Document 6536554

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Document 6536554

Sample Solutions of Assignment 3 for MAT3270A:2.8-3.2
Note: Any problems to the sample solutions, email Mr. Yao Xiao
a
(yxiaomath.cuhk.edu.hk)
directly.
September,2013
1. Determine if the following equation is exact. If it is exact, find out
the solutions
0
0
(a). (x4 + 4y) + (4x − 3y 8 )y = 0
(b). (x + y) + (2x − y)y = 0
0
(c). (2xy 2 + 2y) + (2x2 y + 2x)y = 0 (d). (x log y + xy)dx + (y log x +
xy)dy = 0
(e). (x2 +yx2 )3/2 dx +
ydy
3
(x2 +y 2 ) 2
=0
Answer: (a). Since ∂y (x4 +4y) = 4 and ∂x (4x−3y 8 ) = 4, the equation
1
1
is exact. And x5 + 4xy − y 9 = C is the first integral.
5
3
(b). Since ∂y (x + y) = 1 and ∂x (2x − y) = 2, the equation is not
exact.
(c). Since ∂y (2xy 2 + 2y) = 4xy + 2 and ∂x (2x2 y + 2x) = 4xy + 2, the
equation is exact. And x2 y 2 + 2xy = C is the first integral.
y
x
(d). Since ∂y (x log y + xy) = + x and ∂x (y log x + xy) = + y, the
y
x
equation is not exact.
3
2xy
3
2xy
(e). Since ∂y ( (x2 +yx2 )3/2 ) = −
and ∂x ( 2 y 2 3 ) = −
,
2
2
5/2
2
2
(x
+y
)
2 (x + y )
2 (x + y 2 )5/2
1
= C is the first integral.
the equation is exact. And − 2
(x + y 2 )1/2
2. Using the integrating factors given below to solve the corresponding ODEs.
0
1
xy 3
cos y+2e−x cos x
dy
y
(a). x2 y 3 + x(1 + y 2 )y = 0, µ(x, y) =
(b). ( siny y − 2e−x sin x)dx +
2
= 0, µ(x, y) = yex
(c). (3x + y6 )dx + ( xy + 2 xy )dy = 0, µ(x, y) = xy
1
2
Answer: (a). Multiply µ(x, y) =
1
xy 3
on both sides of equation, one
has
1 + y2 0
y =0
y3
2
1
d( x2 − 2 + ln y) = 0
2
y
2
1 2
x − 2 + ln y = C
2
y
x+
⇒
⇒
(b). Multiply µ(x, y) = yex on both sides of equation, one has
(ex sin y − 2ysin x)dx + (ex cos y + 2 cos x)dy = 0
⇒
d(ex sin y + 2y cos x) = 0
⇒
ex sin y + 2y cos x = C
(c). Multiply µ(x, y) = xy on both sides of equation, one has
⇒
⇒
(3x2 y + 6x)dx + (x3 + 2y 2 )dy = 0
2
d(x3 y + 3x2 + y 3 ) = 0
3
2
x3 y + 3x2 + y 3 = C
3
3. Find out the integrating factors of the following ODEs and solve
the corresponding ODEs
(a). (3x2 y+2xy+y 3 )dx+(x2 +y 2 )dy = 0
(c). ydx+(2xy−e−2y )dy = 0
(b). dx+( xy −sin y)dy = 0
(d). ex dx+(ex cot y+2y csc y)dy = 0
My − Nx
Answer: (a). Since My −Nx = 3x2 +3y 2 = 3N , µx =
µ = 3µ.
