ODE Homework 3
3.2. Solutions of Linear Homogeneous Equations; the Wronskian
1. Verify that the functions y1 (t) = et and y2 (t) = tet are solutions
of the differential equation
y 00 − 2y 0 + y = 0
Do they constitute a fundamental set of solutions?
[§3.2 #25]
Sol. Observe that
y100 − 2y10 + y1 = et − 2et + et = 0
y200 − 2y20 + y2 = (2et + tet ) − 2(et + tet ) + tet = 0
which shows that y1 , y2 are solutions of the equation. Furthermore
t
e
tet t
t
= e2t + te2t − te2t = e2t 6≡ 0
W (e , te ) = t t
e e + tet Hence y1 , y2 forms a fundamental set of solutions.
2. Verify that the functions y1 (x) = x and y2 (x) = sin x are solutions of the differential equation
(1 − x cot x)y 00 − xy 0 + y = 0,
0<x<π
Do they constitute a fundamental set of solutions?
[§3.2 #27]
Sol. Observe that
(1 − x cot x)y100 − xy10 + y1 = 0 − x + x = 0
(1 − x cot x)y200 − xy20 + y2 = −(1 − x cot x) sin x − x cos x + sin x
= − sin x + x cos x − x cos x + sin x = 0
which shows that y1 , y2 are solutions of the equation. Furthermore
x sin x = x cos x − sin x
W (x, sin x) = 1 cos x Since π2 cos π2 − sin π2 = −1 6= 0, so W (x, sin x) 6≡ 0.Hence y1 , y2
forms a fundamental set of solutions.
3. Find the Wronskian of two solutions of the differential equation
t2 y 00 − t(t + 2)y 0 + (t + 2)y = 0
with out solving the equation
[§3.2 #29]
Sol. Rewrite the equation as the form
t(t + 2) 0 t + 2
y 00 −
y + 2 y=0
t2
t
Then by Abel’s formula, the Wronskian of two solutions is
Z t(t + 2) W (t) = c exp
dt = c exp t + 2 ln |t| = c t2 et
2
t
for some constant c.
4. Find the Wronskian of two solutions of the differential equation
x2 y 00 + xy 0 + (x2 + ν 2 ) = 0
Bessel’s equation
with out solving the equation
[§3.2 #31]
Sol. Rewrite the equation as the form
1
x2 + ν 2
y 00 + y 0 +
y=0
x
x2
Then by Abel’s formula, the Wronskian of two solutions is
Z 1 c
dx = c exp − ln |x| =
W (x) = c exp −
x
x
for some constant c.
5. If y1 and y2 are a fundamental set of solutions of
t2 y 00 − 2y 0 + (3 + t)y = 0
and if W (y1 , y2 )(2) = 3, find the value of W (y1 , y2 )(6).
[§3.2 #35]
Sol. Rewrite the equation as the form
3+t
2
y 00 − 2 y 0 + 2 y = 0
t
t
Then by Abel’s formula, the Wronskian of two solutions is
Z 2 2
2
W (y1 , y2 )(t) = c exp
dt
=
c
exp
−
= ce− t
2
t
t
for some constant c. Since W (y1 , y2 )(2) = 3, we have that
ce−1 = 3 ⇒ c = 3e. That is, the Wronskian of y1 and y2 is
W (y1 , y2 )(t) = 3e
t−2
t
Thus W (y1 , y2 )(6) = 3e
6−2
6
2
= 3e 3 .
