If y = Peax + Qebx, show that d²y/dx² − (a + b)dy/dx + aby = 0

y = Peax + Qebx — (1)
dy/dx = aPeax + bQebx — (2)
d²y/dx² = a²Peax + b²Qebx — (3)
Multiplying (1) by ab, we getaby = abPeax + abQebx — (4)
Multiplying (2) by (a + b), we get(a + b)dy/dx = (a + b)(aPeax + bQebx) = (a²Peax + b²Qebx) + (abPeax + abQebx)
Or, (a²Peax + b²Qebx) − (a + b)dy/dx + (abPeax + abQebx) = 0
Or, d²y/dx² − (a + b)dy/dx + aby = 0