ODE Homework 4
3.5. Nonhomogeneous Equations; Method of Undetermined
Coefficients
1. Find the general solution of the differential equation
y 00 + 9y = t2 e3t + 6
[§3.5 #5]
Sol. The characteristic equation for the homogeneous problem
is r2 + 9 = 0, with complex roots r = ±3i. Hence the general
solution for the homogeneous problem is
yc (t) = c1 cos 3t + c2 sin 3t.
Let L[y] := y 00 + 9y. and let g1 (t) = 6, g2 (t) = t2 e3t . Assume
that the particular solution Y (t) of the problem L[y] = t2 e3t + 6
is of the form Y (t) = (At2 + Bt + C)e3t + D. Substituting Y
into the equation, we get
18At2 e3t +(12A+18B)te3t +(2A+6B +18C)e3t +9D = t2 e3t +6
Then we obtain the system of equations
18A = 1, 12A + 18B = 0, 2A + 6B + 18C = 0, 9D = 6
1
1
1
which implies A = 18
, B = − 27
, C = 162
, D = 23 . Thus
1 2 3t
1
1 3t
2
3t
Y (t) = 18 t e − 27 te + 162 e + 3 . Therefore, the general
solution for the equation is
y(t) = yc (t) + Y (t)
= c1 cos 3t + c2 sin 3t +
1
1 3t 2
1 2 3t
t e − te3t +
e +
18
27
162
3
2. Find the general solution of the differential equation
y 00 + y = 3 sin 2t + t cos 2t
[§3.5 #8]
Sol. The characteristic equation for the homogeneous problem
is r2 + 1 = 0, with complex roots r = ±i. Hence the general
solution for the homogeneous problem is
yc (t) = c1 cos t + c2 sin t.
Let L[y] := y 00 + y. Assume that the particular solution Y (t)
of the problem L[y] = 3 sin 2t + t cos 2t is of the form Y (t) =
(At + B) cos 2t + C sin 2t. Substituting Y into the equation, we
get
−3At cos 2t − 3B cos 2t − (4A + 3C) sin 2t = 3 sin 2t + t cos 2t
Then we obtain the system of equations
−3A = 1,
−3B = 0,
−4A − 3C = 3
which implies A = − 31 , B = 0, C = − 59 . Thus Y (t) =
− 31 t cos 2t − 59 sin 2t. Therefore, the general solution for the
equation is
y(t) = yc (t) + Y (t)
5
1
= c1 cos t + c2 sin t − t cos 2t − sin 2t
3
9
3. Find the general solution of the differential equation
3
y 00 − y 0 − 2y = 3 cosh 2t = (e2t + e−2t )
2
t
−t
Hint: cosh t = e +e
.
2
[§3.5 #12]
Sol. The characteristic equation for the homogeneous problem
is r2 − r − 2 = 0, with distinct real roots r = −1, 2. Hence the
general solution for the homogeneous problem is
yc (t) = c1 e−t + c2 e2t
Let L[y] := y 00 − y 0 − 2y. Assume that the particular solution
Y (t) of the problem L[y] = 3 cosh 2t is of the form Y (t) =
Ate2t + Be−2t . Substituting Y into the equation, we get
3
3
3Ae2t + 4Be−2t = 3 cosh 2t = e2t + e−2t
2
2
Then we obtain the system of equations
3
3
3A = , 4B =
2
2
1
3
which implies A = 2 , B = 8 . Thus Y (t) = 12 te2t + 38 e−2t .