6. Exact Equations. The equation P (x)y 00 + Q(x)y 0 + R(x)y = 0
is said to be exact if it can be written in the form [P (x)y 0 ]0 +
[f (x)y]0 = 0, where f (x) is to be determined in terms of P (x), Q(x),
and R(x). The latter equation can be integrated once immediately, resulting in a first order linear equation for y that can be
solved. By equating the coefficients of the preceding equations
and then eliminating f (x), the necessary condition for exactness is P 00 (x) − Q0 (x) + R(x) = 0. It can be shown that this is
also a sufficient condition. Determine whether the equation
xy 00 − (cos x)y 0 + (sin x)y = 0,
x>0
is exact. If so,, solve the equation
[§3.2 #44]
Sol. By letting P (x) = x, Q(x) = − cos x, R(x) = sin x, then
P 00 − Q0 + R = 0 − sin x + sin x = 0
which shows that the equation is exact. Assume that the equation can be rewrite as the form (xy 0 )0 + [f (x)y]0 = 0, then
xy 00 − (cos x)y 0 + (sin x)y = xy 00 + y 0 + f (x)y 0 + f 0 (x)y
which implies that f (x) = − cos x − 1. Thus the equation can
be reformulate as the form
(xy 0 )0 − ((cos x + 1)y)0 = 0
Integrating on both side, we get xy 0 − (cos x + 1)y = c1 , that
is, y 0 − cos xx+1 y = cx1 , which is a 1st order linear equation with
integrating factor
Z x cos s + 1 µ(x) = exp −
ds
s
The general solution for the equation is
Z x
c1
µ(s)
c2
y(x) =
ds +
µ(x)
s
µ(x)
R x cos τ +1 Z x
Z x cos s + 1 exp s
dτ
τ
= c1
ds + c2 exp
ds
s
s
for some constants c1 , c2 .
7. The Adjoint Equation. If a second order linear homogeneous
equation is not exact, it can be made exact by multiplying by an
appropriate integrating factor µ(x). Thus we require that µ(x)
be such that µ(x)P (x)y 00 + µ(x)Q(x)y 0 + µ(x)R(x)y = 0 can be
written in the form [µ(x)P (x)y 0 ]0 + [f (x)y]0 = 0. By equating
coefficients in these two equations and eliminating f (x), the
function µ must satisfy
P µ00 + (2P 0 − Q)µ0 + (P 00 − Q0 + R)µ = 0.
This equation is known as the adjoint of the original equation
and is important in the advanced theory of differential equations. In general, the problem of solving the adjoint differential
equation is as difficult as that of solving the original equation,
so only occasionally is it possible to find an integrating factor
for a second order equation. Find the adjoint of the differential
equation
(1 − x2 )y 00 − 2xy 0 + α(α + 1)y = 0.
Legendre’s equation
[§3.2 #48]
Sol. By letting P (x) = 1 − x2 , Q(x) = −2x, R(x) = α(α + 1),
then
P 00 − Q0 + R = −2 + 2 + α(α + 1) = α(α + 1)
which shows that the equation is not exact if α 6= 0, −1. Furthermore, 2P 0 − Q = −4x + 2x = −2x, then the adjoint of the
equation is
(1 − x2 )µ00 − 2xµ0 + α(α + 1)µ = 0
which is the given differential equation itself.
3.3. Complex Roots of the Characteristic Equation
8. Find the general solution of the differential equation
9y 00 + 9y 0 − 4y = 0
[§3.3 #14]
Sol. The characteristic equation is
9r2 + 9r − 4 = 0
Thus we get r = − 34 , 13 . Therefore the general solution is of the
form
4t
t
y(t) = c1 e− 3 + c2 e 3
9. Find the solution of the initial value problem.
y 00 + 4y 0 + 5y = 0,
y(0) = 1, y 0 (0) = 0
Sketch the graph of the solution and describe its behavior for
increasing t.
[§3.3 #18]
Sol. The characteristic equation is
r2 + 4r + 5 = 0
Thus we get r = −2 ± i. Therefore the general solution is of
the form
y(t) = c1 e−2t cos t + c2 e−2t sin t
Thus y 0 (t) = (c2 − 2c1 )e−2t cos t − (c1 + 2c2 )e−2t sin t. Since
y(0) = 1, y 0 (0) = 0, so
c1 = 1
c2 − 2c1 = 0
and we have that c1 = 1, c2 = 2. Hence the solution of the
initial value problem is
√
y(t) = e−2t cos t + 2e−2t sin t = 5e−2t sin(t + θ)
where θ = sin−1 √15 . Note that both e−2t → 0 as t → ∞. Hence
lim y(t) = 0. The graph of solution is as follows
t→∞
10. Consider the initial value problem
y 00 + 2y 0 + 6y = 0,
y(0) = 2, y 0 (0) = α ≥ 0.
