Given, y=aex+bxex+cx2ex . . . (i) On differentiating w.r.t. x, we get ‌y′=aex+b(xex+ex)+c(x2ex+2xex) ⇒y′=aex+bxex+cx2ex+bex+2cxex ⇒y′=y+bex+2cxex . . . (ii) Again differentiating w.r.t. x, we get ‌y′′=y′+bex+2c(xex+ex) ‌⇒‌‌y′′=y′+bex+2cxex+2cex ‌⇒‌‌y′′=2y′−y+2cex . . . (iii) [from Eq. (ii)] Again differentiating w.r.t. x, we get ‌y′′′=2y′′−y′+2cex ⇒‌‌y′′′=2y′′−y′+(y′′−2y′+y) . ⇒‌‌‌ [from Eq. (iii)] ‌ ⇒‌‌y′′′−3y′′+3y′−y=0