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If $y=a \cos 2 x+b \sin 2 x$, then
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The correct answer is:
$\frac{d^{2} y}{d x^{2}}+4 y=0$
The equation of circle touching $\mathrm{y}$ -axis at origin is $(\mathrm{x}-\alpha)^{2}+(\mathrm{y}-0)^{2}=\alpha^{2}$
$\Rightarrow \mathrm{x}^{2}+\alpha^{2}-2 \alpha \mathrm{x}+\mathrm{y}^{2}=\alpha^{2}$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \alpha \mathrm{x}=0$
$\Rightarrow x+\frac{y^{2}}{x}-2 \alpha=0$
Differentiating, $1+\frac{x \cdot 2 y \cdot \frac{d y}{d x}-y^{2}}{x^{2}}=0$
$\Rightarrow x^{2}+2 x y \cdot \frac{d y}{d x}-y^{2}=0$
$\Rightarrow \mathrm{x}^{2}+\alpha^{2}-2 \alpha \mathrm{x}+\mathrm{y}^{2}=\alpha^{2}$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \alpha \mathrm{x}=0$
$\Rightarrow x+\frac{y^{2}}{x}-2 \alpha=0$
Differentiating, $1+\frac{x \cdot 2 y \cdot \frac{d y}{d x}-y^{2}}{x^{2}}=0$
$\Rightarrow x^{2}+2 x y \cdot \frac{d y}{d x}-y^{2}=0$
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