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EXERCISE 1.5 · Q193

Q.y=x+tan⁡xy=x+\tan x, show that cos⁡2x⋅d2ydx2−2y+2x=0\cos^2x\cdot\dfrac{d^2y}{dx^2}-2y+2x=0

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Given y=x+tan⁡xy=x+\tan x.

Step 1: y1=1+sec⁡2xy_1=1+\sec^2x.

Step 2: y2=2sec⁡x⋅sec⁡xtan⁡x=2sec⁡2xtan⁡xy_2=2\sec x\cdot\sec x\tan x=2\sec^2x\tan x.

Step 3 — multiply by cos⁡2x\cos^2x: cos⁡2x⋅y2=cos⁡2x⋅2sec⁡2xtan⁡x=2tan⁡x\cos^2x\cdot y_2=\cos^2x\cdot2\sec^2x\tan x=2\tan x (since cos⁡2xsec⁡2x=1\cos^2x\sec^2x=1). …

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