Sample Solutions of Assignment 2 for MAT3270B: 2.6-2.4

Transcription

Sample Solutions of Assignment 2 for MAT3270B: 2.6-2.4
Note: We amend 6(b),10.(c) and 10.(f) in this Sample solutions.
1. Determine if the following equation is exact. If it is exact, find out
the solution.
0
(a). (x4 + 4y) + (4x − 3y 8 )y = 0
0
(b). (x + y) + (2x − y)y = 0
0
(c). (2xy 2 + 2y) + (2x2 y + 2x)y = 0
(d). (x log x + xy)dx + (y log x + xy)dy = 0
(e).
x
3
(x2 +y 2 ) 2
dx +
ydy
3
(x2 +y 2 ) 2
=0
(a). Answer: The original ODE can be rewritten as
(x4 + 4y)dx + (4x − 3y 8 )dy = 0
∂M (x, y)
=4
∂y
∂N (x, y)
=4
∂x
(x,y)
(x,y)
So this ODE is exact for ∂M∂y
= ∂N∂x
. Thus there is a ψ(x, y) such
that
∂ψ
= M (x, y) = x4 + 4y
∂x
Integrating above equation, we obtain
1
ψ(x, y) = x5 + 4xy + f (y)
5
Setting ψy = N gives
0
f (y) = −3y 8
1
2
then we get
f (y) =
−1 9
y
3
Hence,
1
1
ψ(x, y) = x5 + 4xy − y 9
5
3
and the solutions of original equation are given by
1 5
1
x + 4xy − y 9 = c
5
3
(b). Answer: The original ODE can be rewritten as
(x + y)dx + (2x − y)dy = 0
∂M (x, y)
=1
∂y
∂N (x, y)
=2
∂x
(x,y)
(x,y)
6= ∂N∂x
.
So this ODE is not exact for ∂M∂y
(c). Answer: The original ODE can be rewritten as
(2xy 2 + 2y)dx + (2x2 y + 2x)dy = 0
∂M (x, y)
= 4xy + 2
∂y
∂N (x, y)
= 4xy + 2
∂x
(x,y)
(x,y)
= ∂N∂x
that
∂ψ
= M (x, y) = 2xy 2 + 2y
∂x
ψ(x, y) = x2 y 2 + 2xy + f (y)
0
f (y) = 0
then we get
f (y) = constant
3
Hence,
ψ(x, y) = x2 y 2 + 2xy + constant
x2 y 2 + 2xy = c
(d). Answer: The original ODE can be rewritten as
(x log x + xy)dx + (y log x + xy)dy = 0
∂M (x, y)
=x
∂y
y
∂N (x, y)
= +y
∂x
x
∂M (x,y)
(x,y)
So this ODE is not exact for ∂y 6= ∂N∂x
.
(e). Answer:
∂M (x, y)
−3xy
=
5
∂y
(x2 + y 2 ) 2
∂N (x, y)
−3xy
=
5
∂x
(x2 + y 2 ) 2
So this ODE is exact for
∂M (x,y)
∂y
=
∂N (x,y)
.
∂x
Thus there is a ψ(x, y) such
that
x
∂ψ
= M (x, y) =
3
∂x
(x2 + y 2 ) 2
−1
ψ(x, y) =
1 + f (y)
(x2 + y 2 ) 2
0
f (y) = 0
then we get
f (y) = constant
Hence,
ψ(x, y) =
−1
(x2
1
+ y2) 2
+ constant
4
−1
(x2
1
+ y2) 2
=c
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 +
= 0, µ(x, y) = yex
2
(c). (3x + y6 )dx + ( xy + 2 xy )dy = 0, µ(x, y) = xy
(a). Answer: Multiplying the original ODE by µ, we get the following equation
1 + y2
dy
y3
Integrating both sides of the above equation, and the solutions of the
−xdx =
original equation are given by
log |y| −
1
x2
+
=c
2y 2
2
(b). Answer: Multiplying the original ODE by µ = yex , we get the
following equation
(ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0
∂M (x, y)
= ex cos y − 2 sin x
∂y
∂N (x, y)
= ex cos y − 2 sin x
∂x
(x,y)
(x,y)
= ∂N∂x
that
∂ψ
= M (x, y) = ex sin y − 2y sin x
∂x
5
ψ(x, y) = ex sin y + 2y cos x + f (y)
0
f (y) = 0
then we get
f (y) = constant
Hence,
ψ(x, y) = ex sin y + 2y cos x + constant
ex sin y + 2y cos x = c
(c). Answer: Multiplying the original ODE by µ = xy, we get the
following equation
(3x2 y + 6x)dx + (x3 + 2y 2 )dy = 0
∂M (x, y)
= 3x2
∂y
∂N (x, y)
= 3x2
∂x
(x,y)
(x,y)
= ∂N∂x
that
∂ψ
= M (x, y) = 3x2 y + 6x
∂x
ψ(x, y) = x3 y + 3x2 + f (y)
0
f (y) = 2y 2
then we get
2
f (y) = y 3
3
6
Hence,
2
ψ(x, y) = x3 y + 3x2 + y 3
3
2
x3 y + 3x2 + y 3 = c
3
3. Show that if
Nx −My
xM −yN
= R, where R depends on the quality xy
only, then the different equation M (x, y)dx + N (x, y)dy = 0 has an
integrating factor of the form µ(xy). Using this to solve the following
ODE.
