MHT CET · Maths · Differential Equations
The differential equation of all parabolas having vertex at the origin and axis along positive \(\mathrm{Y}\)-axis is
- A \(x^2 \frac{d y}{d x}-y=0\)
- B \(x \frac{d y}{d x}+2 y=0\)
- C \(x \frac{d y}{d x}+y=0\)
- D \(2 x \frac{d y}{d x}-y=0\)
Answer & Solution
Correct Answer
(D) \(2 x \frac{d y}{d x}-y=0\)
Step-by-step Solution
Detailed explanation
We have \(y^2=4 a x\)
\(\therefore 2 y \frac{d y}{d x}=4 a \quad \Rightarrow a=\left(\frac{y}{2}\right) \frac{d y}{d x}\)
\(\therefore y^2=4\left(\frac{y}{2}\right)\left(\frac{d y}{d x}\right) x \Rightarrow y^2=2 x y \frac{d y}{d x}\)
\(\therefore 2 x \frac{d y}{d x}-y=0\)
\(\therefore 2 y \frac{d y}{d x}=4 a \quad \Rightarrow a=\left(\frac{y}{2}\right) \frac{d y}{d x}\)
\(\therefore y^2=4\left(\frac{y}{2}\right)\left(\frac{d y}{d x}\right) x \Rightarrow y^2=2 x y \frac{d y}{d x}\)
\(\therefore 2 x \frac{d y}{d x}-y=0\)
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