Therefore, the general solution for the equation is
y(t) = yc (t) + Y (t)
1
3
= c1 e−t + c2 e2t + te2t + e−2t
2
8
4. Find the solution of the initial value problem
y 00 − 2y 0 + y = tet + 4,
y(0) = 1, y 0 (0) = 1
[§3.5 #14]
Sol. The characteristic equation for the homogeneous problem
is r2 − 2r + 1 = 0, with repeated real roots r = 1. Hence the
general solution for the homogeneous problem is
yc (t) = c1 et + c2 tet
Let L[y] := y 00 − 2y 0 + y. Assume that the particular solution
Y (t) of the problem L[y] = tet + 4 is of the form Y (t) = t2 (At +
B)et + 4. Substituting Y into the equation, we get
2Bet + 6Atet + 4 = tet + 4
Then we obtain the system of equations
6A = 1,
2B = 0
which implies A = 61 , B = 0. Thus Y (t) = 16 t3 et + 4. Therefore,
the general solution for the equation is
y(t) = yc (t) + Y (t)
1
= c1 et + c2 tet + t3 et + 4
6
Then y 0 (t) = (c1 + c2 )et + c2 tet + 21 t2 et + 16 t3 et . Since y(0) =
1, y 0 (0) = 1, we have that
c1 + 4 = 1,
c1 + c2 = 1
which shows that c1 = −3, c2 = 4. Hence the solution for the
initial value problem is
1
y(t) = −3et + 4tet + t3 et + 4
6
5. Find the solution of the initial value problem
y 00 + 4y = 2 sin 2t,
y(0) = 2, y 0 (0) = −1
[§3.5 #17]
Sol. The characteristic equation for the homogeneous problem
is r2 + 4 = 0, with complex roots r = ±2i. Hence the general
solution for the homogeneous problem is
yc (t) = c1 cos 2t + c2 sin 2t
Let L[y] := y 00 + 4y. Assume that the particular solution Y (t)
of the problem L[y] = 2 sin 2t is of the form Y (t) = At cos 2t +
Bt sin 2t. Substituting Y into the equation, we get
4B cos 2t − 4A sin 2t = 2 sin 2t
Then we obtain the system of equations
−4A = 2,
4B = 0
which implies A = − 21 , B = 0. Thus Y (t) = − 12 t cos 2t. Therefore, the general solution for the equation is
y(t) = yc (t) + Y (t)
1
= c1 cos 2t + c2 sin 2t − t cos 2t
2
1
0
Then y (t) = 2c2 − 2 cos 2t − 2c1 sin 2t + t sin 2t. Since y(0) =
2, y 0 (0) = −1, we have that
1
c1 = 2, 2c2 − = −1
2
which shows that c1 = 2, c2 = − 41 . Hence the solution for the
initial value problem is
1
1
y(t) = 2 cos 2t − sin 2t − t cos 2t
4
2
6. Consider the equation
y 00 − 3y 0 − 4y = 2e−t
(i)
Have known that y1 (t) = e−t and y2 (t) = e4t are solutions of the
corresponding homogeneous equation. Adapting the method
of reduction of order, seek a solution of the nonhomogeneous
equation of the form Y (t) = v(t)y1 (t) = v(t)e−t , where v(t) is
to be determined.
(a) Substitute Y (t), Y 0 (t), and Y 00 (t) into Eq. (i) and show
that v(t) must satisfy v 00 − 5v 0 = 2.
(b) Let w(t) = v 0 (t) and show that w(t) must satisfy w0 −5w =
2. Solve this equation for w(t).
(c) Integrate w(t) to find v(t) and then show that
1
2
Y (t) = − te−t + c1 e4t + c2 e−t
5
5
The first term of the right side is the desired particular
solution of the nonhomogeneous equation. Note that it is
a product of t and e−t .
[§3.5 #27]
Sol.
(a) By setting Y (t) = v(t)y1 (t) = v(t)e−t and substituting it
into the equation, we get
v 00 y1 + 2v 0 y10 − 3v 0 y1 + v(y100 − 3y10 − 4y1 ) = 2y1
Since y1 = e−t which is the solution of the homogeneous
problem, also, we have y10 = −y1 . Thus, the above equation
becomes
v 00 y1 − 2v 0 y1 − 3v 0 y1 = 2y1
which shows that v 00 − 5v 0 = 2.
(b) Let w(t) = v 0 (t), the w satisfies the equation w0 − 5w = 2.
By solving this linear equation, we have that
2
w(t) = c1 e5t −
5
for some constant c1 .
