MathematicsEasy63×since 2002Q3689If cos−1(y2)=loge(x5)5, ∣y∣<2{\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2cos−1(2y)=loge(5x)5,∣y∣<2, then :Ax2y′′+xy′−25y=0{x^2}y'' + xy' - 25y = 0x2y′′+xy′−25y=0Bx2y′′−xy′−25y=0{x^2}y'' - xy' - 25y = 0x2y′′−xy′−25y=0Cx2y′′−xy′+25y=0{x^2}y'' - xy' + 25y = 0x2y′′−xy′+25y=0Dx2y′′+xy′+25y=0{x^2}y'' + xy' + 25y = 0x2y′′+xy′+25y=0Check answerSkip