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Math 220: Lecture 17
§ 4.5
Professor Nicholls
Department of Mathematics, Statistics,
and Computer Science
University of Illinois at Chicago
Math 220: Lecture 17
Professor Nicholls (UIC)
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Chapter 4: Linear Second Order Equations
For second order ODE we can only solve linear equations, and we
only have a general method for constant coefficients.
We have seen in Sections 4.1–4.3 how to solve the homogeneous
problem.
We now extend the method to accommodate the inhomogeneous
problem
ay 00 (t) + by 0 (t) + cy (t) = f (t),
a, b, c ∈ R.
We propose two techniques:
The Method of Undetermined Coefficients (MUC) [§ 4.4, 4.5, today]
The Method of Variation of Parameters (VOP) [§ 4.6]
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4.5: The Superposition Principle and Undetermined
Coefficients Revisited
Recall, we are considering the linear, constant coefficient,
inhomogeneous second order ODE
ay 00 (t) + by 0 (t) + cy (t) = f (t),
a, b, c ∈ R.
The “Method of Undetermined Coefficients” is based upon a good
guess: The inhomogeneous problem has solution which “looks
like” the inhomogeneity!
This only works for f which are (combinations of):
Polynomials,
Exponentials,
Sines and Cosines.
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The Superposition Principle
Theorem 1: Let y1 be a solution to the differential equation
ay 00 + by 0 + cy = f1 (t),
and y2 be a solution to
ay 00 + by 0 + cy = f2 (t).
Then for any constants k1 and k2 the function k1 y1 + k2 y2 is a solution
to the differential equation
ay 00 + by 0 + cy = k1 f1 (t) + k2 f2 (t).
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Example # 1
Example #1: Recall from last time that a particular solution of
y 00 − 3y 0 + 2y = 2t
is y1 = t + 3/2, and a particular solution of
y 00 − 3y 0 + 2y = e−t
is y2 = (1/6)e−t . Thus, a particular solution of
y 00 − 3y 0 + 2y = 2(2t) − 3e−t
is
y = 2y1 − 3y2 = 2t + 3 − (1/2)e−t .
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Existence and Uniqueness Theorem
Theorem 2: For any real numbers a 6= 0, b, c, t0 , Y0 , Y1 , suppose yp (t)
is a particular solution to
ay 00 + by 0 + cy = f (t)
in an interval I containing t0 and that y1 (t) and y2 (t) are linearly
independent solutions to the associated homogeneous equation
ay 00 + by 0 + cy = 0
in I. Then there exists a unique solution in I to the initial value problem
ay 00 + by 0 + cy = f (t),
y (t0 ) = Y0 ,
y 0 (t0 ) = Y1 ,
and it is given by
y (t) = yp (t) + c1 y1 (t) + c2 y2 (t),
for the appropriate choice of the constants c1 , c2 .
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Method of Undetermined Coefficients
To find a particular solution to the differential equation
ay 00 + by 0 + cy = Ct m ert ,
where m is a nonnegative integer, use the form
yp (t) = t s (Am t m + . . . + A1 t + A0 ) ert ,
with
1
s = 0 if r is not a root of the associated characteristic equation;
2
s = 1 if r is a simple root of the associated characteristic equation;
3
s = 2 if r is a double root of the associated characteristic equation.
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Method of Undetermined Coefficients, cont.
To find a particular solution to the differential equation
(
Ct m eαt cos(βt)
00
0
ay + by + cy =
Ct m eαt sin(βt)
for β 6= 0, use the form
yp (t) = t s (Am t m + . . . + A1 t + A0 ) eαt cos(βt)
+ t s (Bm t m + . . . + B1 t + B0 ) eαt sin(βt),
with
1
s = 0 if (α + iβ) is not a root of the associated characteristic
equation;
2
s = 1 if (α + iβ) is a simple root of the associated characteristic
equation.
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Example # 2
Example #2: (NSS 4.5 # 9): Decide whether the method of
undetermined coefficients together with superposition can be applied
to find a particular solution of the given equation. Do not solve the
equation.
y 00 − y 0 + y = (et + t)2 .
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Work Space
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Solution: Example # 2
Solution #2: At first it doesn’t look like we can use MUC, however, if
we expand the right hand side
f (t) = (et + t)2 = e2t + 2tet + t 2 ,
we see that we can write
f (t) = f1 (t) + f2 (t) + f3 (t) = e2t + 2tet + t 2 ,
and for each of these fj (t) we can use the MUC:
yp,1 (t) = Ae2t ,
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yp,2 (t) = (Bt + C)et ,
Math 220: Lecture 17
yp,3 (t) = Dt 2 + Et + F .
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Example # 3
Example #3: (NSS 4.5 # 13): Decide whether the method of
undetermined coefficients together with superposition can be applied
to find a particular solution of the given equation. Do not solve the
equation.
2y 00 + 3y 0 − 4y = 2t + sin2 (t) + 3.
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Work Space
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Solution: Example # 3
Solution #3: At first it doesn’t look like we can use MUC, however, if
we use the trigonometric identity
sin2 (t) = (1/2)(1 − cos(2t))
we see that we can write
f (t) = f1 (t) + f2 (t) = 2t + (7/2) − (1/2) cos(2t),
and for each of these fj (t) we can use the MUC:
yp,1 (t) = At + B,
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yp,2 (t) = C cos(2t) + D sin(2t).