(c) Since v 0 = w, so
Z
2
1
v(t) = w(t)dt = c1 e5t − t + c2
5
5
for some constant c2 . Hence
2
1
Y (t) = v(t)e−t = − te−t + c1 e4t + c2 e−t
5
5
7. In this problem we indicate an alternative procedure for solving
the differential equation
y 00 + by 0 + cy = (D2 + bD + c)y = g(t)
(i)
where b and c are constants, and D denotes differentiation with
respect to t. Let r1 and r2 be the zeros of the characteristic
polynomial of the corresponding homogeneous equation. These
roots may be real and different, real and equal, or conjugate
complex numbers.
(a) Verify that Eq. (i) can be written in the factored form
(D − r1 )(D − r2 )y = g(t)
where r1 + r2 = −b and r1 r2 = c.
(b) Let u = (D − r2 )y. Then show that the solution of Eq.
(i) can be found by solving the following two first order
equations:
(D − r1 )u = g(t),
(D − r2 )y = u(t).
[§3.5 #33]
Proof.
(a) Since D is a linear operator,
(D2 + bD + c)y = D2 y − (r1 + r2 )Dy + r1 r2 y = D(Dy) − r1 Dy − r2 Dy + r1 r2 y
= D(Dy) − D(r1 y) − r2 (Dy − r1 y) = D(Dy − r1 y) − r2 (Dy − r1 y)
= D(D − r1 )y − r2 (D − r1 )y = (D − r1 )(D − r2 )y
(b) Let u = (D − r2 )y, then the equation becomes
(D − r1 )u = g(t)
Since (D − r1 )u = u0 − r1 u = g(t), then we can solve u(t)
as the solution of the linear equation which is of the form
Z t
er1 (t−s) g(s)ds + c1 er1 t
u(t) =
for some constant c1 . Now consider the other linear equation (D − r2 )y = y 0 − r2 y = u(t). By solving this equation,
we get
Z t
er2 (t−s) u(s)ds + c2 er2 t
y(t) =
Z t
Z s
r2 (t−s)
=
e
er1 (s−τ ) g(τ )dτ + c1 er1 s ds + c2 er2 t
Z Zt s
Z t
r2 t+(r1 −r2 )s−r1 τ
=
e
g(τ )dτ ds + c1
er2 t+(r1 −r2 )s ds + c2 er2 t
8. Use the method in problem 7. to solve the differential equation
2y 00 + 3y 0 + y = t3 + 3 sin t
[§3.5 #36]
Sol. Note that
2y 00 + 3y 0 + y = (2D2 + 3D + 1)y = (2D + 1)(D + 1)y
Let u(t) = (D + 1)y. Then (2D + 1)u = g(t), that is,
2u0 + u = t3 + 3 sin t
By solving the linear equation of u, we have that
Z
t
− 2t 1
u(t) = e ·
(t3 + 3 sin t)e 2 dt
2
t
6
3
= c1 e− 2 − cos t + sin t + t3 − 6t2 + 24t − 48
5
5
for some constant c1 . By the fact that (D + 1)y = y 0 + y = u(t),
we have that
Z
−t
y(t) = e · et u(t)dt
Z t
3
6
−t
=e
et c1 e− 2 − cos t + sin t + t3 − 6t2 + 24t − 48 dt
5
5
t
3
9
= 2c1 e− 2 + c2 e−t −
cos t −
sin t + t3 − 9t2 + 42t − 90
10
10
3.6. Variation of Parameters
9. Find the general solution of the differential equation
π
y 00 + 9y = 9 sec2 3t, 0 < t <
6
[§3.6 #8]
Sol. The characteristic equation for the homogeneous problem
is r2 + 9 = 0, with complex roots r = ±3i. Hence the general
solution for the homogeneous problem is
yc (t) = c1 cos 3t + c2 sin 3t
Let y1 (t) = cos 3t, y2 (t) = sin 3t, then W (y1 , y2 )(t) = 3. The
particular solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in
which
Z
Z
y2 (t)g(t)
u1 (t) = −
dt = −3 sec 3t tan 3tdt = − sec 3t
W (t)
Z
Z
y1 (t)g(t)
dt = 3 sec 3tdt = ln(sec 3t + tan 3t)
