Math 3400
Homework 2
Section 2.1
# 13
y 0 − y = 2te2t ,
In this case
y(0) = 1.
t
Z
Rt
−1 ds) = e
µ(t) = exp(
−1 ds
= e−t .
The equation becomes
(e−t y)0 = 2tet ,
and integration by parts gives
−t
Z
e y=
t
2ses ds = 2tet − 2et + C.
Thus
y(t) = 2te2t − 2e2t + Cet .
The initial condition gives
y(0) = −2 + C = 1,
C = 3,
y(t) = 2te2t − 2e2t + 3et .
# 15
ty 0 + 2y = t2 − t + 1,
y(1) = 1/2.
Putting the equation in standard form gives
2
y 0 + y = t − 1 + 1/t.
t
In this case
Rt
µ(t) = e
2/s ds
2
= elog(t ) = t2 .
The equation becomes
(t2 y)0 = t3 − t2 + t,
so that
y(t) = t2 /4 − t/3 + 1/2 + Ct−2 .
The initial condition gives
y(1) = 1/4 − 1/3 + 1/2 + C = 1/2,
y(t) = t2 /4 − t/3 + 1/2 +
C = 1/12,
1 −2
t .
12
# 19
4
y 0 + y = e−t /t3 ,
t
In this case
Rt
µ(t) = e
y(−1) = 0.
4/s ds
1
= t4 .
The equation becomes
(t4 y)0 = te−t ,
so that after integrating by parts
y(t) = −t−3 e−t − t−4 e−t + Ct−4 .
The initial condition gives
y(−1) = C = 0, ,
y(t) = −e−t (1/t3 + 1/t4 ).
# 28
2
y 0 + y = 1 − t/2.
3
In this case
Rt
µ(t) = e
2/3 ds
= e2t/3 ,
and the equation may be rewritten as
d 2t/3
t
(e y) = e2t/3 − e2t/3 .
dt
2
Solving for y in terms of y0 we find
y(t) =
21 3t
− + (y0 − 21/8)e−2t/3 .
8
4
To find a solution which touches but does not cross the t−axis, we want a t0 such that
y(t0 ) = 0 and y 0 (t0 ) = 0. Going back to the differential equation, this gives
0 + 0 = 1 − t/2,
or t = 2. Then we find
0 = y(2) =
21
3
21
− 2 + (y0 − )e−4/3 .
8
4
8
Some algebra leads to
y0 =
21 9 4/3
− e ' −1.6429.
8
8
Section 2.2
# 2.
dy
x2
=
dx
y(1 + x3 )
may be rewritten as
−
x2
dy
+y
= 0.
3
1+x
dx
which has the form
M (x) + N (y)
2
dy
= 0.
dx
Integration gives
Z
x
−
or
s2
ds +
1 + s3
Z
y
s ds = C,
1
− log|1 + x3 | + y 2 /2 = C.
3
# 4.
dy
3x2 − 1
=
.
dx
3 + 2y
Try the ’slicker’ (equivalent) method
(3 + 2y)dy = (3x2 − 1)dx.
Integration gives
3y + y 2 = x3 − x + C.
# 8.
x2
dy
=
.
dx
1 + y2
Back to the original approach,
−x2 + (1 + y 2 )
dy
= 0.
dx
Integration gives
−x3 /3 + y + y 3 /3 = C.
# 12.
dr
r2
= ,
dθ
θ
With the basic approach, the equation is
r(1) = 2.
1 dr
1
= 0.
− + 2
θ r dθ
The general solution is
Z r
1
1
−
ds +
ds = C,
s
s2
and we can solve the initial value problem with
Z θ
Z r
1
1
−
ds +
ds = 0,
2
1 s
2 s
Z
θ
which gives
r
1 1
− log(θ) − s = − log(θ) + − = 0.
2 r
2
−1 Solving for r(θ) gives
r=
which is defined if 0 < θ <
√
1
2
=
,
1/2 − log(θ)
1 − 2 log(θ)
e.
3