Exercises 2
More exercises are available in Elementary Differential Equations. If you have a problem to solve any of
them, feel free to come to office hour.
1
Problem 1
Find the solution of the given initial value problem:
1. y 0 − y = 2te2t , y(0) = 1
2. ty 0 + 2y = t2 − t + 1, y(1) = 0.5, t > 0
3. y 0 − 2y = e2t , y(0) = 2
4. t3 y 0 + 4t2 y = e−t , y(−1) = 0, t < 0
1.1
Solution
1. First, we want to compute µ(t):
Z µ(t) = exp − dt ,
(1)
= exp (−t) .
Then, we need to compute
Rt
t0
e−t 2te2t dt, with t0 = 0:
Z t
Z t
t t
−t
2t
e 2te dt = 2 e t 0 − 2
et dt,
t0
0
= 2 et t − et + 1 .
(2)
Thus, the general solution is:
y(t) = et 2 et t − et + 1 + c .
(3)
Using the initial condition y(0) = 1, we have:
2(−1 + 1) + c = 1.
(4)
y(t) = 3et + 2et (t − 1).
(5)
Therefore, the solution is:
2. First, we rewrite the equation using the fact that t > 0:
1
2
y0 + y = t − 1 + .
t
t
(6)
Then, we compute µ(t):
Z
µ(t) = exp
2
dt ,
t
= exp (2 ln t)
= t2 .
1
(7)
Then, we need to compute
Rt
t2 t − 1 + 1t dt, with t0 = 1:
Z t Z
1
t2 t − 1 +
dt =
t3 − t2 + t dt,
t
t0
t0
4
t
t
t3
t2
=
− +
,
4
3
2 t0
t0
=
(8)
t4
t3
t2
5
− + − .
4
3
2
12
Thus, the general solution is:
1
y(t) = 2
t
t4
t3
t2
5
− + −
+c .
4
3
2
12
(9)
Using the initial condition y(1) = 0.5, we get:
c = 0.5.
(10)
Therefore, the solution is:
y(t) =
3t4 − 4t3 + 6t2 + 1
.
12t2
(11)
3. First, we compute µ:
dt ,
Z
µ(t) = exp −2
(12)
= exp (−2t) .
Then, we need to compute
Rt
t0
e−2t e2t dt, with t0 :
Z
t
e−2t e2t dt =
t0
Z
t
dt,
t0
(13)
= t.
Thus, the general solution is:
y(t) = e2t (t + c).
(14)
Using the initial condition y(0) = 2, we get:
c = 2.
(15)
y(t) = e2t (t + 2).
(16)
4
e−t
y0 + y = 3 .
t
t
(17)
Therefore, the solution is:
4. First, we rewrite the equation:
We compute µ:
Z
µ(t) = exp
4
dt ,
t
(18)
4
=t .
Now, we get:
Z
t
−1
t
t
tet dt = − te−t −1 − e−t −1 ,
= −e
2
−t
(t + 1) .
(19)
The general solution is:
y(t) =
−e−t (t + 1) + c
.
t4
(20)
Using the initial condition y(−1) = 0, we get:
c = 0.
(21)
−e−t (t + 1)
.
t4
(22)
Therefore, the solution is:
y(t) =
2
Problem 2
Solve the initial value problem and find the critical value a0 , such that the behavior of the solution changes
when t is large. Describe the behavior of the solution corresponding to the initial value a0 .
1. y 0 − 12 y = 2 cos t, y(0) = a
2. 3y 0 − 2y = e−πt/2 , y(0) = a
2.1
Solution
1. First, we compute µ:
Z 1
dt ,
µ(t) = exp −
2
t
= exp −
.
2
Then, we need to compute
Z
t
e
− 2t
Rt
t0
(23)
t
e− 2 2 cos t dt with t0 = 0:
h
it Z t
t
− 2t
2 cos t dt = −4 cos te
−
4 sin te− 2 dt,
0
0
= −4 cos te
− 2t
0
h
+ 4 + 8e
− 2t
sin t
it
Z
−8
0
(24)
t
e
− 2t
cos tdt.
