MHT CET · Maths · Differential Equations
The differential equation of all circles which pass through the origin and whose centres lie on \(y\) -axis is
- A \(\left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y=0\)
- B \(\left(x^{2}-y^{2}\right) \frac{d y}{d x}+2 x y=0\)
- C \(\left(x^{2}-y^{2}\right) \frac{d y}{d x}-x y=0\)
- D \(\left(x^{2}-y^{2}\right) \frac{d y}{d x}+x y=0\)
Answer & Solution
Correct Answer
(A) \(\left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y=0\)
Step-by-step Solution
Detailed explanation
Equation of a circle is
\(x^{2}+(y-a)^{2}=a^{2}\)
\(2 x+2 y \frac{d y}{d x}-2 a \frac{d y}{d x}=0\)
Form Eqs. (i) and (ii)
\(\frac{d y}{d x} =\frac{2 x y}{x^{2}-y^{2}}\)
\(\Rightarrow \left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y =0\)
\(x^{2}+(y-a)^{2}=a^{2}\)
\(2 x+2 y \frac{d y}{d x}-2 a \frac{d y}{d x}=0\)
Form Eqs. (i) and (ii)
\(\frac{d y}{d x} =\frac{2 x y}{x^{2}-y^{2}}\)
\(\Rightarrow \left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y =0\)
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