ODE Homework 5
4.2. Homogeneous Equations with Constant Coefficients
1. Find the general solution of the differential equation
y (6) + y = 0
[§4.2 #16]
6
Sol. The
characteristic
equation
is r + 1 = 0. Thus we get
r = cos π+2kπ
+ i sin π+2kπ
, k = 0, 1, 2, 3, 4, 5, that is, r =
6
6
√
± 3±i
,
2
±i. Therefore the general solution is of the form
√ 3
t
t −√3 t
t
y(t) = c1 cos t+c2 sin t+e 2 t c3 cos +c4 sin +e 2 t c5 cos +c6 sin
2
2
2
2
2. Find the general solution of the differential equation
y (6) − 3y (4) + 3y 00 − y = 0
[§4.2 #17]
Sol. The characteristic equation is
r6 − 3r4 + 3r2 − 1 = (r + 1)3 (r − 1)3 = 0
Thus we get r = ±1 with multiplicity 3 respectively. Therefore
the general solution is of the form
y(t) = c1 + c2 t + c3 t2 et + c4 + c5 t + c6 t2 e−t
3. Find the general solution of the differential equation
y (5) − 8y 00 = 0
[§4.2 #20]
Sol. The characteristic equation is
r5 − 8r2 = r2 (r − 2)(r2 + 2r + 4) = 0
√
Thus we get r = 2, −1 ± 3 i and 0 with multiplicity 2. Therefore the general solution is of the form
√
√ y(t) = c1 + c2 t + c3 e2t + e−t c4 cos 3 t + c5 sin 3 t
4. Find the general solution of the differential equation
y 000 − 5y 00 + 3y 0 + y = 0
[§4.2 #23]
Sol. The characteristic equation is
r3 − 5r2 + 3r + 1 = (r − 1)(r2 − 4r − 1) = 0
√
Thus we get r = 1, 2 ± 5. Therefore the general solution is
of the form
y(t) = c1 et + c2 e(2−
√
5)t
√
+ c3 e(2+
5)t
5. Find the solution of the initial value problem, and plot its graph.
y (4) + y = 0;
y(0) = 0, y 0 (0) = 0, y 00 (0) = −1, y 000 (0) = 0
How does the solution behave as t → ∞?
[§4.2 #30]
Sol. The characteristic equation is
r4 + 1 = 0
π+2kπ
+
i
sin
, k = 0, 1, 2, 3, that
Thus we √
get r √= cos π+2kπ
4
4
2
2
is, r = ± 2 ± 2 i. Therefore the general solution is of the form
√
√ √
√ √ √ 2
2
2
2
2
2
t
t
−
y(t) = e 2 c1 cos
t+c2 sin
t +e 2 c3 cos
t+c4 sin
t
2
2
2
2
Since y(0) = 1, y 0 (0) = 0, y 00 (0) = −1, y 000 (0) = 0, we have the
following linear system,
c1 + c3 = 0,
c1 + c2 − c3 + c4 = 0
c2 − c4 = −1, c1 − c2 − c3 − c4 = 0
and we have that c1 = 0, c2 = − 12 , c3 = 0, c4 = 21 . Hence the
solution of the initial value problem is
√
√
2
2
1 − √2 t
1 √2 t
y(t) = e 2 sin
t − e 2 sin
t
2
2
2
2
The graph of y(t) is as follows
Moreover, y(t) oscillates about 0 with increasing amplitude. 6. Find the solution of the initial value problem, and plot its graph.
y 000 − y 00 + y 0 − y = 0;
y(0) = 2, y 0 (0) = −1, y 00 (0) = −2
How does the solution behave as t → ∞?
[§4.2 #33]
Sol. The characteristic equation is
r3 − r2 + r − 1 = 0
Thus we get r = 1, ±i. Therefore the general solution is of the
form
y(t) = c1 et + c2 cos t + c3 sin t
Since y(0) = 2, y 0 (0) = −1, y 00 (0) = −2, we have the following
linear system,
c1 + c2 = 2, c1 + c3 = −1, c1 − c2 = −2
and we have that c1 = 0, c2 = 2, c3 = −1. Hence the solution
of the initial value problem is
√
y(t) = 2 cos t − sin t = 5 cos(t + θ)
where θ = cos−1
√2 .
5
The graph of y(t) is as follows
Moreover, it is clear that y(t) oscillate about 0, with an amplitude of 2.
7. Consider the equation y (4) − y = 0.
(a) Use Abel’s formula to find the Wronskian of the fundamental set of solutions of the given equation.
