Exercises 4

Exercises 4
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1
Problem 1
Determine (without solving the problem) an interval in which the solution of the given initial value problem
us certain to exist:
1. (t − 3)y 0 + (ln t)y = 2t, y(1) = 2
2. y 0 + (tan t)y = sin t, y(π) = 0
3. (4 − t2 )y 0 + 2ty = 3t2 , y(1) = −3
1.1
Solution
1. We need to rewrite:
(t − 3)y 0 + (ln t)y = 2t,
as
y0 +
2t
ln t
y=
.
t−3
t−3
(1)
(2)
Thus, we have 0 < t < 3.
2.
π
2
<t<
3π
2 .
3. We need to rewrite:
(4 − t2 )y 0 + 2ty = 3t2 ,
(3)
32
2t
y
=
.
4 − t2
4 − t2
(4)
as
y0 +
Thus, we have −2 < t < 2.
2
Problem 2
State where in the ty-plane the hypotheses of Theorem 2 are satisfied:
1. y 0 =
t−y
2t+5y
2. y 0 =
ln |ty|
1−t2 +y 2
3.
dy
dt
=
1+t2
3y−y 2
1
2.1
Solution
• f and
∂f
∂y
to be continuous ⇒ 2t + 5y < 0 or 2t + 5y > 0.
• y 6= 0 and t 6= 0 because of ln |ty| and 1 − t2 + y 2 > 0 or 1 − t2 + y 2 < 0.
• y 6= 0, y 6= 3.
3
Problem 3
Solve the given initial value problem and determine how the interval in which the solution exists depends on
the initial value y0 :
1. y 0 = −4 yt , y(0) = y0
2. y 0 + y 3 = 0, y(0) = y0
3.1
Solution
1. We have successively, if y0 6= 0:
t
y 0 = −4 ,
y
ydy = −4dt,
(5)
(6)
2
y
= −2t2 + c.
2
(7)
y02
c.
=
(8)
q
y = ± y02 − 2t2 ,
(9)
Using the initial condition, we can find c:
Thus, we finally get:
with |t| <
|y0 |
2 .
2. y = 0 if y0 = 0 ⇒ −∞ < t < ∞. If y0 6= 0, we have successively:
dy
+ y 3 = 0,
dt
dy
− 3 = dt,
y
1 −2
− y = t + c,
2
1
y2 =
.
2t + k
(10)
(11)
(12)
(13)
Using the initial condition, we get:
y02 =
1
.
k
(14)
Thus, we finally have:
y2 =
1
2t +
1
y02
,
(15)
y02
=
,
2ty02 + 1
,
y = ±p
Therefore, we have − 2y12 < t < ∞
0
2
y0
2ty02 + 1
.
(16)
4
Problem 4
Show that if y = φ(t) is a solution of y 0 + p(t)y = 0, then y = cφ(t) is a also a solution for any value of the
constant c.
4.1
Solution
We have:
(cφ(t))0 + p(t)(cφ(t)) = cφ(t)0 + cp(t)φ(t),
= c (φ(t)0 + p(t)φ(t)) ,
(17)
= 0.
Thus, cφ(t) is a solution of y 0 + p(t)y = 0.
5
Problem 5
Solve the initial value problem:
y 0 + p(t)y = 0,
(18)
y(0) = 1,
(19)
where
(
p(t) =
5.1
2, 0 ≤ t ≤ 1,
1,
t > 1.
(20)
Solution
We have basically two initial value problems. The initial condition for the second problem is obtained by
forcing the continuity of the solution. First, we need to solve:
y 0 + 2y = 0,
(21)
y = c1 e−2t .
(22)
y = e−2t .
(23)
y 0 + y = 0,
(24)
Using the initial condition y(0) = 1, we get:
Now, we need to solve:
−t
y = c2 e .
(25)
Using the continuity of the solution, we have the following condition:
y(1) = e−2 .
(26)
c2 e−1 = e−2 ,
(27)
c = e−1 .
(28)
y = e−(t+1) .
(29)
Thus, we get:
Thus, we finally have:
So the final solution is:
(
y(t) = e−2t ,
y(t) = e
0 ≤ t ≤ 1,
−(t+1)
3
,
t > 1.
(30)
6
Problem 6
The next equations involve equation of the form dy
dt = f (y). Sketch the graph of f (y) versus y, determine
the critical (equilibrium) points, and classify each one as asymptotically stable of unstable:
1.
dy
dt
= y(y − 1)(y − 2), y0 ≥ 0
2.
dy
dt
= e−y − 1, −∞ < y0 < ∞
6.1
Solution
1. y = 1 is asymptotically stable, y = 0 and y = 2 are asymptotically unstable.
2. y = 0 is asymptotically stable.
7
Problem 7
Another equation that has been used to model population growth is the Gompertz equation:
K
dy
= ry ln
,
dt
y
(31)
where r and K are positive constants.
