TS EAMCET · Maths · Differential Equations
The differential equation corresponding to the family of ellipses \(\frac{x^2}{a^2}+\frac{y^2}{4}=1\), where ' \(a\) ' is an arbitrary constant is
- A \(x y \frac{d y}{d x}=4-y^2\)
- B \(x y \frac{d y}{d x}=4-x^2\)
- C \(x y \frac{d y}{d x}=x^2-4\)
- D \(x y \frac{d y}{d x}=y^2-4\)
Answer & Solution
Correct Answer
(D) \(x y \frac{d y}{d x}=y^2-4\)
Step-by-step Solution
Detailed explanation
\( \frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \) \( \frac{1}{a^2} = -\frac{y}{4x} \frac{dy}{dx} \) \( x^2 \left( -\frac{y}{4x} \frac{dy}{dx} \right) + \frac{y^2}{4} = 1 \) \( -xy \frac{dy}{dx} + y^2 = 4 \) \( xy \frac{dy}{dx} = y^2 - 4 \)
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