HW 2 – Due Sep 5, Wed
1. Find the given integral.
Z
(a)
cos2 dx. Hint: use the double angle formula cos 2x = 2 cos2 x − 1
cos 2x + 1
cos2 x =
2
Z
cos 2x + 1
1 1
dx =
sin 2x + x + C
2
2 2
Z
cos x
. Then use the method of substitution.
(b)
cot xdx. Hint: write cot x =
sin x
Z
cos x
=
dx
sin x
Z
1
=
d sin x,
because cos xdx = d sin x
sin x
Z
1
du, (u = sin x)
=
u
= ln |u| + C = ln | sin x| + C
Z
(c)
sin x · cos xdx. (Hint: method of substitution)
Z
=
sin xd sin x,
(cos xdx = d sin x)
Z
=
=
Z
(d)
udu,
(u = sin x)
u2
sin2 x
+C =
+C
2
2
x
dx. (Hint: method of substitution)
1 + x2
1
Let u = 1 + x2 , du = 2xdx, xdx = du
2
Z
1 1
√ · du
=
u 2
Z
1
1 u1/2
−1/2
=
u
du = ·
+C
2
2
1/2
√
= u1/2 + C = 1 + x2 + C
√
1
Z
(e)
arcsin xdx.
Z
arcsin xdx = x · arcsin x −
Hint: by integration-by-part,
Z
Z
xd(arcsin x).
Z
arcsin xdx = x · arcsin x −
xd[arcsin x]
Z
= x · arcsin x −
x
√
dx
1 − x2
1
Let u = 1 − x2 , du = −2xdx, xdx = − du
2
Z
Z
√
√
x
1
√
dx = −
u−1/2 du = − u + C = − 1 − x2 + C
2
1 − x2
Z
√
arcsin xdx = x · arcsin x + 1 − x2 + C
1. Find the general solution of the given differential equation, and use it to determine how
solutions behave as t → ∞.
(a) y 0 + y = 5 sin 2t
p(t) = 1, g(t) = 5 sin 2t
Z
R
p(t)dt = t, µ(t) = e p(t)dt = et
Z
Z
µ(t) · g(t)dt = 5 et sin 2tdt
sin(2t) − 2 cos(2t)
· et = [sin(2t) − 2 cos(2t)]et
=5
5
[sin(2t) − 2 cos(2t)]et + C
y=
et
(b) (1 + t2 )y 0 + 4ty = (1 + t2 )−2
2
Divide (1 + t2 ) from both sides, ⇒ y 0 +
4t
,
p(t) =
1 + t2Z
Z
p(t)dt =
4t
y = (1 + t2 )−3
2
1+t
g(t) = (1 + t2 )−3
4t
dt
1 + t2
Let u = 1 + t2 , du = 2tdt, 4tdt = 2du
Z
1
· 2du = 2 ln |u| = 2 ln |1 + t2 |
=
u
R
2
µ(t) = e p(t)dt = e2 ln |1+t | = (1 + t2 )2
Z
Z
1
µ(t)g(t)dt =
dt = arctan t
1 + t2
arctan t + C
y=
1 + t2
(c) (sin t)y 0 + (cos t)y = et
cos t
et
y=
Divide sin x from both sides: y +
sin t
sin t
0
et
cos t
, g(t) =
p(t) =
sin t Z
sinZt
Z
cos t
1
p(t)dt =
dt =
d sin t = ln sin t
sin t
sin t
R
µ(t) = e p(t)dt = eln sin t = sin t
Z
Z
Z
et
dt = et dt = et + C
µ(t)g(t)dt = sin t ·
sin t
et + C
y=
sin t
2. Find the solution of the following initial value problem, and describe its behavior for
large t.
3
(a) y 0 + 3y = 3 + 10 cos t,
y(0) = 0.
p(t) = 3, g(t) = 3 + 10 cos t
Z
Z
p(t)dt = 3dt = 3t
R
µ(t) = e p(t)dt = e3t
Z
Z
Z
Z
3t
3t
µ(t)g(t)dt = e (3 + 10 cos t)dt = 3 e dt + 10 e3t cos tdt
sin t + 3 cos t
+ C = e3t + sin t + 3 cos t + C
12 + 32
e3t + sin t + 3 cos t + C
y=
e3t
= e3t + 10 ·
e0 + sin 0 + 3 cos 0 + C
=0
e0
1 + 3 + C = 0 ⇒ C = −4
e3t + sin t + 3 cos t − 4
y=
e3t
π
sin t
, y −
= 1, t < 0.
(b) ty 0 + 2y =
t
2
y(0) = 0,
⇒
2
y0 + y =
t
2
p(t) = ,
t
Z
sin t
t2
sin t
g(t) = 2
t
Z
2
p(t)dt =
dt = 2 ln t = ln t2
t
R
2
µ(t) = e p(t)dt = eln t = t2
Z
Z
sin t
µ(t)g(t)dt = t2 · 2 dt = − cos t + C
t
− cos t + C
y=
t2
π
− cos − π2 + C
= 1, ⇒
=1
y −
2
(π/2)2
π 2
⇒ C=
2
− cos t + (π/2)2
y=
t2
4
3. Solve the given differential equation.
(a) y 0 = (1 − 2x)y 2 ,
y0 =
1
y(0) = − .
6
dy
= (1 − 2x)y 2
dx
dy
= (1 − 2x)dx
y2
Z
Z
1
dy = (1 − 2x)dx
y2
1
− = x − x2 + C
y
1
1
y(0) = −
⇒ −
= 0 − 02 + C
6
−(1/6)
C=6
1
− = x − x2 + 6, implicit solution
y
1
y=−
, explicit solution
x − x2 + 6
(b) y 2 (1 − x2 )1/2 dy = arcsin xdx,
1
dx = d arcsin x.
y(0) = 1. Hint: √
1 − x2
arcsin x
y 2 dy = √
dx, ⇒ y 2 dy = arcsin xd arcsin x
2
1Z− x
Z
Z
2
y dy = arcsin xd arcsin x = udu, u = arcsin x
y3
(arcsin x)2
=
+C
3
2
(arcsin 1)2
13
=
+ C,
y(0) = 1 ⇒
3
2
1
π
= +C
3
4
1 π
C= −
3 4
arcsin(1) =
5
π
2