u2 (t) =
W (t)
since 0 < t < π6 . Hence Y (t) is of the form
Y (t) = sin 3t ln(sec 3t + tan 3t) − 1
and the general solution for the equation is
y(t) = c1 cos 3t + c2 sin 3t + sin 3t ln(sec 3t + tan 3t) − 1
10. Find the general solution of the differential equation
y 00 − 2y 0 + y =
et
1 + t2
[§3.6 #10]
Sol. The characteristic equation for the homogeneous problem
is r2 − 2r + 1 = 0, with repeated real roots r = 1. Hence the
general solution for the homogeneous problem is
yc (t) = c1 et + c2 tet
Let y1 (t) = et , y2 (t) = tet , then W (y1 , y2 )(t) = e2t . The particular solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in
which
Z
Z
√
y2 (t)g(t)
t
u1 (t) = −
dt = −
1 + t2
dt
=
−
ln
W (t)
1 + t2
Z
Z
t
y1 (t)g(t)
dt =
u2 (t) =
dt = tan−1 t
W (t)
1 + t2
Hence Y (t) is of the form
√
Y (t) = −et ln 1 + t2 + tet tan−1 t
and the general solution for the equation is
√
y(t) = c1 et + c2 tet − et ln 1 + t2 + tet tan−1 t
11. Verify the given functions y1 and y2 satisfy the corresponding
homogeneous equation; then find a particular solution of the
nonhomogeneous equation.
ty 00 − (1 + t)y 0 + y = t2 e2t ,
t > 0;
y1 (t) = 1 + t, y2 (t) = et
[§3.6 #16]
Sol. Observe that y10 = 1, y100 = 0 and y20 = y200 = et , so
ty100 − (1 + t)y10 + y1 = −1 − t + 1 + t = 0
ty200 − (1 + t)y20 + y2 = et (t − 1 − t + 1) = 0
which shows that y1 , y2 are solutions for the corresponding homogeneous equation. Also, we have W (y1 , y2 )(t) = tet . Rewrite
the equation as y 00 − 1+t
y 0 + 1t y = te2t . Then the particular sot
lution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in which
Z
Z
y2 (t)g(t)
1
u1 (t) = −
dt = − e2t dt = − e2t
W (t)
2
Z
Z
y1 (t)g(t)
u2 (t) =
dt = (t + 1)et dt = tet
W (t)
Hence we get
1
1
Y (t) = te2t − e2t
2
2
12. Verify the given functions y1 and y2 satisfy the corresponding
homogeneous equation; then find a particular solution of the
nonhomogeneous equation.
x2 y 00 −3xy 0 +4y = x2 ln x,
x > 0;
y1 (x) = x2 , y2 (x) = x2 ln x
[§3.6 #17]
Sol. Observe that y10 = 2x, y100 = 2 and y20 = x + 2x ln x, y200 =
3 + 2 ln x, so
x2 y100 − 3xy10 + 4y1 = 2x2 − 6x2 + 4x2 = 0
x2 y200 − 3y20 + 4y2 = 3x2 + 2x2 ln x − 3x2 − 6x2 ln x + 4x2 ln x = 0
which shows that y1 , y2 are solutions for the corresponding homogeneous equation. Also, we have W (y1 , y2 )(x) = x3 . Rewrite
the equation as y 00 − x3 y 0 + x42 y = ln x. Then the particular solution is given by Y (x) = u1 (x)y1 (x) + u2 (x)y2 (x), in which
Z
Z
y2 (x)g(x)
(ln x)2
1
u1 (x) = −
dx = −
dx = − (ln x)3
W (x)
x
3
Z
Z
y1 (x)g(x)
ln x
1
u2 (x) =
dx =
dx = (ln x)2
W (x)
x
2
Hence we get
1
Y (x) = x2 (ln x)3
6
13. Verify the given functions y1 and y2 satisfy the corresponding
homogeneous equation; then find a particular solution of the
nonhomogeneous equation.
x2 y 00 + xy 0 + (x2 − 0.25)y = g(x),
x > 0;
y1 (x) = x−1/2 sin x, y2 (x) = x−1/2 cos x
where g(x) is an arbitrary continuous function.