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Example # 4
Example #4: (NSS 4.5 # 18): Find a general solution to the differential
equation.
y 00 − 2y 0 − 3y = 3t 2 − 5.
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Work Space
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Solution: Example # 4
Solution #4: First, we find the homogeneous solution by considering
the homogeneous equation
y 00 − 2y 0 − 3y = 0.
We seek a solution of the form y = ert which gives us the
characteristic equation
r 2 − 2r − 3 = 0.
We can use the Quadratic Formula or simply factor
(r − 3)(r + 1) = 0,
and deduce the roots r1 = 3, r2 = −1. From these we find the
homogeneous solution:
yh (t) = C1 e3t + C2 e−t .
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Example # 4, cont.
Solution: Now, we need to find a particular solution to
y 00 − 2y 0 − 3y = 3t 2 − 5.
We use the Method of Undetermined Coefficients and seek a solution
that “looks like” the right hand side:
yp (t) = At 2 + Bt + C.
Computing
yp0 = 2At + B,
yp00 = 2A,
we insert this into the ODE
2A − 2(2At + B) − 3(At 2 + Bt + C) = 3t 2 − 5.
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Example # 4, cont.
Matching the powers of t we find
−3A = 3,
−4A − 3B = 0,
2A − 2B − 3C = −5,
and thus A = −1, B = 4/3, C = 1/9. Putting this all together we find
the general solution:
y = yh + yp = C1 e3t + C2 e−t − t 2 + (4/3)t + (1/9).
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Example # 5
Example #5: (NSS 4.5 # 25): Find the solution to the initial value
problem.
z 00 (x) + z(x) = 2e−x ,
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z(0) = 0,
Math 220: Lecture 17
z 0 (0) = 0.
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Work Space
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Solution: Example # 5
Solution #5: First, we find the homogeneous solution by considering
the homogeneous equation
z 00 + z = 0.
We seek a solution of the form z = erx which gives us the
characteristic equation
r 2 + 1 = 0.
We can use the Quadratic Formula
q
r = −0 ± 02 − 4(1)(1) /(2(1)) = ±i,
and deduce the roots r1 = i, r2 = −i. From these we find the
homogeneous solution:
zh (x) = C1 cos(x) + C2 sin(x).
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Example # 5, cont.
Solution: Now, we need to find a particular solution to
z 00 (x) + z(x) = 2e−x .
We use the Method of Undetermined Coefficients and seek a solution
that “looks like” the right hand side:
zp (x) = Ae−x .
Computing
zp0 = −Ae−x ,
zp00 = Ae−x ,
we insert this into the ODE
Ae−x + Ae−x = 2e−x .
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Example # 5, cont.
Matching the exponential we find A = 1. Putting this all together we
find the general solution:
z = zh + zp = C1 cos(x) + C2 sin(x) + e−x .
Finally, we need to consider the initial conditions. First
0 = z(0) = C1 + 1
so that C1 = −1, and second
0 = z 0 (0) = C2 − 1
so that C2 = 1. Thus, the unique solution is
z(x) = − cos(x) + sin(x) + e−x .
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Example # 6
Example #6: (NSS 4.5 # 31): Determine the form of a particular
solution for the differential equation. Do not solve.
y 00 + y = sin(t) + t cos(t) + 10t .
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Work Space
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Solution: Example # 6
Solution #6: First, we solve the homogeneous solution so that we
know how to modify the “guess,” if necessary. The homogeneous
equation is
y 00 + y = 0,
which has characteristic equation
r 2 + 1 = 0,
and roots r1 = i and r2 = −i. This has solutions
y1 (t) = cos(t),
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y2 (t) = sin(t).
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Example # 6, cont.
Consider the right hand side
f (t) = sin(t) + t cos(t) + 10t ,
which we rewrite as
t
f (t) = sin(t) + t cos(t) + eln(10 ) = sin(t) + t cos(t) + eln(10)t .
So, we would guess
yp (t) = (At + B) cos(t) + (Ct + D) sin(t) + Eeln(10)t ,
however, cos(t) and sin(t) are homogeneous solutions so we multiply
the trigonometric terms by one power of t:
yp (t) = t(At + B) cos(t) + t(Ct + D) sin(t) + Eeln(10)t .
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Example # 7
Example #7: (NSS 4.5 # 32): Determine the form of a particular
solution for the differential equation. Do not solve.
y 00 − y = e2t + te2t + t 2 e2t .
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Work Space
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Solution: Example # 7
Solution #7: First, we solve the homogeneous solution so that we
know how to modify the “guess,” if necessary. The homogeneous
equation is
y 00 − y = 0,
which has characteristic equation
r 2 − 1 = 0,
and roots r1 = 1 and r2 = −1. This has solutions
y1 (t) = et ,
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y2 (t) = e−t .
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Example # 7, cont.
Consider the right hand side
f (t) = e2t + te2t + t 2 e2t ,
which we rewrite as
f (t) = (1 + t + t 2 )e2t .
So, we guess
yp (t) = (At 2 + Bt + C)e2t .
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