0
Let’s rewrite equation (24), using:
Z
t
t
e− 2 2 cos t dt,
X=
(25)
0
we get successively:
t
t
X = 4 − 4 cos te− 2 + 8e− 2 sin t − 4X,
t
4 4
8 t
− cos te− 2 + e− 2 sin t.
5 5
5
After using the initial condition, we finally have:
t
4
4
8
y(t) =
+ a e 2 − cos t + sin t.
5
5
5
X=
(26)
(27)
(28)
The critical a is a0 = − 45 . For this value, y oscillates when t → ∞. Otherwise, y goes to infinity.
2. First, we need to rewrite the equation:
2
e−πt/2
y0 − y =
.
3
3
3
(29)
We compute µ:
2
µ(t) = e− 3 t .
(30)
Then, we get:
Z
t
2
e− 3 t e−
πt
2
dt = −
0
6 −t(2/3+π/2)
e
−1 .
4 + 3π
(31)
When t = 0, y = a:
−
6
c = a.
4 + 3π
(32)
Thus, finally get:
6
y(t) = −
4 + 3π
2
4 + 3π
t
t
−π
2
3
−
.
e
a+1 e
6
(33)
6
When a = − 4+3π
, y goes to 0.
3
Problem 3
Consider the initial value problem:
1
y 0 + y = 2 cos t, y(0) = −1.
2
Find the coordinates of the first local maximum point of the solution for t > 0.
3.1
(34)
Solution
First, we compute µ:
1
µ(t) = e 2 t .
(35)
Then, we compute:
Z
0
t
h t
it Z t t
1
e 2 t cos t dt = 4e 2 cos t +
4e 2 sin tdt,
0
0
Z t
t
t
t
2
2
= 4e cos t − 4 + 8e sin t − 8
e 2 cos tdt.
(36)
0
Let’s rewrite equation (36), using:
Z
t
t
e 2 2 cos t dt,
X=
(37)
0
we get successively:
t
t
X = 4e 2 cos t − 4 + 8e sin t − 4X,
2
4 t
4 8 t
X = e 2 cos t − + e 2 sin t.
5
5 5
(38)
(39)
Using the initial condition, we get:
4
8
9 t
cos t + sin t − e− 2 .
5
5
5
Now, we to compute the first local maximum, we need to compute:
y(t) =
(40)
dy
= 0,
dt
(41)
t
4
8
9
= − sin t + cos t + e− 2 .
5
5
10
Solving this equation (use Matlab) gives:
t = 1.364312,
(42)
y = 0.820082.
(43)
4
4
Problem 4
Find the value of y0 for which the solution of the initial value problem:
y 0 − y = 1 + 3 sin t, y(0) = y0
(44)
µ = e−t .
(45)
remains finite as t → ∞.
4.1
Solution
First, we compute µ:
Then, we have:
Z
t
e−t (1 + 3 sin t) dt =
0
Z
t
e−t dt + 3
0
Z
0
t
Z
t
e−t sin tdt.
(46)
0
t
e−t dt = −e−t 0 ,
(47)
= 1 − e−t .
Z t
Z t
−t
t
−t
e sin tdt = −e sin t 0 +
e−t cos tdt,
0
0
Z t
−t
t
−t
= −e sin t − e cos t 0 −
e−t sin tdt,
(48)
0
Using the same method than before, we get:
Z t
1
−e−t sin t − e−t cos t + 1 .
e−t sin tdt =
2
0
(49)
Thus, we finally have:
3
3
y(t) = − sin t − cos t +
2
2
5
+ a et − 1.
2
(50)
We want y to stay finite → we need a = − 52 . Therefore, we need to have:
3
5
y(0) = y0 = − − 1 = − .
2
2
5
(51)