1. Sketch the graph of f (y) versus y, find the critical points, and determine whether each is asymptotically
stable or unstable.
2. For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is concave down.
3. For each y in 0 < y ≤ K, show that
given by the logistic equation.
7.1
dy
dt
as given by the Gompertz equation is never less than
dy
dt
as
Solution
1. y = 0 is unstable, y = K is asymptotically stable.
2. Concave up for 0 < y ≤ K/e, concave down for K/e ≤ y < K.
3. We want to show that:
K
y
y
≥r 1−
y,
K
y
K
≥ 1−
,
ln
y
K
ry ln
where 0 < y ≤ K. Let’s use x =
y
K,
with 0 < x ≤ 1:
1
ln
≥ (1 − x),
x
− ln x ≥ (1 − x).
(32)
(33)
(34)
(35)
We know that for |x − 1| ≤ 1, we have:
ln x = (x − 1) −
(x − 1)2
(x − 1)3
(x − 1)4
+
−
− ...
2
3
4
(36)
Thus, we have:
− ln x = (1 − x) +
(x − 1)2
(x − 1)3
(x − 1)4
−
+
+ . . . ≥ (1 − x).
2
3
4
4
(37)
8
Problem 8
Determine whether each of the equations is exact. If it is exact, find the solution.
1. (2x + 3) + (2y − 2)y 0 = 0
2. (3x2 − 2xy + 2)dx + (6y 2 − x2 + 3)dy = 0
3. (ex sin y − 2y sin x) dx + (ex cos y + 2 cos x) dy = 0
4. (yexy cos(2x) − 2exy sin(2x) + 2x) dx + (xexy cos(2x) − 3) dy = 0
8.1
Solution
1. We need to check that
∂(2x+3)
∂y
=
∂(2y−2)
:
∂x
∂(2x + 3)
= 0,
∂y
∂(2y − 2)
= 0.
∂x
Thus, the equation is exact. Now, we can solve the problem:
Z
(2x + 3)dx = x2 + 3x + h(y),
∂(x2 + 3x + h(y))
= h0 (y) = 2y − 2,
∂y
Z
(2y − 2)dy = y 2 − 2y − c.
(38)
(39)
(40)
(41)
(42)
So the solution is:
y 2 − 2y + x2 + 3x = c.
(43)
2. We need to compute:
∂(3x2 − 2xy + 2)
= −2x,
∂y
∂(6y 2 − x2 + 3)
= −2x.
∂x
Thus, the equation is exact. Now, we can solve the problem:
Z
3x2 − 2xy + 2 dx = x3 − x2 y + 2x + h(y),
∂ x3 − x2 y + 2x + h(y)
= −x2 + h0 (y) = 6y 2 − x2 + 3.
∂y
Z
6y 2 + 3 dy = 2y 3 + 3y − c.
(44)
(45)
(46)
(47)
(48)
So, the solution is:
2y 3 + 3y + x3 − x2 y + 2x = c.
5
(49)
3. We need to compute:
∂ (ex sin y − 2y sin x)
= ex cos y − 2 sin x,
∂y
∂ (ex cos y + 2 cos x)
= ex cos y − 2 sin x.
∂x
Thus, the equation is exact. Now, we can solve the problem:
Z
(ex sin y − 2y sin x) dx = ex sin y + 2y cos x + h(y),
∂ (ex sin y + 2y cos x + h(y))
= ex cos y + 2 cos x + h0 (y) = ex cos y + 2 cos x.
∂y
Z
0dy = −c.
(50)
(51)
(52)
(53)
(54)
So, the solution is:
ex sin y + 2y cos x = c.
(55)
4. We need to compute:
∂ (yexy cos(2x) − 2exy sin(2x) + 2x)
= exy cos(2x) + xyexy cos(2x) − 2xexy sin(2x),
∂y
∂ (xexy cos(2x) − 3)
= exy cos(2x) + xyexy cos(2x) − 2xexy sin(2x).
∂x
Thus, the equation is exact. Now, we can solve the problem:
Z
Z
(yexy cos(2x) − 2exy sin(2x) + 2x) dx = (yexy cos(2x) − 2exy sin(2x)) dx + x2 ,
Z
yexy cos(2x)dx = exy cos(2x) +
Z
2 sin(2x)exy dx,
(56)
(57)
(58)
(59)
Thus, we have:
Z
(yexy cos(2x) − 2exy sin(2x)) dx = exy cos(2x).
(60)
(yexy cos(2x) − 2exy sin(2x) + 2x) dx = exy cos(2x) + x2 + h(y).
(61)
So, we finally get:
Z
∂(exy cos(2x) + x2 + h(y)
= xexy cos(2x) + h0 (y) = xexy cos(2x) − 3.
∂y
Z
− 3dy = −3y − c.