(b) Determine the Wronskian of the solutions et , e−t , cos t,
and sin t.
(c) Determine the Wronskian of the solutions cosh t, sinh t, cos t
and sin t.
[§4.2 #38]
Sol.
(a) According to the Abel’s formula and the equation, the
Wronskian of the fundamental set of solutions is
Z
W (t) = c exp − 0dt = c
for some constant c.
(b)
t −t
W (e , e , cos t, sin t)(t) = et
e−t
cos t
sin t
t
e −e−t − sin t
cos t
et
e−t − cos t − sin t
t
e −e−t
sin t − cos t
= −8
(c)
W (cosh t, sinh t, cos t, sin t)(t) = cosh t
sinh t
cosh t
sinh t
sinh t
cos t
sin t
cosh t − sin t
cos t
sinh t − cos t − sin t
cosh t
sin t − cos t
4.3. The Method of Undetermined Coefficients
8. Determine the general solution of the differential equation
y 000 + y 00 + y 0 + y = e−t + 4t
[§4.3 #3]
Sol. The characteristic equation for the homogeneous problem
is r3 + r2 + r + 1 = (r + 1)(r2 + 1) = 0, with roots r = −1, ±i.
Hence the general solution for the homogeneous problem is
yc (t) = c1 e−t + c2 cos t + c3 sin t
=4
Let L[y] := y 000 +y 00 +y 0 +y. Assume that the particular solution
Y (t) of the problem L[y] = e−t + 4t is of the form Y (t) =
Ate−t + Bt + C. Substituting Y into the equation, we get
2Ae−t + Bt + B + C = e−t + 4t
Then we obtain the system of equations
2A = 1, B = 4, B + C = 0
which implies A = 21 , B = 4, C = −4. Thus Y (t) = 12 te−t +
4t − 4. Therefore, the general solution for the equation is
y(t) = yc (t) + Y (t)
1
= c1 e−t + c2 cos t + c3 sin t + te−t + 4t − 4
2
9. Determine the general solution of the differential equation
y (4) + 2y 00 + y = 3 + cos 2t
[§4.3 #6]
Sol. The characteristic equation for the homogeneous problem
is r4 +2r2 +1 = (r2 +1)2 = 0, with roots r = ±i with multiplicity
2 respectively. Hence the general solution for the homogeneous
problem is
yc (t) = (c1 + c2 t) cos t + (c3 + c4 t) sin t
Let L[y] := y (4) + 2y 00 + y. Assume that the particular solution
Y (t) of the problem L[y] = 3 + cos 2t is of the form Y (t) =
A + B cos 2t + C sin 2t. Substituting Y into the equation, we
get
9B cos 2t + 9C sin 2t + A = 3 + cos 2t
Then we obtain the system of equations
A = 3, 9B = 1, 9C = 0
which implies A = 3, B = 91 , C = 0. Thus Y (t) = 91 cos 2t + 3.
Therefore, the general solution for the equation is
y(t) = yc (t) + Y (t)
= (c1 + c2 t) cos t + (c3 + c4 t) sin t +
1
cos 2t + 3
9
10. Find the general solution of the differential equation.
1
3
y 000 − 3y 00 + 2y 0 = t + et ; y(0) = 1, y 0 (0) = − , y 00 (0) = −
4
2
Then plot a graph of the solution.
[§4.3 #10]
Sol. The characteristic equation for the homogeneous problem
is r3 − 3r2 + 2r = r(r − 1)(r − 2) = 0, with roots r = 0, 1, 2.
Hence the general solution for the homogeneous problem is
yc (t) = c1 + c2 et + c3 e2t
Let L[y] := y 000 − 3y 00 + 2y 0 . Assume that the particular solution
Y (t) of the problem L[y] = t + et is of the form Y (t) = Atet +
Bt2 + Ct. Substituting Y into the equation, we get
−Aet + 4Bt + 2C − 6B = t + et
Then we obtainA = −1, B = 41 , C = 43 . Thus Y (t) = −tet +
1 2
t + 34 t. Therefore, the general solution for the equation is
4
y(t) = yc (t) + Y (t)