6
x2
y dy
(3x + ) + ( + 3 )
=0
y
y
x dx
R
Answer: Let ω(z) = e
R(Z)dz
, which is well defined for R depends on
the quality xy only. Then setting µ(x, y) = ω(xy) and multiplying the
original ODE by µ(x, y), we get the following equation
M (x, y)µ(x, y)dx + N (x, y)µ(x, y)dy = 0
Set M = M µ, N = N µ
R
∂M (x, y)
= e R(Z)dz (My + M R(Z)x), Z = xy
∂y
R
∂N (x, y)
= e R(Z)dz (Nx + N R(Z)y), Z = xy
∂x
Nx −My
= R, where R
xM −yN
∂N (x,y)
. So the original
∂x
If
depends on the quality xy only, then
∂M (x,y)
∂y
=
ODE has an integrating factor of the form of
µ(xy).
Now, we solve the special equation with method above. It is easy to
get R(xy) =
1
xy
R
and ω(z) = e
1
z
=z
7
Multiplying the original ODE by µ = ω(xy) = xy, we get the following equation
(3x2 y + 6x)dx + (x3 + 3y 2 )dy = 0
∂M (x, y)
= 3x2
∂y
∂N (x, y)
= 3x2
∂x
(x,y)
(x,y)
= ∂N∂x
that
∂ψ
= M (x, y) = 3x2 y + 6x
∂x
ψ(x, y) = x3 y + 3x2 + f (y)
0
f (y) = 3y 2
then we get
f (y) = y 3
Hence,
ψ(x, y) = x3 y + 3x2 + y 3
x3 y + 3x2 + y 3 = c
4. Finding 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
(b). dx + ( xy − sin y)dy = 0
(c). ydx + (2xy − e−2y )dy = 0
8
(d). ex dx + (ex cot y + 2y csc y)dy = 0
(a). Answer:
My −Nx
N
∂M (x, y)
= 3x2 + 2x + 3y 2
∂y
∂N (x, y)
= 2x
∂x
= 3 is a function of x only, then there is an integrating factor µ
that also depends on x only and satisfies the following equation
dµ
= 3µ
dx
Integrating above equation, we get µ = e3x . Multiplying the original
ODE by µ = e3x , we get the following equation
(3e3x x2 y + 2e3x xy + e3x y 3 )dx + (e3x x2 + e3x y 2 )dy
∂M (x, y)
= 3e3x x2 + 2e3x x + 3e3x y 2
∂y
∂N (x, y)
= 3e3x x2 + 2e3x x + 3e3x y 2
∂x
So the resulting ODE is exact for
∂M (x,y)
∂y
=
∂N (x,y)
.