(62)
(63)
So, the solution is:
exy cos(2x) + x2 − 3y = c.
9
(64)
Problem 9
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor.
Then, solve the equation.
1. x2 y 3 + x(1 + y 2 )y 0 = 0, µ(x, y) = xy1 3
−x
2. siny y − 2e−x sin x dx + cos y+2ey cos x dy = 0, µ(x, y) = yex
6
9.1
Solution
1. First, we should notice that y = 0 is a solution. If y 6= 0, we have:
∂ x 1 + y2
∂ x2 y 3
2 2
= 3x y ,
= 1 + y2 .
∂y
∂x
and thus, the equation is not exact. If we use
1
xy 3
(65)
as integrating factor, we get:
∂ x2 y 3
= 0,
∂y xy 3
∂ x(1 + y 2 )
= 0.
∂x xy 3
Thus, the new equation is exact. Now, we can solve the problem:
Z 2 3
x y
x2
dx
=
+ h(y).
xy 3
2
2
∂ x2 + h(y)
1 + y2
= h0 (y) =
.
∂y
y3
Z 1
1
1
+
dy = − 2 + ln y − C
3
y
y
2y
(66)
(67)
(68)
(69)
(70)
So, the solution is:
1
x2
− 2 + ln y = c.
2
2y
(71)
2. We have:
∂(y −1 sin y − 2e−x sin x)
= −y −2 sin y + y −1 cos y,
∂y
∂ cos y + 2e−x cos x
2 −x
e cos x − e−x sin x .
=
∂x
y
y
(72)
(73)
Thus, the equation is not exact. If we use yex as integrating factor, we get:
∂
y −1 sin y − 2e−x sin x yex = ex cos y − 2 sin x,
∂y
∂
(cos yex + 2 cos x) = ex cos y − 2 sin x.
∂x
Thus, the new equation is exact. Now, we can solve the problem:
Z
(ex sin y − 2y sin x) dx = ex sin y + 2y cos x + h(y).
∂ x
(e sin y + 2y cos x + h(y)) = ex cos y + 2 cos x + h0 (y) = ex cos y + 2 cos x.
∂y
Z
0.dx = −c.
(74)
(75)
(76)
(77)
(78)
So, the solution is:
ex sin y + 2y cos x = c.
7
(79)
10
Problem 10
Find an integrating factor and solve the given equation.
1. 3x2 y + 2xy + y 3 dx + x2 + y 2 dy = 0
2. dx + xy − sin y dy = 0
10.1
Solution
1. We look for µ such that µ only depends on x. Thus, we have:
dµ
M y − Nx
=
µ.
dx
N
(80)
dµ
3x2 + 2x + 3y 2 − 2x
µ,
=
dx
x2 + y 2
= 3µ.
(81)
Here, we get:
Therefore, µ = e3x . Let’s check that the new equation is exact:
∂
∂y
∂
∂x
3x2 y + 2xy + y 3 e3x = 3x2 e3x + 2xe3x + 3y 2 e3x ,
(82)
x2 + y 2 e3x = 3x2 e3x + 2xe3x + 3y 2 e3x .
(83)
Thus, the equation is exact. Now, we can solve the problem:
Z
y3
3x2 ye3x + 2xye3x + y 3 e3x dx = yx2 e3x + e3x + h(y),
3
where we have used:
Z
1 3 3x
y e ,
3
Z
Z
2ye3x
2y 3x
2xye3x dx =
−
e dx,
3
3
2y 3x 2y 3x
=
e − e ,
3
9
Z
y 3 e3x dx =
(84)
(85)
(86)
3x2 ye3x dx = x2 ye3x − 2xye3x dx,
= x2 ye3x −
2y 3x 2 3x
e − ye .
3
9
Now, we can get h(y):
y3
∂
yx2 e3x + e3x + h(y) = x2 e3x + y 2 e3x + h0 (y) = x2 e3x + y 2 e3x .
∂y
3
Z
0dy = −c.
(87)
(88)
(89)
Thus, the solution of the problem is:
y3
e3x x2 y +
= c.
3
8
(90)
2. We look for µ such that µ only depends on y. Thus, we have:
dµ
Nx − M y
=
µ.
dx
M
(91)
dµ
µ
= .
dy
y
(92)
Here, we get:
Therefore, µ = y. Let’s check that the new equation is exact:
∂
(x − y sin y) = 1.
∂x
∂y
= 1,
∂y
Thus, the equation is exact. Now, we can solve the problem:
Z
ydx = yx + h(y).
∂yx + h(y)
= x + h0 (y) = x − y sin y.
∂y
(93)
(94)
(95)
Thus, h is given by:
Z
h(y) = −
y sin ydy,
Z
= y cos y − cos ydy,
(96)
= y cos y − sin y − c.
Thus, the solution of the problem is:
yx + y cos y − sin y = c.
9
(97)

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