1
3
= c1 + c2 et + c3 e2t − tet + t2 + t
4
4
1
3
0
00
Since y(0) = 1, y (0) = − 4 , y (0) = − 2 , we have that
1
1
3
3
= − , c2 + 4c3 − = −
4
4
2
2
which shows that c1 = 1, c2 = c3 = 0. Hence the solution for
the initial value problem is
1
3
y(t) = −tet + t2 + t + 1
4
4
The graph of y(t) is as follows
c1 + c2 + c3 = 1,
c2 + 2c3 −
11. Determine a suitable form of Y (t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.
y 000 − y 0 = te−t + 2 sin t
[§4.3 #14]
Sol. The characteristic equation for the homogeneous problem
is r3 − r = r(r − 1)(r + 1) = 0, with roots r = 0, ±1. Hence the
general solution for the homogeneous problem is
yc (t) = c1 + c2 et + c3 e−t
Let L[y] := y 000 − y 0 . If Y (t) is the particular solution of the
problem L[y] = te−t + 2 sin t, then it can be assumed as the
form
Y (t) = t(At + B)e−t + C cos t + D sin t
12. Determine a suitable form of Y (t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.
y (4) + 2y 000 + 2y 00 = 3et + 2te−t + e−t sin t
[§4.3 #16]
Sol. The characteristic equation for the homogeneous problem
is r4 + 2r3 + 2r2 = r2 (r2 + 2r + 2) = 0, with roots r = 0 with
multiplicity 2, and r = −1 ± i. Hence the general solution for
the homogeneous problem is
yc (t) = c1 + c2 t + e−t (c3 cos t + c4 sin t)
Let L[y] := y (4) + 2y 000 + 2y 00 . If Y (t) is the particular solution
of the problem L[y] = 3et + 2te−t + e−t sin t, then it can be
assumed as the form
Y (t) = Aet + (Bt + C)e−t + te−t (D cos t + E sin t)
13. Determine a suitable form of Y (t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.
y (4) − y 000 − y 00 + y 0 = t2 + 4 + t sin t
[§4.3 #17]
Sol. The characteristic equation for the homogeneous problem
is r4 − r3 − r2 + r = r(r + 1)(r − 1)2 = 0, with roots r = 0, −1
and r = 1 with multiplicity 2. Hence the general solution for
the homogeneous problem is
yc (t) = c1 + c2 e−t + (c3 + c4 t)et
Let L[y] := y (4) − y 000 − y 00 + y 0 . If Y (t) is the particular solution
of the problem L[y] = t2 + 4 + t sin t, then it can be assumed as
the form
Y (t) = t(At2 + Bt + C) + (Dt + E) sin t + (F t + G) cos t
14. Determine a suitable form of Y (t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.
y (4) + 4y 00 = sin 2t + tet + 4
[§4.3 #18]
Sol. The characteristic equation for the homogeneous problem
is r4 +4r2 = r2 (r2 +4) = 0, with roots r = 0 with multiplicity 2,
and r = ±2i. Hence the general solution for the homogeneous
problem is
yc (t) = c1 + c2 t + c3 cos 2t + c4 sin 2t
Let L[y] := y (4) + 4y 00 . If Y (t) is the particular solution of the
problem L[y] = sin 2t + tet + 4, then it can be assumed as the
form
Y (t) = At2 + (Bt + C)et + t(D cos 2t + E sin 2t)
4.4. The Method of Variation of Parameters
15. Use the method of variation of parameters to determine the
general solution of the differential equation
y 000 + y 0 = tan t,
0<t<π
[§4.4 #1]
Sol. The characteristic equation for the homogeneous problem
is r3 + r = r(r2 + 1) = 0, with roots r = 0, ±i. Hence the
general solution for the homogeneous problem is
yc (t) = c1 + c2 cos t + c3 sin t
Let y1 (t) = 1, y2 (t) = cos t, y3 (t) = sin t, then W (y1 , y2 , y3 )(t) =