(a) Find the solution y(t) of this problem.
(b) Find α so that y = 0 when t = 1.
(c) Find, as a function of α, the smallest positive value of t for
which y = 0.
(d) Determine the limit of the expression found in part (c) as
α → ∞.
[§3.3 #25]
Sol.
(a) The characteristic equation is
r2 + 2r + 6 = 0
√
Thus we get r = −1 ± 5 i. Therefore the general solution
is of the form
√
√
y(t) = c1 e−t cos 5 t + c2 e−t sin 5 t
√
√
Thus y 0 (t) = ( 5 c2 − c1 )e−2t cos t − ( 5 c1 + c2 )e−2t sin t.
Since y(0) = 2, y 0 (0) = α, so
c1 = 2
√
5 c2 − c1 = α
√ . Hence the solution of
and we have that c1 = 2, c2 = α+2
5
the initial value problem is
√
√
α+2
y(t) = 2e−t cos 5 t + √ e−t sin 5 t
5
√
√
√ e−1 sin 5 = 0. That is,
(b) If y(1) = 0, then 2e−1 cos 5 + α+2
5
√
√
√
√
α+2
2 cos 5 = − √ sin 5 ⇒ α = −2 5 cot 5−2 ≈ 1.50878
5
√
√
√ sin 5 t = 0, that is,
(c) If y(t) = 0, then 2 cos 5 t + α+2
5
√
√
2 5
tan 5 t = −
α+2
√
2 5
Note that α ≥ 0, then α > −2 and we have that − α+2
< 0.
Thus, the time t = t(α) for which y = 0, is
√
2 5 kπ
1
−1
t = √ tan
−
+√
α+2
5
5
for some k ∈ N. Hence the smallest positive value of t for
which y = 0 is
2√5 π
2√5 π
1
1
−1
−1
t(α) = √ tan
−
+ √ = √ − √ tan
α+2
α+2
5
5
5
5
(d)
π
2√5 1
lim t(α) = lim √ − √ tan−1
α→∞
α→∞
α+2
5
5
√
π
π
1
2 5
−1
= √ − √ lim tan
=√
α+2
5
5 α→∞
5
11. Consider the initial value problem
y 00 + 2ay 0 + (a2 + 1)y = 0,
y(0) = 1, y 0 (0) = 0.
(a) Find the solution y(t) of this problem.
(b) For a = 1 find the smallest T such that |y(t)| < 0.1 for
t > T.
(c) Repeat part (b) for a = 41 , 12 , and 2.
(d) Using the results of parts (b) and (c), plot T versus a and
describe the relation between T and a.
[§3.3 #26]
Sol.
(a) The characteristic equation is
r2 + 2ar + (a2 + 1) = 0
Thus we get r = −a ± i. Therefore the general solution is
of the form
y(t) = c1 e−at cos t + c2 e−at sin t
Thus y 0 (t) = (c2 − ac1 )e−2t cos t − (c1 + ac2 )e−2t sin t. Since
y(0) = 1, y 0 (0) = 0, so
c1 = 1
c2 − ac1 = 0
and we have that c1 = 1, c2 = a. Hence the solution of the
initial value problem is
√
y(t) = e−at cos t + ae−at sin t = 1 + a2 e−at sin(t + θ)
1
where θ = sin−1 √1+a
.