∂x
Thus there is a
ψ(x, y) such that
∂ψ
= M (x, y) = 3e3x x2 y + 2e3x xy + e3x y 3
∂x
1
ψ(x, y) = e3x x2 y + e3x y 3 + f (y)
3
0
f (y) = 0
then we get
f (y) = constant
Hence,
1
ψ(x, y) = e3x x2 y + e3x y 3
3
9
1
e3x x2 y + e3x y 3 = c
3
(b). Answer:
My −Nx
−M
1
y
=
is a function of y only, then there is
an integrating factor µ that also depends on y only and satisfies the
following equation
dµ
=
dy
Integrating above equation, we get
1
µ
y
µ = y. Multiplying the original
ODE by µ = y, we get the following equation
ydx + (x − y sin y)dy = 0
∂M (x, y)
=1
∂y
∂N (x, y)
=1
∂x
So the resulting ODE is exact for
∂M (x,y)
∂y
=
∂N (x,y)
.
∂x
ψ(x, y) such that
∂ψ
= M (x, y) = y
∂x
ψ(x, y) = xy + f (y)
0
f (y) = −y sin y
then we get
f (y) = y cos y − sin y
Hence,
ψ(x, y) = xy + y cos y − sin y
xy + y cos y − sin y = c
Thus there is a
10
(c). Answer:
My −Nx
−M
=
1−2y
−y
is a function of y only, then there is
following equation
dµ
1 − 2y
=
µ
dy
−y
Integrating above equation, we get µ = y1 e2y . Multiplying the original
ODE by µ, we get the following equation
1
e2y dx + (2xe2y − )dy = 0
y
∂M (x, y)
= 2e2y
∂y
∂N (x, y)
= 2e2y
∂x
(x,y)
So the resulting ODE is exact for ∂M∂y
=
∂N (x,y)
.
∂x
Thus there is a
ψ(x, y) such that
∂ψ
= M (x, y) = e2y
∂x
ψ(x, y) = xe2y + f (y)
0
f (y) =
1
−y
then we get
f (y) = − log y
Hence,
ψ(x, y) = xe2y − log y
xe2y − log y = c
(d). Answer:
My −Nx
−M
= cot y is a function of y only, then there is
11
following equation
dµ
= cot yµ
dy
Integrating above equation, we get µ = sin y. Multiplying the original
sin yex dx + (ex cos y + 2y)dy = 0
∂M (x, y)
= ex cos y
∂y
∂N (x, y)
= ex cos y
∂x
(x,y)
=
∂N (x,y)
.
∂x
Thus there is a
ψ(x, y) such that
∂ψ
= M (x, y) = sin yex
∂x
ψ(x, y) = sin yex + f (y)
0
f (y) = 2y
then we get
f (y) = y 2
Hence,
ψ(x, y) = sin yex + y 2
sin yex + y 2 = c
5. State the region in the ty-plane where the (existence and uniqueness)
hypotheses of Theorem 2.4.2 are satisfied.
12
(a).
(b).
dy
dt
dy
dt
=
=
log |ty|
1−t2 +y 2
1+t2
3y−y 2
(a). Answer: Setting f (t, y) =
log |ty|
1−t2 +y 2
If ty > 0 and 1 − t2 + y 2 6= 0 then
1
(1 − t2 + y 2 ) − 2y log ty
∂f
y
=
∂y
(1 − t2 + y 2 )2
Obviously f and
∂f
∂y
are continuous if ty > 0 and 1 − t2 + y 2 =
6 0.
If ty < 0 and 1 − t2 + y 2 6= 0 then
1
(1 − t2 + y 2 ) − 2y log −ty
∂f
y
=
∂y
(1 − t2 + y 2 )2
Obviously f and
∂f
∂y
are continuous if ty > 0 and 1 − t2 + y 2 =
6 0.
Hence, if t 6= 0, y 6= 0 and 1 − t2 + y 2 6= 0, the hypotheses of Theorem
2.4.2 are satisfied.
(b). Answer: Setting f (t, y) =
1+t2
3y−y 2
If 3y − y 2 6= 0 then
∂f
(−3 + 2y)(1 + t2 )
=
∂y
(3y − y 2 )2
Obviously f and
∂f
∂y
are continuous if 3y − y 2 6= 0.
Hence, if 3y − y 2 6= 0, the hypotheses of Theorem 2.4.2 are satisfied.
6. Verify if the following functions satisfy the Lipschitz condition
|f (t, y1 ) − f (t, y2 )| ≤ L|y1 − y2 |, (t, y1 ) ∈ R, (t, y2 ) ∈ R
at the given domain R.
