22 Chapter 0 Ordinary Differential Equations
Thus, we are left with a pair of simultaneous equations,
v′ 1 u 1 +v′ 2 u 2 =0, (12′)v′ 1 u′ 1 +v′ 2 u′ 2 =f(t), (15)in the unknownsv′ 1 andv′ 2. The determinant of this system is
∣∣
∣∣
∣
u 1 u 2
u′ 1 u′ 2∣∣
∣∣
∣=W(t), (16)the Wronskian ofu 1 andu 2. Since these were to be independent solutions of
Eq. (10), their Wronskian is nonzero, and we may solve forv′ 1 (t)andv′ 2 (t)and
hence forv 1 andv 2.
Example.
Use variation of parameters to solve the nonhomogeneous equation
d^2 u
dt^2+u=cos(ωt).Assume a solution in the form
up(t)=v 1 cos(t)+v 2 sin(t),because sin(t)and cos(t)are independent solutions of the corresponding ho-
mogeneous equationu′′+u=0. The assumption of Eq. (12) is
v 1 ′cos(t)+v′ 2 sin(t)= 0. (17)Then our equation reduces to the following, corresponding to Eq. (15):
−v 1 ′sin(t)+v 2 ′cos(t)=cos(ωt). (18)Now we solve Eqs. (17) and (18) simultaneously to find
v 1 ′=−sin(t)cos(ωt), v′ 2 =cos(t)cos(ωt). (19)Theseequationsaretobeintegratedtofindv 1 andv 2 ,andthenup(t).
Finally, we note thatv 1 (t)andv 2 (t)can be found from Eqs. (12) and (15)
in general:
v 1 ′=−u^2 f
W,v′ 2 =u^1 f
W. (20)
Integrating these two equations, we find that
v 1 (t)=−∫ u
2 (t)f(t)
W(t)dt,v 2 (t)=∫ u
1 (t)f(t)
W(t)dt. (21)