AP EAMCET · Maths · Differential Equations
If \(a\) and \(b\) are arbitrary constants, then the differential equation having \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) as its general solution is
- A \(\left(\frac{d^2 y}{d x^2}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3\)
- B \(\left(x^2-y^2\right) \frac{d^2 y}{d x^2}-2 x y \frac{d y}{d x}-y=0\)
- C \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\)
- D \(x^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}-2 y=0\)
Answer & Solution
Correct Answer
(C) \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\)
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) On differentiating, \[ \frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x}=0 \quad \Rightarrow \quad \frac{y}{x} \frac{d y}{d x}=\frac{-b^2}{a^2} \] Again differentiating w.r.t. \(x\)…
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