Answer to Question #154228 in Differential Equations for Jishnu

Question

Show that y=x+tanx satisfies the differential equation cos2x d2y/dx²-2y+2x=0

Answer 1

$y = x + \tan{x}\\ y' = 1 + \sec^2{x}\\ y'' = 2\sec^2{x}\tan{x}\\ \begin{aligned} \cos^2{x}\cdot y'' - 2y + 2x &= 2\cos^2{x}\sec^2{x}\tan{x} \\&- 2x - 2\tan{x} + 2x \\&= 2\tan{x} - 2x - 2\tan{x} + 2x &= 0 \end{aligned}$