[§3.6 #19]
Sol. Observe that
2x cos x − sin x 00 −4x2 sin x − 4x cos x + 3 sin x
, y1 =
2x3/2
4x5/2
2
−2x sin x − sin x 00 −4x cos x + 4x sin x + 3 cos x
y20 =
, y2 =
2x3/2
4x5/2
y10 =
so
x2 y100 + xy10 + (x2 − 0.25)y1
−4x2 sin x − 4x cos x + 3 sin x 2x cos x − sin x (4x2 − 1) sin x
√
√
√
=
+
+
=0
4 x
2 x
4 x
x2 y200 + xy20 + (x2 − 0.25)y2
−4x2 cos x + 4x sin x + 3 cos x −2x sin x − cos x (4x2 − 1) cos x
√
√
√
=
+
+
=0
4 x
2 x
4 x
which shows that y1 , y2 are solutions for the corresponding
homogeneous equation. Also, we have W (y1 , y2 )(x) = − x1 .
2
Rewrite the equation as y 00 + x1 y 0 + x −0.25
y = g(x)
. Then the
x2
x2
particular solution is given by Y (x) = u1 (x)y1 (x) + u2 (x)y2 (x),
in which
Z x
Z x
g(s) cos s
y2 (s)g(s)
√
ds =
u1 (x) = −
ds
2
s W (s)
s s
Z x
Z x
y1 (s)g(s)
g(s) sin s
√
u2 (x) =
ds = −
ds
2
s W (s)
s s
Hence we get
Z
Z
sin x x g(s) cos s
cos x x g(s) sin s
√
√
Y (x) = √
ds − √
ds
x
s s
x
s s
Z x
g(x)(sin x cos s − cos x sin s)
1
√
ds
=√
x
s s
Z x
1
g(x) sin(x − s)
√
=√
ds
x
s s
14. The method of reduction of order can also be used for the nonhomogeneous equation
y 00 + p(t)y 0 + q(t)y = g(t),
(i)
provided one solution y1 of the corresponding homogeneous
equation is known. Let y = v(t)y1 (t) and show that y satisfies Eq. (i) if v is a solution of
y1 (t)v 00 + [2y10 (t) + p(t)y1 (t)]v 0 = g(t).
(ii)
Equation (ii) is a first order linear equation for v 0 . Solving this
equation, integrating the result, and then multiplying by y1 (t)
lead to the general solution of Eq. (i).
[§3.6 #28]
Proof. Let y = v(t)y1 (t), then y 0 = y1 v 0 + y10 v, y 00 = y1 v 00 +
2y10 v 0 + y100 v. By substituting y into Eq. (i) and the fact that
y1 is a solution of the corresponding homogeneous equation, we
obtain
y1 v 00 + (2y10 + py1 )v 0 = g
which shows that v is a solution of Eq. (ii). Rewrite the above
2y 0 +py
equation as the form v 00 + 1y1 1 v 0 = yg1 and use the integrating
R x 2y10 (s)+p(s)y1 (s) factor µ(x) = exp
ds , we have that
y1 (s)
Z x
Z
s 2y 0 (τ ) + p(τ )y (τ ) g(s)
1
1
exp
dτ ds
v 0 (x) =
y1 (s)
y
(τ
)
1
x
Thus
Z xZ s
Z z 2y 0 (z) + p(z)y (z) g(τ )
1
1
v(x) =
exp
dz dτ ds
y1 (τ )
y1 (z)
s
Therefore the general solution of Eq. (i) is of the form
Z xZ s
Z z 2y 0 (z) + p(z)y (z) g(τ )
1
1
y(x) = y1 (x)
exp
dz dτ ds
y1 (τ )
y1 (z)
s
15. Use the method outlined in problem 14. to solve the differential
equation
(1 − t)y 00 + ty 0 − y = 2(t − 1)2 e−t ,
0 < t < 1;
y1 (t) = et
[§3.6 #32]
Sol. Write the equation as the form
t 0
1
y 00 +
y −
y = 2(1 − t)e−t
1−t
1−t
1
t
00
It is easy to see that y1 + 1−t
y10 − 1−t
y1 = 0. Set y = y1 v, then
v satisfies the equation
tet 0