√
(a). f (t, y) = t y, R = [−2, 2] × [1, 10],
√
(b). f (t, y) = t y, R = [−2, 2] × [0, 10],
(c). f (t, y) = (sin t)(sin y), R = (−∞, +∞) × (−∞, +∞).
13
(a). Answer:
√
√
|f (t, y1 ) − f (t, y2 )| = |t y1 − t y2 |
|t|
|y1 − y2 |, if y1 + y2 6= 0
y1 + y2
√
√
In R = [−2, 2]×[1, 10], we have y1 +y2 6= 0, y1 + y2 ≥ 2, and |t| ≤ 2.
√
So let L = 2, we can get
=√
|f (t, y1 ) − f (t, y2 )| ≤ L|y1 − y2 |,
(t, y1 ) ∈ [−2, 2] × [1, 10], (t, y2 ) ∈ [−2, 2] × [1, 10].
(b). Answer: Assume that there exists L, which satisfy the condition
√
√
|f (t, y1 ) − f (t, y2 )| = |t y1 − t y2 | ≤ L|y1 − y2 |,
(t, y1 ) ∈ R = [−2, 2] × [0, 10], (t, y2 ) ∈ R = [−2, 2] × [0, 10]
Obviously L > 0. Let t = 32 , y1 =
1
, y2
4L2
=
1
,
16L2
then
3
√
√
|t y1 − t y2 | =
8L
L|y1 − y2 | =
3
16L
So we have
√
√
|t y1 − t y2 | > L|y1 − y2 |
This is a contradiction. So f does not satisfy the Lipschitz condition
in R.
(c). Answer: |f (t, y1 ) − f (t, y2 )| = | sin t|| sin y1 − siny2 |
= 2| sin t|| sin
y1 − y2
y1 + y2
y1 − y2
|| cos
| ≤ 2| sin
| ≤ |y1 − y2 |
2
2
2
for | sin x| ≤ |x|. So f satisfies the Lipschitz condition in R.
14
7. Solve the following ODE and state the interval of Existence in terms
of y0 .
(a). y 0 = 2ty 2 , y(0) = y0
0
(b). y =
t2
, y(0)
y(1+t2 )
= y0
dy
= 2tdt, y 6= 0
y2
Integrating the above equation, we get
−1
y(t) = 2
t +c
y0
−1
From y(0) = y0 , then c = y0 and y(t) = 1−y
2
0t
Since y(t) 6= 0, then y0 6= 0
1 − y0 t2 6= 0, then t 6= ± √1y0
Hence, the interval of existence is (− √1y0 , √1y0 )
t2 dt
,y =
6 0
1 + t2
Integrating the above equation, we get
1 2
y = t − tan−1 t + c
2
From y(0) = y0 , then c = 12 y0 2 . Substituting to the equation,
ydy =
y 2 = 2t − 2 tan−1 t + y0 2
i.e.
q
y=
2t − 2 tan−1 t + y0 2
Let f (t) = 2t − 2 tan−1 t + y0 2 , then
0
f (t) = (
2t2
) > 0, if t 6= 0;
1 + t2
0
Obviously, f (0) = 0. y(0) = y02 ≥ 0.
Then there exists t0 ∈ (−∞, 0) such that f (t0 ) = 0. Therefore the
15
interval of existence is (t0 , +∞).
8. Show that
(
(t − C)2
0
y(t) =
0
for t ≥ C ≥ 0
for t ≤ C
(1)
1
satisfies y = 2y 2 , y(0) = 0 for any constant C ≥ 0. So we don’t have
uniqueness, why?
Answer: Obviously,
0
y =
0
(
2(t − C)
0
for t ≥ C ≥ 0
for t < C
(2)
1
1
It is easy to check y = 2y 2 and y(0) = 0. The function f (t, y) = 2y 2 is
continuous everywhere, but
∂f
∂y
=y
−1
2
is not continuous when y = 0. So
Theorem 2.4.2 does not apply to this problem and we can not assure
the uniqueness of the solution.
9. Solve the following Bernoulli equations and state the Interval of
Existence.
