JEE Mains · Maths · STD 12 - 9. differential equations
The differential equation of the family of curves, \(x^{2}=4 b(y+b), b \in R,\) is
- A \(\mathrm{x}\left(\mathrm{y}^{\prime}\right)^{2}=\mathrm{x}+2 \mathrm{yy}^{\prime}\)
- B \(\mathrm{x}\left(\mathrm{y}^{\prime}\right)^{2}=2 \mathrm{yy}^{\prime}-\mathrm{x}\)
- C \(\mathrm{xy}^{\prime \prime}=\mathrm{y}^{\prime}\)
- D \(\mathrm{x}\left(\mathrm{y}^{\prime}\right)^{2}=\mathrm{x}-2 \mathrm{yy}^{\prime}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{x}\left(\mathrm{y}^{\prime}\right)^{2}=\mathrm{x}+2 \mathrm{yy}^{\prime}\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{x}=4 \mathrm{by}^{\prime} \Rightarrow \mathrm{y}^{\prime}=\frac{2 \mathrm{x}}{4 \mathrm{b}}\) Required D.E. is \(\mathrm{x}^{2}=\frac{2 \mathrm{x}}{\mathrm{y}^{\prime}} \mathrm{y}+\left(\frac{\mathrm{x}}{\mathrm{y}^{\prime}}\right)^{2}\)…
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