et v 00 + 2et +
v = 2(1 − t)e−t
1−t
2−t 0
that is, v 00 + 1−t
v = 2(1−t)e−2t . By using the integrating factor
R 2−t et
µ(t) = exp
dt = 1−t
, since 0 < t < 1. We can solve v 0 as
1−t
Z
1 − t t es
0
v (t) = t
·2(1−s)e−2s ds = 2(t−1)e−2t +c1 (1−t)e−t
e
1−s
for some constant c1 . Hence
1
v(t) =
− t e−2t + c1 te−t + c2
2
for some constant c2 . Therefore, the general solution for the
equation is
1
t
y(t) = y1 (t)v(t) = c2 e + c1 t +
− t e−t
2
4.1. General Theory of nth Order Linear Equation
16. In this problem we show how to generalize Abel’s theorem to
higher order equation. We first outline the procedure for the
third order equation
y 000 + p1 (t)y 00 + p2 (t)y 0 + p3 (t)y = 0
Let y1 , y2 and y3 be solutions of this equation on an interval I.
(a) If W = W (y1 , y2 , y3 ), show that
y1 y2 y3 W 0 = y10 y20 y30 y1000 y2000 y3000 Hint: The derivative of a 3-by-3 determinant is the sum of
three 3-by-3 determinants obtained by differentiating the
first, second, and third rows, respectively.
(b) Substitute for y1000 , y2000 , and y3000 from the differential equation; multiply the first row by p3 , multiply the second row
by p2 , and add these to the last row to obtain
W 0 = −p1 (t)W
(c) Show that
h
W (y1 , y2 , y3 )(t) = c exp −
Z
i
p1 (t)dt
it follows that W is either always zero or nowhere zero on
I.
(d) Generalize this argument to the nth order equation
y (n) + p1 (t)y (n−1) + · · · + pn (t)y = 0
with solutions y1 , · · · , yn . That is, establish Abel’s formula
h Z
i
W (y1 , · · · , yn )(t) = c exp − p1 (t)dt
for this case.
[§4.1 #20]
Proof.
(a) Note that
y1 y2 y3 X
0
00
0
yσ(3)
yσ(1) yσ(2)
W (y1 , y2 , y3 ) = y1 y20 y30 =
00
00
00
y1 y2 y3 σ∈S3
Then
d X
0
00
W =
yσ(1) yσ(2) yσ(3)
dt σ∈S
3
X
0
0
00
00
00
0
000
=
yσ(1)
yσ(2)
yσ(3)
+ yσ(1) yσ(2)
yσ(3)
+ yσ(1) yσ(2)
yσ(3)
0
σ∈S3
= = y10 y20 y30 y1 y2 y3 y1 y2 y3
y10 y20 y30 + y100 y200 y300 + y10 y20 y30
y100 y200 y300 y100 y200 y300 y1000 y2000 y3000
y1 y2 y3 y10 y20 y30 y1000 y2000 y3000 (b)
0
p2 p3 W = = = p3 y1 p3 y2 p3 y3 p2 y10 p2 y20 p2 y30 y3000 y2000
y1000
p3 y 1
p3 y 2
p3 y 3
0
0
p2 y 1
p2 y 2
p2 y30
000
0
000
0
000
y1 + p2 y1 + p3 y1 y2 + p2 y2 + p3 y2 y3 + p2 y30 + p3 y3
p3 y1
p3 y 2
p3 y3 p2 y10
p2 y20
p2 y30 = −p1 p2 p3 W
00
00
−p1 y1 −p1 y2 −p1 y300 As long as the coefficients p2 , p3 are not zero, we obtain
W 0 = −p1 W . For if p2 ≡ 0, p3 6≡ 0 then yi000 + p1 yi00 + p3 yi =
0, i = 1, 2, 3. Thus
p3 y1 p3 y2 p3 y3 p3 y1
p
y
p
y
3
2
3
3
0
0
0
0
0
0
0
y2
y3 = y1
y2
y3 = −p1 p3 W
p3 W = y1
y1000
y2000
y3000 −p1 y100 −p1 y200 −p1 y300 We also have W 0 = −p1 W . Similar result can be concluded
for case p2 6≡ 0, p3 ≡ 0 and p2 ≡ p3 ≡ 0.
