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1.1
Nonhomogeneous Equations
Generalities
Now we consider the nonhomogeneous equation
y 00 + p(t)y 0 + q(t)y = f (t)
(1.1)
together with the corresponding homogeneous equation
y 00 + p(t)y 0 + q(t)y = 0.
(1.2)
Theorem 1.1. If z1 and z2 satisfy the nonhomogeneous equation (1.1), then z1 − z2 satisfies
the associated homogeneous equation (1.2).
Suppose y1 and y2 are a basis of solutions to (1.2) and z is any solution of (1.1). Then
every solution φ of (1.1) can be written as
φ(t) = z(t) + c1 y1 (t) + c2 y2 (t).
Proof. If z1 and z2 satisfy the nonhomogeneous equation (1.1), then the linearity of differentiation gives
(z1 − z2 )00 + p(t)(z1 − z2 )0 + q(t)(z1 − z2 )
= (z100 + p(t)z10 + q(t)z1 ) − (z200 + p(t)z20 + q(t)z2 ) = f (t) − f (t) = 0,
so z1 − z2 satisfies the associated homogeneous equation (1.2).
Suppose z is any solution to (1.1). If y1 and y2 are solutions to (1.2) , then putting
φ(t) = z(t) + c1 y1 (t) + c2 y2 (t)
into the equation shows that (1.1) is satisfied. If φ and z are any solutions to (1.1), then
since φ(t) − z(t) is a solution of (1.2) and y1 and y2 are a basis of solutions to (1.2) ,
φ(t) − z(t) = c1 y1 (t) + c2 y2 (t)
for some constants c1 , c2 , and so
φ(t) = z(t) + c1 y1 (t) + c2 y2 (t).
1.2
The method of undetermined coefficients
This method is sometimes called the annihilator method.
In some cases an elementary approach lets us solve (1.1). Consider the example
y 00 + y = 1.
Differentiate both sides to get
y 000 + y 0 = 0.
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Third order equations have 3 independent solutions. N -th order equations have N independent solutions. When the coefficients are constant, guessing exponential solutions
works.(When there are repeated roots, an extended method is required.)
Look for solutions of the form er t to get
r3 + 3r = 0,
giving r = 0, ±i. The solution of the original equation must have the form
y = c1 eit + c2 e−it + c3 .
Plug this into the original nonhomogeneous equation to get c3 = 1. A particular solution is
yp (t) = 1,
and the general solution is
y(t) = 1 + c1 eit + c2 e−it .
As the examples grow more complicated, it helps to modify the notation. Use D to denote
differentiation. Consider the example
y 00 − 3y 0 + 2y = 1 + 2t.
Write it as
D2 y − 3Dy + 2y = 1 + 2t.
Apply D2 to get
D2 (D2 y − 3Dy + 2y) = 0.
The characteristic polynomial, obtained by looking for solutions ert is
r2 (r2 − 3r + 2) = r2 (r − 2)(r − 1) = 0,
which just replaces D by r. The solution y must have the form
y(t) = c1 e2t + c2 et + c3 + c4 t.
Putting this into the nonhomogeneous equation yields
−3c4 + 2(c3 + c4 t) = 1 + 2t,
or
c4 = 1,
c3 = 2.
Thus a particular solution yp is
yp (t) = 2 + t,
and the general solution is
y(t) = c1 e2t + c2 et + 2 + t.
Another example is
y 00 − 3y 0 + 2y = e−t + 1,
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or
(D − 2)(D − 1)y = e−t + 1.
Applying D knocks out the 1, and D + 1 knocks out e−t , so
D(D + 1)(D − 2)(D − 1)y = 0,
and
y = c1 + c2 e−t + c3 et + c4 e2t .
The general solution is
1 1 −t
+ e + c3 et + c4 e2t .
2 6
y(t) =
A physically important example is
y 00 + 4y = sin(2t).
This is forcing at the natural frequency.
(D2 + 4)y = sin(2t),
(D2 + 4)2 y = (D + 2i)2 (D − 2i)2 y = 0.
Then
y(t) = c1 e2it + c2 te2it + c3 e−2it + c4 te−2it ,
or
y(t) = c1 cos(2t) + c2 t cos(2t) + c3 sin(2t) + c4 t sin(2t).
After some algebra the general solution is
y(t) = c1 cos(2t) + c2 sin(2t) −
t
cos(2t).
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For the method to work we want f (t) to be a linear combination of terms of the form
t e . If roots are repeated with multiplicity m, then terms up to tm−1 ert may be needed.
k rt
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