Calculus- Early Transcendentals, 2021a

10.7. Second Order Linear Equations -
Variation of Parameters 399
Example 10.32: Variation of Parameters
The differential equation y ′′ − 5y ′ + 6y = e t sint can be solved using the method of undetermined
coefficients, though we have not seen any examples of such a solution. Again, we will solve it by
variation of parameters.
Solution. The equations to be solved are
u ′ e 2t + v ′ e 3t = 0
2u ′ e 2t + 3v ′ e 3t = e t sint.
If we multiply the first equation by 2 and subtract it from the second equation we get
v ′ e 3t = e t sint
v ′ = e −3t e t sint = e −2t sint
v = − 1 5 (2sint + cost)e−2t .
Then substituting we get
u ′ = −e −2t v ′ e 3t = −e −2t e −2t sin(t)e 3t = −e −t sint
u = 1 2 (sint + cost)e−t .
The particular solution is
ue 2t + ve 3t = 1 2 (sint + cost)e−t e 2t − 1 5 (2sint + cost)e−2t e 3t
= 1 2 (sint + cost)et − 1 (2sint + cost)et
5
= 1
10 (sint + 3cost)et ,
and the solution to the differential equation is Ae 2t + Be 3t + e t (sint + 3cost)/10.
♣
Example 10.33: Solving a DE
The differential equation y ′′ − 2y ′ + y = e t /t 2 is not of the form amenable to the method of undetermined
coefficients. Solve it using variation of parameters.
Solution. The solution to the homogeneous equation is Ae t + Bte t and so the simultaneous equations are
Subtracting the equations gives
u ′ e t + v ′ te t = 0
u ′ e t + v ′ te t + v ′ e t = et
t 2 .
v ′ e t = et
t 2