(c) Since W 0 (t) = −p1 (t)W (t), by solving this separation equation, we have that
h Z
i
W (y1 , y2 , y3 )(t) = c exp − p1 (t)dt
Accordingly, if c = 0 then W ≡ 0 and W is nowhere zero
on I if c 6= 0.
(d) It can be shown, by mathematical induction, that
y1
y
·
·
·
y
2
n
0
0
y10
···
yn y2
..
..
..
W 0 (y1 , · · · , yn )(t) = .
.
.
(n−2) (n−2)
(n−2) y1
(n) y2 (n) · · · yn (n) y
y2
· · · yn 1
Base on the reasoning in (b), it follows that
p2 p3 · · · pn W 0 = −p1 p2 p3 · · · pn W
Therefore, we get W 0 = −p1 (t)W .
17. Use Abel’s formula to find the Wronskian of a fundamental set
of solutions of the differential equation
t3 y (4) + t2 y 000 + ty 00 − 2y = 0
[§4.1 #24]
Sol. Rewrite the equation as the form
1
1
2
y (4) + y 000 + 2 y 00 − 3 y = 0
t
t
t
Then the Wronskian W (t) of a fundamental set of solutions is
h Z dt i c
=
W (t) = c exp −
t
t
18. Show that if y1 is a solution of
y 000 + p1 (t)y 00 + p2 (t)y 0 + p3 (t)y = 0,
then the substitution y = y1 (t)v(t) leads to the following second
order equation for v 0 :
y1 v 000 + (3y10 + p1 y1 )v 00 + (3y100 + 2p1 y10 + p2 y1 )v 0 = 0
[§4.1 #26]
Proof. Since y = y1 v, then y 0 = y1 v 0 + y10 v, y 00 = y1 v 00 + 2y10 v 0 +
y100 v and y 000 = y1 v 000 + 3y10 v 00 + 3y100 v 0 + y1000 v. Substituting into the
equation, we obtain
y1 v 000 +(3y10 +p1 y1 )v 00 +(3y100 +2p1 y10 +p2 y1 )v 0 +(y1000 +p1 y100 +p2 y10 +p3 y1 )v = 0
Since y1 is a solution of the equation, we have that v must
satisfies
y1 v 000 + (3y10 + p1 y1 )v 00 + (3y100 + 2p1 y10 + p2 y1 )v 0 = 0
19. Use the method of reduction of order to solve the differential
equation
t2 (t + 3)y 000 − 3t(t + 2)y 00 + 6(1 + t)y 0 − 6y = 0,
y1 (t) = t2 , y2 (t) = t3
t > 0;
[§4.1 #28]
Sol. Write the equation in the form
3(t + 2) 00 6(1 + t) 0
6
y 000 −
y + 2
y − 2
y=0
t(t + 3)
t (t + 3)
t (t + 3)
Assume that y(t) is a solution of the equation with y(t) =
y1 (t)v(t) = t2 v(t), then v(t) satisfies the equation
3(t + 2) 00 3(t + 2) 2 6(1 + t) 0
t2 v 000 + 6t−t2
v + 6−4t·
+t · 2
v =0
t(t + 3)
t(t + 3)
t (t + 3)
That is, by letting w = v 00 , we have that
3(t + 4)
w=0
w0 +
t(t + 3)
which can be solved that
t+3
v 00 (t) = w(t) = c1 4
t
Hence
t+1
v(t) = c1 2 + c2 t + c3
2t
Hence the general solution for the equation is of the form
c1
y(t) = t2 v(t) = (t + 1) + c2 t3 + c3 t2
2