N
Hence we can choose µ = e3x . Multiple µ on both side of ODE, we
3
have
e3x (3x2 y + 2xy + y 3 )dx + e3x (x2 + y 2 )dy = 0
1
⇒
d(e3x (x2 y + y 3 )) = 0
3
1
⇒
e3x (x2 y + y 3 ) = C
3
1
1
My − Nx
1
(b). Since My − Nx = − = − M , µy = −
µ = µ. Hence
y
y
M
y
we can choose µ = y. Multiple µ on both side of ODE, we have
ydx + (x − y sin y)dy = 0
⇒
d(xy − sin y + y cos y) = 0
⇒
xy − sin y + y cos y = C
2y − 1
My − Nx
M , µy = −
µ=
y
M
2y − 1
e2y
µ. Hence we can choose µ =
. Multiple µ on both side of
y
y
ODE, we have
(c). Since My − Nx = 1 − 2y = −
⇒
e2y
e2y
)ydx + ( )(2xy − e−2y )dy = 0
y
y
1
e2y dx + (2xe2y − )dy = 0
y
2y
d(e x − ln y) = 0
⇒
e2y x − ln y = C
(
⇒
My − Nx
µ=
M
cot yµ. Hence we can choose µ = sin y. Multiple µ on both side of
(d). Since My − Nx = −ex cot y = − cot yM , µy = −
ODE, we have
ex sin ydx + (ex cos y + 2y)dy = 0
⇒
d(ex sin y + y 2 ) = 0
⇒
ex sin y + y 2 = C
4
4. Solve the following ODEs
0
(a). y =
0
2x+y
, y(0)
3+3y 2 −x
2 2x
(c). xy + y − y e
=0
0
(b). y =
= 0, y(0) = 0
(e). xdy − ydx = 2x2 y 2 dy, y(1) = −2
0
(g). y =
y3
, y(0)
1−2xy 2
=1
y
t−y
0
(d). y = ex+y
(f).
dy
dx
(h). (2y + 1)dx +
= − 2xy+1
x2 +2y
x2 −y
dy
x
=0
Answer: (a). Original ODE can be written as
(2x + y)dx + (x − 3y 2 − 3)dy = 0
Since My − Nx = 0, the equation is exact. Then we have
d(x2 + xy − y 3 − 3y) = 0
⇒
x2 + xy − y 3 − 3y = y 3 (0) − 3y(0) = 0
(b). Original ODE can be written as
ydt + (y − t)dy = 0
Since My − Nt = 2 =
choose µ =
2
My − Nt
2
M ,µy = −
µ = − µ. Hence we can
y
M
y
1
. Multiple µ on both side of ODE, we have
y2
⇒
⇒
(c). Multiply
1
t
dt + (1 − 2 )dy = 0
y
y
t
d(ln y + ) = 0
y
t
ln y + = C
y
1
on both sides of original equation, we have
x2
y − y 2 e2x
1
dx
+
dy = 0
x2
x
1 − 2ye2x 1
2 − 2ye2x
Since My −Nx =
+ 2 =
=
x2
x
x2
2
1
− µ. Hence we can choose µ = 2 . Multiple
y
y
(1)
2
My − Nx
M ,µy = −
µ=
y
M
µ on both side of ODE
5
(1), we have
(
⇒
⇒
1
e2x
1
−
)dx
+
dy = 0
x2 y
x2
xy 2
R e2x
1
d( +
dx) = 0
xy
x2
R e2x
1
+
dx = C
xy
x2
Thus original ODE can’t have a nontrivial solution such that y(x =
0) = 0. Hence y ≡ 0 is the only solution.