√2
√
(b) Note that |y(t)| ≤ 1 + a2 e−at , the inequality 1 + a2 e−at ≤
0.1 implies that
√
1
t ≥ ln(10 1 + a2 )
a
Therefore T ≤
y(t) becomes
1
a
√
ln(10 1 + a2 ). For a = 1, the solution
√
π
y(t) = 2e sin t +
4
By using Newton’s method, to solve |y(t)| = 0.1, we have
that T1 ≈ 1.8763.
(c) For a = 41 , 21 and 2, the corresponding T is
−t
T 1 ≈ 4.30035, T 1 ≈ 7.42844, T2 ≈ 1.51159
2
4
respectively.
(d) The plots of T versus a is as follows
√
Note that the estimate Ta approach the graph a1 ln(10 1 + a2 )
as a gets large.
12. Euler Equations. An equation of the form
d2 y
dy
+ αt + βy = 0,
t > 0,
(ii)
2
dt
dt
where α and β are real constants, is called an Euler equation.
2
dy
(a) Let x = ln t and calculate dy
and ddt2y in terms of dx
and
dt
d2 y
.
dx2
(b) Use the results of part (a) to transform Eq. (ii) into
t2
d2 y
dy
+
(α
−
1)
+ βy = 0.
(iii)
dx2
dx
Observe that Eq. (iii) has constant coefficients. If y1 (x)
and y2 (x) form a fundamental set of solutions of Eq. (iii),
then y1 (ln t) and y2 (ln t) form a fundamental set of solutions of Eq. (ii).
[§3.3 #34]
Sol.
(a) Let x = ln t, then dx
= 1t . Thus
dt
dy
dy dx
1 dy
=
=
dt
dx dt
t dx
2
dy
d 1 dy 1 dy 1 d dy d dy
=
=− 2
+
=
dt2
dt dt
dt t dx
t dx t dt dx
1 d dy 1 dy
1 d2 y
1 dy
+ 2
=− 2
+ 2 2
=− 2
t dx t dx dx
t dx t dx
(b) By (a), Eq. (ii) becomes
1 dy
1 d2 y 1 dy
2
+ 2 2 + αt ·
+ βy = 0
t − 2
t dx t dx
t dx
d2 y
dy
⇒
+ (α − 1) + βy = 0
2
dx
dx
13. Solve the equation
t2 y 00 + 7ty 0 + 10y = 0
[§3.3 #41]
Sol. By letting x ln t, the above problem shows that the equation t2 y 00 + 7ty 0 + 10y = 0 can be transformed to the equation
ÿ + 6ẏ + 10y = 0
d
where˙ = dx
. The characteristic equation for the new differential
equation is
r2 + 6r + 10 = 0
Thus we get r = −3 ± i. Therefore the general solution is of
the form
y(x) = c1 e−3x cos x + c2 e−3x sin x
Since x = ln t, hence the general solution for the original equation is
y(t) = c1 t−3 cos ln t + c2 t−3 sin ln t
3.4. Repeated Roots; Reduction of Order
14. Find the general solution of the differential equation
y 00 − 6y 0 + 9y = 0
[§3.4 #6]
Sol. The characteristic equation is
r2 − 6r + 9 = (r − 3)2 = 0
Thus we get r = 3 with multiplicity 2. Therefore the general
solution is of the form
y(t) = c1 e3t + c2 te3t
15. Consider the initial value problem
4y 00 + 12y 0 + 9y = 0,
y(0) = 1, y 0 (0) = −4.
(a) Solve the initial value problem and plot its solution for
0 ≤ t ≤ 5.
(b) Determine where the solution has the value zero.
(c) Determine the coordinates (t0 , y0 ) of the minimum point.
(d) Change the second initial condition to y 0 (0) = b and find
the solution as a function of b. Then find the critical value
of b that separates solutions that always remain positive
from those that eventually become negative.
[§3.4 #15]
Sol.