(a). t2 y 0 + 2ty − y 3 = 0, t > 0, y(1) = 1
0
(b). y = 2y − y 2 , y(0) = 1
(a). Answer: (a) The original ODE can be rewritten as
t2 0 2t
y + 2 −1=0
y3
y
Let v =
1
,
y2
then
−2 dy
dv
= 3
dt
y dt
16
1 0
y
y3
Substituting to the equation, we
dv
−
dt
R
v(t) = e
=
−1 dv
2 dt
get the following resulting equation
4
2
v=− 2
t
t
4
dt
t
=
Z
2 R
(e
t2
[−
2
+ ct4
5t
−4
dt
t
)dt]
Hence, the solutions are
y 2 (t) =
From y(1) = 1, then c =
3
5
2
5t
1
+ ct4
and y 2 (t) =
Since y 6= 0, then t 6= 0 and y(t) =
5t
5
q 3t +2
5t
.
3t5 +2
Therefore the interval of
existence is (0, +∞).
Answer: (b). Let v = y1 , then
dv
−1 dy
= 2
dt
y dt
dy
−1 dv
= 2
dt
v dt
The original ODE can be rewritten as
dv
= dt
−2v + 1
Integrating above equation, we get
2
y(t) = 2(t+c)
e
−1
1
Since y(0) = 1, then c = 2 log 3. Substituting to the equation, the
solution is
2
3e2t − 1
1
2t
From 3e − 1 6= 0, then t 6= 2 log 3. Hence the interval of existence is
y(t) =
( 12 log 13 , +∞).
17
10. Solve the following ODEs
0
(a). y =
0
(b). y =
0
2x+y
, y(0)
3+3y 2 −x
y
t−y
2 2x
(c). xy + y − y e
0
(d). y = e
=0
= 0, y(1) = 1
x+y
(e). xdy − ydx = 2x2 y 2 dy, y(1) = −2
(f).
dy
dx
= − 2xy+1
x2 +2y
0
y3
, y(0) = 1
1−2xy 2
2
+ 1)dx + x x−y dy
(g). y =
(h). (2y
=0
(2x + y)dx + (−3 − 3y 2 + x)dy = 0
∂M (x, y)
=1
∂y
∂N (x, y)
=1
∂x
(x,y)
(x,y)
= ∂N∂x
that
∂ψ
= M (x, y) = 2x + y
∂x
ψ(x, y) = x2 + xy + f (y)
0
f (y) = −3 − 3y 2
then we get
f (y) = −3y − y 3
18
Hence,
ψ(x, y) = x2 + xy − 3y − y 3
x2 + xy − 3y − y 3 = c
To satisfy the initial condition we must choose c = 0, so
x2 + xy − 3y − y 3 = 0
is the solution of the given initial value problem.
ydt + (−t + y)dy = 0
My −Nt
−M
=
−2
y
is a function of y only, then there is an integrating factor
µ that also depends on y only and satisfies the following equation
dµ
−2
=
µ
dy
y
Integrating above equation, we get µ =
1
.
t2
Multiplying the original
y
−1
1
dt + (
+ 2 y)dy = 0
2
t
t
t
∂M (t, y)
1
= 2
∂y
t
∂N (t, y)
1
= 2
∂x
t
(t,y)
∂M (t,y)
So the resulting ODE is exact for ∂y = ∂N∂t
. Thus there is a
ψ(t, y) such that
∂ψ
y
= M (t, y) = 2
∂t
t
−y
+ f (y)
ψ(t, y) =
t
y
0
f (y) = 2
t
19
then we get
f (y) =
1 y2
2 t2
Hence,
−y 1 y 2
+ 2
t
2t
ψ(t, y) =
−y 1 y 2
+ 2 =c
t
2t
(c). Answer: The original ODE can be rewritten as
x dy
1
+ ( − e2x ) = 0
2
y dx
y
Let v =
−1
,
y
then
dv
1 dy
= 2
dx
y dx
dy
1 dv
= 2
dx
v dx
So we get the following resulting equation
dv
x
= v − e2x
dx
It can be written as
(v − e2x )dx + xdv = 0
∂M (x, y)
=1
∂v
∂N (x, y)
=1
∂x
(x,v)
(x,v)
= ∂N∂x
. Thus there is a ψ(x, v) such
So this ODE is exact for ∂M∂v
that
∂ψ
= M (x, v) = v − e2x
∂x
1
ψ(x, y) = vx − e2x + f (y)
2
0
f (y) = 0
20
then we get
f (y) = constant
Hence,
1
ψ(x, y) = vx − e2x
2
and the solutions of resulting equation are given by
1
vx − e2x = c.
