JEE Mains · Maths · STD 12 - 9. differential equations
Given that the slope of the tangent to a curve \(y = y(x)\) at any point \((x, y)\) is \(\frac{{2y}}{{{x^2}}}\). If the curve passes through the centre of the circle \(x^2 + y^2 - 2x - 2y = 0\), then its equation is
- A \(x\,{\log _e}\,\left| y \right| = x - 1\)
- B \(x\,{\log _e}\,\left| y \right| = -2(x - 1)\)
- C \(x^2\,{\log _e}\,\left| y \right| = -2(x - 1)\)
- D \(x\,{\log _e}\,\left| y \right| = 2(x - 1)\)
Answer & Solution
Correct Answer
(D) \(x\,{\log _e}\,\left| y \right| = 2(x - 1)\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{2 y}{x^{2}}\) \(\Rightarrow \ln y=-\frac{2}{x}+\ln C\) Passes through \((1,1)\) \(0=-2+\ell n C\) \(\Rightarrow \ln y=\frac{-2}{x}+2\) \(x\ell n\left| y \right| = 2\left( {x - 1} \right)\)
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