,3,4