Show that y=x+tanx satisfies the differential equation cos2x d2y/dx2-2y+2x=0
y=x+tanxy′=1+sec2xy′′=2sec2xtanxcos2x⋅y′′−2y+2x=2cos2xsec2xtanx−2x−2tanx+2x=2tanx−2x−2tanx+2x=0y = x + \tan{x}\\ y' = 1 + \sec^2{x}\\ y'' = 2\sec^2{x}\tan{x}\\ \begin{aligned} \cos^2{x}\cdot y'' - 2y + 2x &= 2\cos^2{x}\sec^2{x}\tan{x} \\&- 2x - 2\tan{x} + 2x \\&= 2\tan{x} - 2x - 2\tan{x} + 2x &= 0 \end{aligned}y=x+tanxy′=1+sec2xy′′=2sec2xtanxcos2x⋅y′′−2y+2x=2cos2xsec2xtanx−2x−2tanx+2x=2tanx−2x−2tanx+2x=0
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