If `y=A sin^2x+Bcos^2x` then prove that `sin2x.y_(2)-2cos2x.y_(1)=0`
where `y_(1)=(dy)/(dx),y_(2)=(d^2y)/(dx^(2))`.

Explanation

To prove that \( \sin 2x \cdot y_{2} - 2 \cos 2x \cdot y_{1} = 0 \) where \( y = A \sin^2 x + B \cos^2 x \), we will follow these steps:

### Step 1: Differentiate \( y \) to find \( y_{1} \)
Given:
\[ y = A \sin^2 x + B \cos^2 x \]
We differentiate \( y \) with respect to \( x \):
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