2
Obviously, the solutions of the original equations are given by
−x 1 2x
− e = c.
y
2
To satisfy the initial condition we must choose c = −1 − 21 e2 , so
−x 1 2x
1
− e + 1 + e2 = 0
y
2
2
x
(d). Answer: The original ODE can be rewritten as
ex dx − e−y dy = 0
Integrating above equation, the solutions of original equation are given
by
ex + e−y = 0
(e). Answer: The original ODE can be rewritten as
(−y)dx + (x − 2x2 y 2 )dy = 0
My −Nx
N
=
−2
x
∂M (x, y)
= −1
∂y
∂N (x, y)
= 1 − 4xy 2
∂x
is a function of x only, then there is an integrating factor
µ that also depends on x only and satisfies the following equation
dµ
−2
=
µ
dx
x
21
1
.
x2
−y
1
dx + ( − 2y 2 )dy = 0
2
x
x
∂M (x, y)
−1
= 2
∂y
x
∂N (x, y)
−1
= 2
∂x
x
∂M (x,y)
(x,y)
So the resulting ODE is exact for ∂y = ∂N∂x
. Thus there is a
ψ(x, y) such that
∂ψ
−y
= M (x, y) = 2
∂x
x
y
ψ(x, y) = + f (y)
x
0
f (y) = −2y 2
then we get
f (y) =
−2 3
y
3
Hence,
y 2 3
− y
x 3
y 2 3
− y = c.
x 3
To satisfy the initial condition we must choose c =
ψ(x, y) =
10
,
3
so
y 2 3 10
− y −
=0
x 3
3
(f). Answer: The original ODE can be rewritten as
(2xy + 1)dx + (x2 + 2y)dy = 0
∂M (x, y)
= 2x
∂y
22
∂N (x, y)
= 2x
∂x
(x,y)
(x,y)
= ∂N∂x
that
∂ψ
= M (x, y) = 2xy + 1
∂x
ψ(x, y) = x2 y + x + f (y)
0
f (y) = 2y
then we get
f (y) = y 2
Hence,
ψ(x, y) = x2 y + x + y 2
x2 y + x + y 2 = c
(g). Answer: The original ODE can be rewritten as
(y 3 )dx + (2xy 2 − 1)dy = 0
My −Nx
−M
=
−1
y
∂M (x, y)
= 3y 2
∂y
∂N (x, y)
= 2y 2
∂x
is a function of y only, then there is an integrating factor
µ that also depends on y only and satisfies the following equation
dµ
−1
=
µ
dy
y
1
.
y
1
y 2 dx + (2xy − )dy = 0
y
23
∂M (x, y)
= 2y
∂y
∂N (x, y)
= 2y
∂x
(x,y)
=
∂N (x,y)
.
∂x
Thus there is a
ψ(x, y) such that
∂ψ
= M (x, y) = y 2
∂x
ψ(x, y) = xy 2 + f (y)
0
f (y) =
−1
y
then we get
f (y) = − log y
Hence,
ψ(x, y) = xy 2 − log y
xy 2 − log y = c.
To satisfy the initial condition we must choose c = 0, so
xy 2 − log y = 0
(h). Answer:
My −Nx
N
=
1
x
∂M (x, y)
=2
∂y
y
∂N (x, y)
=1+ 2
∂x
x
is a function of x only, then there is an integrating factor
µ that also depends on x only and satisfies the following equation
dµ
1
= µ
dx
x
24
Integrating above equation, we get µ = x. Multiplying the original
(2xy + x)dx + (x2 − y)dy = 0
∂M (x, y)
= 2x
∂y
∂N (x, y)
= 2x
∂x
(x,y)
=
∂N (x,y)
.
∂x
ψ(x, y) such that
∂ψ
= M (x, y) = 2xy + x
∂x
1
ψ(x, y) = x2 y + x2 + f (y)
2
0
f (y) = −y
then we get
f (y) =
Hence,
−1 2
y
2
1
1
ψ(x, y) = x2 y + x2 − y 2
2
2
1
1
x2 y + x2 − y 2 = c.
2
2
Thus there is a

Sample Solutions of Assignment 2 for MAT3270B: 2.6-2.4

Transcription

Similar documents