(a) The characteristic equation is
4r2 + 12r + 9 = (2r + 3)2 = 0
Thus we get r = − 23 with multiplicity 2. Therefore the
general solution is of the form
3t
3t
y(t) = c1 e− 2 + c2 te− 2
3t
3t
Thus y 0 (t) = c2 − 3c21 e− 2 − 3c22 te− 2 .
1, y 0 (0) = −4, so
Since y(0) =
c1 = 1
3
c2 − c1 = −4
2
and we have that c1 = 1, c2 = − 52 . Hence the solution of
the initial value problem is
3t
5t 3t
2 − 5t − 3t
y(t) = e− 2 − e− 2 =
e 2
2
2
The graph of solution for 0 ≤ t ≤ 5 is as follows
3t
e− 2 = 0 ⇒ 2−5t
= 0 ⇒ t = 25 .
(b) Let y(t) = 0, then 2−5t
2
2
3t
3t
3t
(c) Observe that y 0 (t) = − 25 e− 2 − 6−15t
e− 2 = 15t−16
e− 2 .
4
4
Then y 0 (t) = 0 if and only if t = t0 = 16
. Note that
15
15 − 8
16
78−45t − 3t
00
00
y (t) = 8 e 2 and y 15 = 4 e 5 > 0. Hence y(t) has
8
16
minimum y0 = y 15
= − 35 e− 5 at t = t0 . That is, the min8
16
imum point for the graph of y(t) is (t0 , y0 ) = 15
, − 53 e− 5 .
(d) If the initial data becomes y(0) = 1, y 0 (0) = b, then the
solution for the initial value problem becomes
y(t) =
3t
2b + 3 − 3t
(3 + 2b)t + 2 − 3t
e 2 = e− 2 +
te 2
2
2
Then we can see that the solution y(t) remains positive for
all t if and only if 2b+3
≥ 0, therefore, the critical value is
2
3
b = −2.
16. Consider the initial value problem
16y 00 + 24y 0 + 9y = 0,
y(0) = a > 0, y 0 (0) = −1.
(a) Solve the initial value problem.
(b) Find the critical value of a that separates solutions that
become negative from those that are always positive.
[§3.4 #18]
Sol.
(a) The characteristic equation is
16r2 + 24r + 9 = (4r + 3)2 = 0
Thus we get r = − 34 with multiplicity 2. Therefore the
general solution is of the form
3t
3t
y(t) = c1 e− 4 + c2 te− 4
3t
3t
Thus y 0 (t) = c2 − 3c41 e− 4 − 3c42 te− 4 .
a, y 0 (0) = −1, so
Since y(0) =
c1 = a
3
c2 − c1 = −1
4
. Hence the solution of
and we have that c1 = a, c2 = 3a−4
4
the initial value problem is
3t
3a − 4 − 3t
(3a − 4)t + 4a − 3t
y(t) = ae− 4 +
te 4 =
e 4
4
4
3t
3t
(b) Since y(t) = ae− 4 + 3a−4
te− 4 . The solution remains pos4
itive for all t if and only if 3a−4
≥ 0, so the critical value is
4
4
a = 3.
17. (a) Consider the equation y 00 + 2ay 0 + a2 y = 0. Show that the
roots of the characteristic equation are r1 = r2 = −a, so
that one solution of the equation is e−at .
(b) Use Abel’s formula to show that the Wronskian of any two
solutions of the given equation is
W (t) = y1 (t)y20 (t) − y10 (t)y2 (t) = c1 e−2at ,
where c1 is a constant.
(c) Let y1 (t) = e−at and use the result of part (b) to obtain
a differential equation satisfied by a second solution y2 (t).
By solving this equation, show that y2 (t) = te−at .
[§3.4 #20]
Sol.
(a) The characteristic equation is
r2 + 2ar + a2 = (r + a)2 = 0
Thus the root of characteristic equation are r1 = r2 = −a,
and thus e−at is a solution of the equation.
(b) By Abel’s formula, the Wronskian is
Z
W (t) = c1 exp − 2adt = c1 e−2at
for some constant c1 . Without loss of generality, we may
take c1 = 1.