(d). Original ODE can be written as
ex+y dx − dy = 0
My − Nx
µ = −µ. Hence we can
M
choose µ = e−y . Multiple µ on both side of ODE, we have
Since My − Nx = ex+y = M ,µy = −
ex dx − e−y dy = 0
⇒
d(ex + e−y ) = 0
⇒
ex + e−y = C
(e). Original ODE can be written as
ydx + (2x2 y 2 − x)dy = 0
My − Nx
2
2
µ = − µ. Hence
Since My − Nx = 2 − 4xy 2 = − N ,µx =
x
N
x
1
we can choose µ = 2 . Multiple µ on both side of ODE, we have
x
y
1
dx + (2y 2 − )dy = 0
2
x
x
2 3 y
⇒
d( y − ) = 0
3
x
2 3 y
2 3
10
⇒
y − = y (1) − y(1) = −
3
x
3
3
(f). Original ODE can be written as
(2xy + 1)dx + (x2 + 2y)dy = 0
6
Since My − Nx = 0, the equation is exact. Then we have
(2xy + 1)dx + (x2 + 2y)dy = 0
⇒
d(x2 y + x + y 2 ) = 0
⇒
x2 y + x + y 2 = C
(g). Original ODE can be written as
y 3 dx + (2xy 2 − 1)dy = 0
Since My − Nx = y 2 =
1
My − Nx
1
M ,µy = −
µ = − µ. Hence we can
y
M
y
1
choose µ = . Multiple µ on both side of ODE, we have
y
1
y 2 dx + (2xy − )dy = 0
y
2
⇒
d(xy − ln y) = 0
⇒
xy 2 − ln y = − ln y(0) = 0
(h). Original ODE can be written as
y
(2y + 1)dx + (x − )dy = 0
x
1
My − Nx
1
y
µ = µ. Hence we can
Since My − Nx = 1 − 2 = N ,µx =
x
x
N
x
choose µ = x. Multiple µ on both side of ODE, we have
⇒
⇒
(2xy + x)dx + (x2 − y)dy = 0
1
1
d(x2 y + x2 − y 2 ) = 0
2
2
1 2 1 2
2
x y+ x − y =C
2
2
5. Transform the given initial value problem into an equivalent problem with the initial point at the origin
(a).
dy
= t2 + y 2 , y(1) = 2,
dt
(b).
dy
= 1 − y 3 , y(−1) = 3
dt
7
Answer: (a)Let t = s + 1, y = w + 2, then
dy
=1
dw
dt
=1
ds
the original problem can be written as
dw
= (s + 1)2 + (w + 2), w(0) = 0.
ds
(b)Let t = s − 1, y = w + 3, then
dy
=1
dw
dt
=1
ds
the original problem can be written as
dw
= 1 − (w + 3)2 , w(0) = 0.
ds
6. Use the method of successive approximations to solve the given
initial value problem: (1) Determine φn (t); (2) Find the limit of {φn }.
0
0
(a).y = 2(y + 1), y(0) = 0, (b)y = y + 1 − t, y(0) = 0.
Answer:
(a) If y = φ(t), then the corresponding integral equation is
Z t
φ(t) =
2(φ(s) + 1)ds
0
If the initial approximation is φ0 (t) = 0, then
Z t
φ1 (t) =
2ds = 2t
0
Z
φ2 (t) =
t
2(2s + 1)ds = 2t + 2t2
0
and
Z
φn (t) =
t
2(φn−1 (s) + 1)ds =
0
n
X
2k tk
1
k!
8
lim φn (t) = e2t − 1
n→∞
(b) If y = φ(t), then the corresponding integral equation is
Z t
φ(t) =
(φ(s) + 1 − s)ds
0
If the initial approximation is φ0 (t) = 0, then
Z t
1
(1 − s)ds = t − t2
φ1 (t) =
2!
0
Z t
1
1
2(1 − s2 )ds = t − t3
φ2 (t) =
2
3!
0
and
Z t
tn+1
2(φn−1 (s) + 1 − s)ds = t −
φn (t) =
(n + 1)!
0
lim φn (t) = t
n→∞
Problem 2,4,5 on page 155
Find the Wronskian of the given pair of functions.
(2). cos t, sin t
(4). x, ee2x s
(5). et sin t, e2t cos t
Answer: The computation is easy, so we just give the final result.
(2). W = 1
(4). W = 2x2 e2x
(5). W = −e4t
Problem 7,9,12 on page 155-156
In the following problems determine the longest interval in which the
given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution.
9
7.ty 00 + 3y = t,
y(2) = 1,
y 0 (2) = 2.
