How to find particular solutions to nonhomogeneous equations
Example: Find the general solution of the equation
(∗)
y 00 − 2y 0 + 2y = tet cos t.
Step 1: The characteristic equation of the corresponding homogeneous equation is
λ2 − 2λ + 2 = 0
with complex roots λ1 = 1 + i and λ2 = 1 − i. The real fundamental solutions of the corresponding
homogeneous problem are therefore given by
y1 (t) = et cos t
and
y2 (t) = et sin t.
Step 2: Following the table 3.6.1 (p. 181) of the book one could set
Z(t) = t(A0 t + A1 )et cos t + t(B0 + B1 )et sin t
and plug Z into (∗) and solve for A0 , A1 , B0 , B1 in order to obtain a particular solution of (∗).
However, there is a more efficient method, which we describe now. We set
Z(t) = t(At + B)e(1+i)t
and plug Z into the ‘complexified’ equation
(∗∗)
z 00 − 2z 0 + 2z = te(1+i)t
and solve for A and B. Note that the real part of the right hand side of (∗∗) is the right hand side
of (∗). Once we have solved for A and B, Re(Z) will be a particular solution of (∗). This approach
has two advantages: First of all the derivatives of Z become less cumbersome and, secondly, there
appear only two undetermined coefficients A, B. We compute
Z 0 (t) = (2At + B)e(1+i)t + (1 + i)(At2 + Bt)e(1+i)t
and
Z 00 (t) = 2Ae(1+i)t + (2 + 2i)(2At + B)e(1+i)t + 2i(At2 + Bt)e(1+i)t .
and plug into (∗∗) to obtain the equation
[2A + (2 + 2i)(2At + B) − 2(2At + B)]e(1+i)t = te(1+i)t ,
which yields A = − 4i and B =
1
4
and consequently
i 2 1
Z(t) = − t + t e(1+i)t .
4
4
The real part of Z,
Y (t) =
is then a particular solution of (∗).
1 t
1
te cos t + t2 et sin t,
4
4