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MISCELLANEOUS EXERCISE 1 (II) · Q248

Q.If log⁡y=log⁡(sin⁡x)−x2\log y=\log(\sin x)-x^2, show that d2ydx2+4xdydx+(4x2+3)y=0\dfrac{d^2y}{dx^2}+4x\dfrac{dy}{dx}+(4x^2+3)y=0

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Differentiate log⁡y=log⁡(sin⁡x)−x2\log y=\log(\sin x)-x^2 w.r.t. xx:

y′y=cot⁡x−2x⟹y′=y(cot⁡x−2x).\frac{y'}{y}=\cot x-2x\quad\Longrightarrow\quad y'=y(\cot x-2x).

Differentiate again (product rule):

y′′=y′(cot⁡x−2x)+y(−csc⁡2x−2).y''=y'(\cot x-2x)+y(-\csc^2x-2).

Substitute y′=y(cot⁡x−2x)y'=y(\cot x-2x) into the first term:

y′′=y(cot⁡x−2x)2+y(−csc⁡2x−2)=y[(cot⁡x−2x)2−csc⁡2x−2].y''=y(\cot x-2x)^2+y(-\csc^2x-2)=y\left[(\cot x-2x)^2-\csc^2x-2\right].

Expand (cot⁡x−2x)2=cot⁡2x−4xcot⁡x+4x2(\cot x-2x)^2=\cot^2x-4x\cot x+4x^2, and use cot⁡2x−csc⁡2x=−1\cot^2x-\csc^2x=-1:

y′′=y[cot⁡2x−4xcot⁡x+4x2−csc⁡2x−2]=y[−1−4xcot⁡x+4x2−2]=y[4x2−3−4xcot⁡x].y''=y\left[\cot^2x-4x\cot x+4x^2-\csc^2x-2\right]=y\left[-1-4x\cot x+4x^2-2\right]=y\left[4x^2-3-4x\cot x\right]. …

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