1. Moreover,
0
cos
t
sin
t
cos t = 1
W1 (t) = 0 − sin t
1 − cos t − sin t 1 0
sin
t
cos t = − cos t
W2 (t) = 0 0
0 1 − sin t 0
cos t 0 W3 (t) = 0 − sin t 0 = − sin t
1 − cos t 1 Let g(t) = tan t. The particular solution is given by Y (t) =
3
P
ui (t)yi (t), in which
i=1
Z
g(t)W1 (t)
dt = tan tdt = ln | sec t|
u1 (t) =
W (t)
Z
Z
g(t)W2 (t)
u2 (t) =
dt = − sin tdt = cos t
W (t)
Z
Z
g(t)W3 (t)
u3 (t) =
dt = − sin t tan tdt = sin t − ln | sec t + tan t|
W (t)
Z
Hence Y (t) is of the form
Y (t) = ln | sec t| − sin t ln | sec t + tan t| + 1
and the general solution for the equation is
y(t) = c1 + c2 cos t + c3 sin t + ln | sec t| − sin t ln | sec t + tan t|
16. Use the method of variation of parameters to determine the
general solution of the differential equation
y 000 − 2y 00 − y 0 + 2y = e4t
[§4.4 #2]
Sol. The characteristic equation for the homogeneous problem
is r3 − 2r2 − r + 2 = (r + 1)(r − 1)(r − 2) = 0, with roots
r = −1, 1, 2. Hence the general solution for the homogeneous
problem is
yc (t) = c1 e−t + c2 et + c3 e2t
Let y1 (t) = e−t , y2 (t) = et , y3 (t) = e2t , then W (y1 , y2 , y3 )(t) =
6e2t . Moreover,
0 et e2t W1 (t) = 0 et 2e2t = e3t
1 et 4e2t e−t 0 e2t W2 (t) = −e−t 0 2e2t = −3et
e−t 1 4e2t e−t et 0 W3 (t) = −e−t et 0 = 2
e−t et 1 Let g(t) = e4t . The particular solution is given by Y (t) =
3
P
ui (t)yi (t), in which
i=1
Z
Z
g(t)W1 (t)
1
1
u1 (t) =
dt =
e5t dt = e5t
W (t)
6
30
Z
Z
1
1
g(t)W2 (t)
u2 (t) =
dt = −
e3t dt = − e3t
W (t)
2
6
Z
Z
1
g(t)W3 (t)
1
dt =
u3 (t) =
e2t dt = e2t
W (t)
3
6
Hence Y (t) is of the form
1
1
1
1
Y (t) = e4t − e4t + e4t = e4t
30
6
6
30
and the general solution for the equation is
1
y(t) = c1 e−t + c2 et + c3 e2t + e4t
30
17. Find the general solution of the differential equation.
π
π
y 000 − y 00 + y 0 − y = sec t, − < t <
2
2
Leave your answer in terms of one or more integrals.
[§4.4 #7]
Sol. The characteristic equation for the homogeneous problem
is r3 − r2 + r − 1 = (r − 1)(r2 + 1) = 0, with roots r = 1, ±i.
Hence the general solution for the homogeneous problem is
yc (t) = c1 et + c2 cos t + c3 sin t
Let y1 (t) = et , y2 (t) = cos t, y3 (t) = sin t, then W (y1 , y2 , y3 )(t) =
2et . Moreover,
0
cos t
sin t cos t = 1
W1 (t) = 0 − sin t
1 − cos t − sin t et 0
sin t t
cos t = et (sin t − cos t)
W2 (t) = e 0
et 1 − sin t et
cos
t
0
t
W3 (t) = e − sin t 0 = −et (sin t + cos t)
et − cos t 1 Let g(t) = sec t. The particular solution is given by Y (t) =
3
P
ui (t)yi (t), in which
i=1
Z
g(t)W1 (t)
1 t −s
e sec sds
u1 (t) =
dt =
W (t)
2
Z
Z
√
g(t)W2 (t)
1
t
u2 (t) =
dt =
(tan t − 1)dt = − − ln cos t
W (t)
2
2
Z
Z
√
g(t)W3 (t)
1
t
u3 (t) =
dt = −
(tan t + 1)dt = − + ln cos t
W (t)
2
2
Z
since − π2 < t < π2 . Hence Y (t) is of the form
Z
√
t
1 t t−s
e sec sds− (cos t+sin t)+ln cos t(sin t−cos t)
Y (t) =
2
2
and the general solution for the equation is
√
t
1
y(t) = c1 e +c2 cos t+c3 sin t− (cos t+sin t)+ln cos t(sin t−cos t)+
2
2
t
18. Given that x, x2 , and 1/x are solutions of the homogeneous
equation corresponding to
x3 y 000 + x2 y 00 − 2xy 0 + 2y = 2x4 ,
determine a particular solution.
