WBJEE · Maths · Differentiation
Given \(\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}^2 x=0\). Changing the independent variable \(x\) to \(z\) by the substitution \(z=\log \tan \frac{x}{2}\), the equation is changed to
- A \(\frac{d^2 y}{d z^2}+\frac{3}{y}=0\)
- B \(2 \frac{d^2 y}{d z^2}+e^y=0\)
- C \(\frac{d^2 y}{d z^2}-4 y=0\)
- D \(\frac{d^2 y}{d z^2}+4 y=0\)
Answer & Solution
Correct Answer
(D) \(\frac{d^2 y}{d z^2}+4 y=0\)
Step-by-step Solution
Detailed explanation
Hint : \(\frac{d z}{d x}=\frac{\frac{1}{2} \sec ^2(x / 2)}{\tan (x / 2)}=\frac{1}{\sin x}=\operatorname{cosec} x\)…
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