DIFFERENTIAL EQUATIONS OF SECOND ORDER 851 



Solving the quadratic equation for a, 

 a* + 2aa + a 2 -- ft 2 + a 2 



ka + a = Va 2 - b 2 

 and a x = - a + Va 2 - b 2 



a a = - a - Va- - b z 

 The form of the solution of the differential equation depends 

 entirely upon the nature of the values of 04 and a 2 that is, upon 

 the relation between the quantities a and b. 



(1) If a - 0, 



Then aj = bi, and a. z = bi 



and y = Ae te +Be- ito 



= C cos bx + D sin bx 



(2) If a =6. 



Then a t and a a are each equal to a, and y= e~ ax (C+ Dx) 

 is the complete solution. For let a and a + h be the values 

 of aj and a a respectively, ft being small, 



Then y = Ae' * + Be ( -+ ft)z 



= e-** (A + 



and if ft is taken to be very small, 



i/ = e-^{(A+B) + Bfta;} 



= e- ax (C+ Da?) 



This result can be proved by direct differentiation. 

 For y = e-+* (C + Da;) 



- = De-^ - ae-^ (C + Da?) 





Then + 2a + a 2 y = - 



P^^* - ^-"(C + Da:)} 

 + a*-(C + Da;) 



1 + a 2 e- <M: (C + Da;) + 



- 2a 2 e-** (C + D) + a a <?- (C + Da;) 



(8) When a< 6, Va 2 - b 2 becomes imaginary, and <Xj and a 2 

 become complex quantities. 



and 



x = a 4- iV6 2 a 2 , or a + dt 

 a. 2 = a t V6 2 a 2 , or a di 