[§4.4 #13]
Sol. Write the equation as y 000 + x1 y 00 −
x>0
2 0
y
x2
+
2
x3
= 2x. Then
Z
t
et−s sec sds
Wronskian W (x, x2 , 1/x) =
0
W1 (x) = 0
1
x
W2 (x) = 1
0
x
W3 (x) = 1
0
6
.
x
Moreover
1
x2
x
1 2x − x2 = −3
2
2
x3
0 x1 2
0 − x12 =
x
1 x23 x2 0 2x 0 = x2
2 1 Let g(x) = 2x. The particular solution is given by Y (x) =
3
P
ui (x)yi (x), in which
i=1
Z
x3
g(x)W1 (x)
dx = − x2 dx = −
u1 (x) =
W (x)
3
Z
Z
2
x2
g(x)W2 (x)
dx =
xdx =
u2 (x) =
W (x)
3
3
Z
Z
g(x)W3 (x)
1
x5
u3 (x) =
dx =
x4 dx =
W (x)
3
15
Z
Hence Y (x) is of the form
Y (t) = −
x4 x4 x4
x4
+
+
=
3
3
15
15
19. Find a formula involving integrals for a particular solution of
the differential equation
y 000 − 3y 00 + 3y 0 − y = g(t)
If g(t) = t−2 et , determine Y (t).
[§4.4 #16]
Sol. The characteristic equation for the homogeneous problem
is r3 − 3r2 + 3r − 1 = (r − 1)3 = 0, with roots r = 1 with
multiplicity 3. Hence the general solution for the homogeneous
problem is
yc (t) = c1 et + c2 tet + c3 t2 et
Let y1 (t) = et , y2 (t) = tet , y3 (t) = t2 et , then W (y1 , y2 , y3 )(t) =
2e3t . Moreover,
t
2 t
0
te
t
e
t
t
= t2 e2t
(2 + t)te
W1 (t) = 0 (1 + t)e
1 (2 + t)et (t2 + 4t + 2)et 2 t et 0
t
e
(2 + t)tet = −2te2t
W2 (t) = et 0
et 1 (t2 + 4t + 2)et t
et
te
0
t
t
W3 (t) = e (1 + t)e 0 = e2t
et (2 + t)et 1 The particular solution is given by Y (t) =
3
P
ui (t)yi (t), in
i=1
which
Z
g(t)W1 (t)
1 t 2 −s
u1 (t) =
s e g(s)ds
dt =
W (t)
2
Z
Z t
g(t)W2 (t)
u2 (t) =
dt = −
se−s g(s)ds
W (t)
Z
Z
g(t)W3 (t)
1 t −s
e g(s)ds
u3 (t) =
dt =
W (t)
2
Z
Hence Y (t) is of the form
Z
Z t
Z
1 t 2 t−s
1 2 t t−s
t−s
Y (t) =
s e g(s)ds − t
se g(s)ds + t
e g(s)ds
2
2
Z
1 t
=
(t − s)2 et−s g(s)ds
2
If g(t) = t−2 et , then
Z
Z
1 t
et
2 t−s −2 s
Y (t) =
(t − s) e s e ds =
2
2
t
(t − s)2
ds = −tet ln t
s2
20. Find a formula involving integrals for a particular solution of
the differential equation
x3 y 000 − 3x2 y 00 + 6xy 0 − 6y = g(x),
x>0
Hint: Verify that x, x2 , and x3 are solutions of the homogeneous
equation.
[§4.4 #17]
Sol. Let y1 (x) = x, y2 (x) = x2 , y3 (x) = x3 . Then
x3 y1000 − 3x2 y100 + 6xy10 − 6y1 = 6x − 6x = 0
x3 y2000 − 3x2 y200 + 6xy20 − 6y2 = −6x2 + 12x2 − 6x2 = 0
x3 y3000 − 3x2 y300 + 6xy30 − 6y3 = 6x3 − 18x3 + 18x3 − 6x3 = 0
which shows that y1 , y2 , y3 are indeed solution of the homogeneous equation. Hence W (y1 , y2 , y3 )(x) = 2x3 . Also,
0 x2 x3 W1 (x) = 0 2x 3x2 = x4
1 2 6x x 0 x3 W2 (x) = 1 0 3x2 = −2x3
0 1 6x x x2 0 W3 (x) = 1 2x 0 = x2
0 2 1 The particular solution is given by Y (x) =
3
P
ui (x)yi (x), in
i=1
which
Z
g(x)W1 (x)
1 x g(t)
u1 (x) =
dx =
dt
W (x)
2
t2
Z
Z x
g(t)
g(x)W2 (x)
u2 (x) =
dx = −
dt
W (x)
t3
Z
Z
g(x)W3 (x)
1 x g(t)
u3 (x) =
dx =
dt
W (x)
2
t4
Hence Y (t) is of the form
Z
Z x
Z
x x g(t)
g(t)
x3 x g(t)
2
dt − x
dt +
dt
Y (t) =
2
t2
t3
2
t4
Z x
x
x2
x3 =
− 3 + 4 g(t)dt
2t2
t
2t
Z x
2
x(x − t)
=
g(t)dt
2t4
Z