(c) Let y1 (t) = e−at , then y10 (t) = −ae−at . By the definition of
the Wronskian and the result of (b), the second solution
y2 (t) must satisfy the differential equation
e−at y20 (t) + ae−at y2 (t) = W (t) = e−2at
That is,
y20 (t) + ay2 (t) = e−at
R
The integrating factor µ(t) = exp adt = eat , we obtain
that
Z
−at
y2 (t) = e
dt = te−at + c2 e−at
for some constant c2 , by taking c2 = 0, we get y2 (t) = te−at .
18. Use the method of reduction of order to find a second solution
of the differential equation
t2 y 00 − 4ty 0 + 6y = 0,
t > 0; , y1 (t) = t2
[§3.4 #24]
Sol. Let y2 (t) = t2 v(t). Substituting into the equation, we get
t2 (2v + 4tv 0 + t2 v 00 ) − 4t(2tv + t2 v 0 ) + 6t2 v = 0
That is, v 00 = 0, which implies that v(t) = at + b for some
constants a, b. Thus, y2 (t) = at3 + bt2 . By setting a = 1, b = 0,
we get y2 (t) = t3 .
19. Use the method of reduction of order to find a second solution
of the differential equation
xy 00 − y 0 + 4x3 y = 0,
x > 0; , y1 (x) = sin x2
[§3.4 #27]
Sol. Let y2 (x) = v(x) sin x2 = vy1 . Substituting into the equation, we get
x(v 00 y1 + 2v 0 y10 + vy100 ) − (v 0 y1 + vy10 ) + 4x3 vy1 = 0
That is,
xy1 v 00 +2xy10 v 0 −y1 v 0 +(xy100 −y10 +4x3 y1 )v = xy1 v 00 +(2xy10 −y1 )v 0 = 0
2y 0 Let w = v 0 , then we have that w0 = x1 − y11 w, which is a
separable equation. Thus
Z
Z dw
1 2y10 =
−
dx ⇒ ln |w| = ln x − 2 ln |y1 | + c0
w
x
y1
which implies that
w(x) = v 0 (x) =
c1 x
c1 x
=
, where c1 = ec0
2
y1
sin2 x2
By integrating w(x), we get
Z
Z
xdx
c1
c1
v(x) = c1
csc2 udu = − cot u + c2
2 2 =
2
2
sin x
2
c1 cos x
=−
+ c2
2 sin x2
for some constants c1 , c2 . Thus, y2 (t) = − c21 cos x2 + c2 sin x2 .
By setting c1 = −2, c2 = 0, we get y2 (t) = cos x2 .
20. Use the method of reduction of order to find a second solution
of the differential equation
(x − 1)y 00 − xy 0 + y = 0,
x > 1; , y1 (x) = ex
[§3.4 #29]
Sol. Let y2 (x) = v(x)ex = vy1 . Substituting into the equation,
we get
(x − 1)(v 00 y1 + 2v 0 y10 + vy100 ) − x(v 0 y1 + vy10 ) + vy1 = 0
That is,
(x − 1)y1 v 00 + (2(x − 1)y10 − xy1 )v 0 = 0
2y 0 x
Let w = v 0 , then we have that w0 = x−1
− y11 w, which is a
separable equation. Thus
Z Z
dw
x
2y10 =
−
dx ⇒ ln |w| = x+ln(x−1)−2 ln |y1 |+c0
w
x − 1 y1
which implies that
w(x) = v 0 (x) =
c1 (x − 1)ex
= c1 (x − 1)e−x , where c1 = ec0
y12
By integrating w(x), we get
Z
v(x) = c1 (x − 1)e−x dx = −c1 xe−x + c2
for some constants c1 , c2 . Thus, y2 (t) = −c1 x+c2 ex . By setting
c1 = −1, c2 = 0, we get y2 (t) = x.