9. t(t − 4)y 00 + 3ty 0 + 5y = 2, y(3) = 0, y 0 (3) = −1
12. (x − 2)y 00 + y 0 + (x − 2)(tan x)y = 0, y(4) = 1, y 0 (4) = 2
Answer: (7). The original solution can written as
3
y 00 + y = 1,
t
hence the longest interval is t > 0. Then the points of discontinuity
of the coefficients are t = 0. Therefore, the longest open interval,
containing the initial point t = 1, in which all the coefficients are
continuous, is 0 < t < ∞.
(9).The original solution can written as
y 00 +
5
2
3t
y0 +
y=
t(t − 4)
t(t − 4)
t(t − 4)
3t
5
2
, q(t) =
, g(t) =
. Then the points
t(t − 4)
t(t − 4)
t(t − 4)
of discontinuity of the coefficients are t = 0 and t = 4. Therefore, the
and p(t) =
longest open interval, containing the initial point t = 3, in which all
the coefficients are continuous, is 0 < t < 4.
(12). The original solution can written as
y 00 +
y0
+ (tan x)y = 0
x−2
Then the only points of discontinuity of the coefficients are x = 2, and x =
1
kπ + π, k ∈ Z. Therefore, the longest open interval, containing the
2
initial point x = 4, in which all the coefficients are continuous, is
3
2 < x < π.
2
Problem14 on page 156
1
Verify that y1 (t) = 1 and y2 (t) = t 2 are solutions of the differential
00
0
1
equation yy + (y )2 = 0 for t > 0. Then show that c1 + c2 t 2 is not, in
10
general, a solution of this equation. Explain why this result does not
contradict Theorem 3.2.2.
Answer: It is easy to verify y1 and y2 are solutions of the differential
00
1
0
equation yy + (y )2 = 0 for t > 0, and y = c1 + c2 t 2 is not a solution(in
general) of this equation.
This result does not contradict Theorem 3.2.2 because this equation is
nonlinear.
Problem 15 on page 156
00
Show that if y = φ(t) is a solution of the differential equation y +
0
p(t)y + q(t)y = g(t), where g(t) is not always zero, the y = cφ(t),
where c is any constant other than 1, is not a solution. Explain why
this result does not contradict the remark following Theorem 3.2.2.
Answer:
00
0
[cφ(t)] + p(t)[cφ(t)] + q(t)[cφ(t)]
00
0
= c[φ(t) + p(t)φ(t) + q(t)φ(t)]
= cg(t) 6= g(t)
if c is a constant other than 1, and g(t) is not always zero.
This result does not contradict Theorem 3.2.2 because this equation is
not homogeneous.
Can y = sin t2 be a solution on an interval containing t = 0 of an
equation y 00 + p(t)y 0 + q(t)y = 0 with continuous coefficients? Explain
your answer.
Answer: By direct computing, we have
y = sin t2 , y 0 = 2t cos t2 , y 00 = −4t2 sin t2 + 2 cos t2
we have
(−4t2 + q(t)) sin t2 + (2tp(t) + 2) cos t2 = 0
(∗)
11
Assume that y = sin t2 is a solution on an interval containing t = 0 and
coefficients are continuous, substitute t = 0 into (*), we get 2 cos 0 = 0
which is not true. So y = sin t2 can not satisfy all the requirements.
If the Wronskian W of f and g is 2e6t , and if f (t) = e3t , find g(t).
Answer:
0
0
0
W = f (t)g (t) − f (t)g(t) = e3t g (t) − 3e3t g(t)
Let W = 2e6t , we get the following equation
0
g − 3g(t) = 2e3t .
From the above equation, g(t) = 2te3t + ce3t .
Proble 18 on page 156
If the Wronskian W of f and g is 2t2 et , and if f (t) = t, find g(t).
Answer:
0
W = f (t)g 0 (t) − f 0 (t)g(t) = tg (t) − g(t)
Let W = 2t2 et , we get the following equation
0
tg − g(t) = 2t2 et .
From the above equation, g(t) = 2tet + Ct, where C be a constant.
If the Wronskian of f and g is t cos t−sin t and if u = f +3g, v = f −2g,
find the Wronskian of u and v.
