Consider the dierential equation (E) y 00 (t) + p(t)y 0 (t) + q(t)y(t) = g(t).
The homogeneous equation associated to (E) is :
(H) y 00 (t) + p(t)y 0 (t) + q(t)y(t) = 0.
The general solution of (E) is given by the sum of the general solution of (H) and a particular solution of
(E).
Method of variation of parameters
If y1 and y2 are two solutions of the homogeneous equation (H) that form a fundamental set of solutions
(i.e. W (y1 , y2 , t) 6= 0), then the function z(t) = λ(t)y1 (t) + µ(t)y2 (t) is a particular solution of (E) if λ and µ
satisfy :
λ0 (t)y1 (t) + µ0 (t)y2 (t)
λ0 (t)y10 (t) + µ0 (t)y20 (t)
=
=
0
g(t).
The functions λ and µ are given explicitly by the formulas :
Z
λ(t) = −
Z
y2 (t)g(t)
dt,
W (y1 , y2 , t)
µ(t) =
y1 (t)g(t)
dt
W (y1 , y2 , t)
Example : Find the general solution of (E) y 00 − 2y 0 + y = t2 et .
Step 1 : Find the general solution of (H) y 00 − 2y 0 + y = 0.
The characteristic equation associated to (H) is r2 − 2r + 1 = 0 which has one double-root r = 1, therefore
the general solution of (H) is given by yH (t) = Aet + Btet for some constants A, B .
Note that we know that the functions y1 (t) = et and y2 (t) = tet form a fundamental set of solutions of (H).
Step 2 : Find a particular solution of (E) y 00 − 2y 0 + y = t2 et .
Using the method of variation of parameters we can nd a particular solution of (E) of the form z(t) =
λ(t)et + µ(t)tet where λ and µ satisfy :
λ0 (t)et + µ0 (t)tet
0
λ (t)et + µ0 (t)(t + 1)et
= 0
= t2 et
(1)
(2).
By subtracting (1) and (2) for instance one obtains µ0 (t)tet − µ0 (t)(t + 1)et = −t2 et , which after simplifying
3
gives the auxiliary dierential equation µ0 (t) = t2 , which we can solve. One gets µ(t) = t3 (up to a constant
0
2
that does not matter here). Using equation (1) for instance, and since µ (t) = t one gets after simplifying the
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auxiliary equation λ0 (t) = −t3 , which we can solve. Indeed, λ(t) = − t4 (up to a constant that does not matter
here). And hence a particular solution of (E) is :
z(t) = −
4t4
3t4
t4 t
t4 t t3 t
e + te = et (
−
)=
e.
4
3
12
12
12
Step 3 : The general solution of (E) is :
t4
y(t) = Aet + Btet + et .
| {z } 12
| {z }
yH (t)
z(t)
Alternative Step 2 : Since the Wronskian W (y1 , y2 , t) = et (et + tet ) − et tet = e2t , using the method of
variation of parameters we can nd a particular solution of (E) of the form z(t) = λ(t)et + µ(t)tet where λ and
µ satisfy :
Z
λ(t) = −
tet t2 et
dt = −
e2t
Z
t3 dt = −
t4
,
4
Z
µ(t) =
et t2 et
dt =
e2t
And hence a particular solution of (E) is z(t) = − t4 et + t3 tet = et ( 4t12 − 3t12 ) =
4
3
4
4
t4 t
12 e
.
Z
t2 dt =
t3
.
3