JEE Mains · Maths · STD 12 - 9. differential equations
The differential equation of the family of circles passing the origin and having center at the line \(y=x\) is :
- A \( \left(x^2-y^2+2 x y\right) d x=\left(x^2-y^2+2 x y\right) d y \)
- B \( \left(x^2+y^2+2 x y\right) d x=\left(x^2+y^2-2 x y\right) d y \)
- C \( \left(x^2-y^2+2 x y\right) d x=\left(x^2-y^2-2 x y\right) d y \)
- D \( \left(x^2+y^2-2 x y\right) d x=\left(x^2+y^2+2 x y\right) d y\)
Answer & Solution
Correct Answer
(C) \( \left(x^2-y^2+2 x y\right) d x=\left(x^2-y^2-2 x y\right) d y \)
Step-by-step Solution
Detailed explanation
\( C \equiv x^2+y^2+g x+g y=0 \) .................(\(1\)) \( 2 x+2 y y^{\prime}+g+g y^{\prime}=0 \) \( g=-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)\) Put in (\(1\)) \( x^2+y^2-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)(x+y)=0 \)…
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