21. The differential equation
y 00 + δ(xy 0 + y) = 0
arises in the study of the turbulent flow of a uniform stream
2
past a circular cylinder. Verify that y1 (x) = exp − δx2 is one
solution and then find the general solution in the form of an
integral.
[§3.4 #32]
Sol. Note that
y10 (x) = −δxe−
δx2
2
, y100 (x) = −δe−
δx2
2
+ δ 2 x2 e −
δx2
2
So
y100 +δ(xy10 +y1 ) = −δe−
δx2
2
+δ 2 x2 e−
δx2
2
−δ 2 x2 e−
δx2
2
+δe−
δx2
2
=0
which shows that y1 (x) is indeed a solution of the equation. Let
y2 (x) = y1 (x)v(x), substituting into the equation, we get
v 00 y1 + 2v 0 y10 + vy100 + δ x(v 0 y1 + vy10 ) + vy1 = 0
That is,
y1 v 00 + (2y10 + δxy1 )v 0 = 0
2y 0
Let w = v 0 , then we have that w0 = −( y11 + δx)w,
separable equation. Thus
Z 0
Z
2y1
dw
=−
+ δx dx ⇒ ln |w| = −2 ln |y1 | −
w
y1
which is a
δ 2
x + c0
2
for some constant c0 , which implies that
w(v) = v 0 (x) = c1 y1−2 e−
δx2
2
= c1 e
δx2
2
, where c1 = ec0
Integrating the above, we obtain
Z x 2
δs
v(x) = c1
e 2 ds
and thus
Z
2
− δx2
y2 (x) = e
c1
x
e
δs2
2
Z
ds = c1
x
e
δ(s2 −x2 )
2
ds
Therefore, the general solution for the equation is
Z x
δ(s2 −x2 )
δx2
y(x) = c1
e 2 ds + c2 e− 2
22. The method in above problem can be extended to second order
equations with variable coefficients. If y1 is a known nonvanishing solution of y00 +p(t)y 0 +q(t)y = 0, show that a second solution
0
y2 satisfies yy21 = W (yy12,y2 ) , where W (y1 , y2 ) is the Wronskian
1
of y1 and y2 . Then use Abel’s formula to determine y2 .
[§3.4 #33]
Sol. Suppose the y2 (t) is the second solution for the equation which linearly independent of y1 , then by definition of the
Wronskian,
y 0 y y 0 − y 0 y
W (y1 , y2 )
2
1 2
1 2
=
=
2
y1
y1
y12
According to Abel’s formula, we have that
Z t
p(s)ds
W (y1 , y2 )(t) = exp −
Hence
Rt
exp − p(s)ds
W (y1 , y2 )t
=
=
y1
y1 (t)2
y1 (t)2
Therefore, we can obtain y2 (t) by
Rs
Z t
exp − p(τ )dτ
ds
y2 (t) = y1 (t)
y1 (s)2
y 0
2
23. Find a second independent solution of the equation
(x − 1)y 00 − xy 0 + y = 0,
x > 1; , y1 (x) = ex
[§3.4 #34]
x
1
Sol. Rewrite the equation as the form y 00 − x−1
y 0 + x−1
y = 0.
According to above problem, the second independent solution
y2 (x) is
R s τ −1+1 Rs τ
Z x
Z x
exp
dτ
exp
dτ
τ −1
τ −1
y2 (x) = y1 (x)
ds = ex
ds
2
2s
y1 (s)
e
Rs
Z x
Z x
1
exp
1
+
dτ
exp
s
+
ln(s
−
1)
+
c
0
τ −1
= ex
ds = ex
ds
e2s
e2s
Z x
Z x
c1 (s − 1)es
x
x
=e
ds = c1 e
(s − 1)e−s ds, where c1 = ec0
e2s
= c1 ex − xe−x + c2 = −c1 x + c1 c2 ex
By taking c1 = −1, c2 = 0, we get y2 (x) = x.