12
Answer:
0
0
= uv − u v
W (u, v)
0
0
0
0
= (f + 3g)(f − 2g ) − (f + 3g )(f − 2g)
0
0
= −5f g + 5f g
= −5W (f, g)
= −5(t cos t − sin t).
Answer:
W (y3 , y4 ) = y3 y40 − y30 y4
= (a1 y1 + a2 y2 )(b1 y10 + b2 y20 ) − (a1 y10 + a2 y20 )(b1 y1 + b2 y2 )
= (a1 b2 − a2 b1 )(y1 y20 − y10 y2 )
= (a1 b2 − a2 b1 )W (y1 , y2 )
Find the fundamental set of solutions specified by Thm 3.2.5 for the
given differential equation and initial point.
y 00 + 5y 0 + 4y = 0,
t0 = 1.
Answer: The characteristic equation is
r2 + 5r + 4 = 0,
we have r = −1 and r = −4. The solution is y1 (t) = e−t , y2 (t) = e−4t .
By computation, we have W (t0 ) = −3e−5 6= 0.
So the set of fundamental solutions is {e−t , e−4t }
13
Answer:
y100 − 4y10 + 4y1 = 4e2t − 8e2t + 4e2t = 0
y200 − 4y20 + 4y2 = 4e2t + 4te2t − 8te2t − 4e2t + 4te2t = 0
Heace y1 y2 are the solutions of the ODE.
W (y1 , y2 ) = y1 y20 − y10 y2
= e4t
6= 0
Hence they constitute the fundamental set of solutions.
Answer:
(a) Since the Wronskian W = 3et , they form a fundamental set of
solutions.
(b) Since they are all linear combinations of y1 and y2 , they are also
solutions of the given equation.
(c) By calculating the respective Wronskians, we see that [y1 , y3 ] and
[y1 , y4 ] [y4 , y5 ] are fundamental set of solutions while the others are not.
Answer:
14
1
v2
Divide by x2 on both sides: y 00 + y 0 + (1 − 2 )y = 0
x
x
By Abel’s theorem:
Z
1
dx]
W (y1 , y2 ) = c exp[−
x
= c exp[− ln |x|]
c
=
|x|
Answer: Convert the ODE into:
y 00 +
p0 q
+ y=0
p
p
By Abel’s Theorem:
p0
dt]
p
= c exp[− ln p(t)]
Z
W (y1 , y2 )(t) = c exp[−
= c/p(t)
Answer:
2
y 00 + y 0 + et y = 0
t
Z
2
dt]
t
= c exp[−2 ln t]
c
= 2
t
W (y1 , y2 )(t) = c exp[−
Since W (y1 , y2 )(1) = c = 3, W (y1 , y2 )(5) =
3
.
25
15
Answer: Since W (y1 , y2 )(t) = c exp −
R
p(t)dt = c0 ,thus p(t) = 0 for
all t.
Answer: Since W (y1 , y2 )(t) = y1 y20 −y10 y2 ,suppose that y1 (t0 ) = y2 (t0 ) =
0,then W (y1 , y2 )(t0 ) = 0. Thus then cannot constitute the fundamental set of solutions.
Answer: If so then exist a point t0 such that y1 (t0 )0 = y2 (t0 )0 = 0,
hence W (y1 , y2 )(t0 ) = 0
Answer:
W (y1 , y2 )(t0 )0 = y1 y200 (t0 ) − y100 (t0 )y2 = 0
On the other hand:
0
Z
W (y1 , y2 )(t0 ) = −c exp[−
p(t)dt]p(t0 ) = 0
Then if Wronskian determinant is not zero then p(t0 ) = 0, For arbitrary
solution φ(t) of the ODE which is the linear combination of y1 , y2 ,at
the point t0 we have:
q(t0 )φ(t0 ) = 0
If y1 (t0 ) = y2 (t0 ) = 0,q(t0 ) is arbitrary, otherwise q(t0 ) has to be zero.

Document 6536554

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