Unformatted text preview: MATH315 2010
(Ordinary Diﬀerential Equations)
Solutions to WrittenAssignment #1
February 6, 2010
Solution of 1(a): One can rewrite the given equation as
y (y 2 − 1)−4/3 dy = xdx,
which is a separable equation, so that a simple integration yields
(y 2 − 1)−1/3 = −1/3(x2 + c).
The initial condition determines c as c = −3(b2 − 1)−1/3 . Therefore
−3
12
2 − 1)−1/3
y = 1 + − x + (b
.
3
Note that we have to choose the + sign since b > 0.
In order to ﬁnd the interval of deﬁnition, we need to know for which values
of x, the expression under the radical is nonnegative. That is, when is
−3
12
2
−1/3
− x + (b − 1)
≥ −1?
3
To this end, we ﬁrst make the following simple observation. For A ∈ R − {0},
A−3 ≥ −1 ⇐⇒ A > 0 or A ≤ −1. Now we consider two cases:
(I) b > 1. Then obviously b2 − 1 > 0. With A = − 1 x2 + (b2 − 1)−1/3 ,
3
note that
3
A > 0 ⇐⇒ x <
;
2 − 1)1/3
(b
and that A ≤ −1 ⇐⇒ x ≥
1 3+ (b2 3
.
− 1)1/3 Therefore, in this case, the interval of deﬁnition is
3
3
x ∈ R : x <
∪ x ∈ R : x ≥ 3 + 2
.
(b2 − 1)1/3
(b − 1)1/3
(II) 0 < b < 1. Then obviously, 0 < 1 − b2 < 1. We can write
1
A = − x2 − b2 − 1−1/3 .
3
In this case, A > 0 cannot happen(why?).
1
1
A ≤ −1 ⇐⇒ − x2 − b2 − 1−1/3 ≤ −1, ⇐⇒ x2 − 1 ≥ −b2 − 1−1/3 ,
3
3
⇐⇒ 12
x ≥ 1 − b2 − 1−1/3 .
3 This implies that
x ≥
3 − 3b2 − 1−1/3 . Solution of 1(b): This is a Bernoulli equation. So, we use the change of
variable u = y −2 to get
1
3
13
− u− 2 u = u− 2 + e2x u− 2
2 or u + 2u = −2e2x . This is a ﬁrst order linear diﬀerential equation with the general solution
1
u = − e2x + ce−2x .
2
Therefore, the general solution for the original equation is seen to be given by
y2 =
Solution of 1(c): With u =
u+x y
x − 1 e2x
2 1
.
+ ce−2x we arrive at du
1
=u+
dx
u or This leads to the general solution
12
u = ln x + c
2
2 udu = dx
.
x or y 2 = 2x2 (ln x + c). By imposing the initial condition y (1) = −4 the constant c is found to be 8.
Thus the explicit solution turns out to be
y = − 2x2 (ln x + 8).
Solution of 1(d): Using the change of variables x = X +
we get the following homogeneous equation 29
7 and y = Y − 38
7 dY
2X + 3Y
=
.
dX
5X + 4Y
Next we put U = Y
X to get
U +X or dU
3U + 2
=
dX
4U + 5 4U + 5
1
dU = − dX
4U 2 + 2U − 2
X
−1/6
14/3
1
+
dU = − dX.
U + 1 4U − 2
X or By integrating we ﬁnd the implicit solution −1/6 ln U + 1 + 7/6 ln 4U − 2 = − ln X  + c.
In terms of x and y , the general solution becomes
y + 38/7
+ 7/6 ln 4 y + 38/7 − 2 = − ln x − 29/7 + c.
−1/6 ln
+ 1
x − 29/7
x − 29/7 Solution of 1(e): First suppose that such integrating factor µ = p(x + y )
exists. We thus have
∂ (M µ)
∂ (N µ)
=
=⇒
∂y
∂x
∂M
∂N
−
p(x + y ) = −(M − N )p (x + y ) =⇒
∂y
∂x
∂M
∂y which means that is a function of x + y only. − ∂N
∂x M −N =− ∂M
∂y − p (x + y )
,
p(x + y ) ∂N
∂x M −N 3 Conversely, assume now that ∂M
∂y − ∂N
∂x
M −N = q (x + y ), for some onevariable
p (t)
function q (t). Take p(t) to satisfy q (t) = −
, and set
p(t)
µ = p(x + y ).
It now easily follows, from the computations above, that µ is an integrating
factor for the original equation.
As an application, we consider the situation where
M = x2 + 2xy − y 2 ,
Since ∂M
∂y − ∂N
∂x M −N = N = y 2 + 2xy − x2 .
2
= q (x + y )
x+y is a function of x + y only, where q (t) = 2 , we can choose µ = (x + y )−2 (since
t
q (t) = − p ((tt)) with p(t) = t−2 .) By multiplying µ to the equation, we arrive at
p
the exact equation
x2 + 2xy − y 2
y 2 + 2xy − x2
dx +
dy = 0,
2
(x + y )
(x + y )2
which has, by integrating, the following implicit solution
F (x, y ) = x2 + y 2
= c.
x+y Solution of 2: Let t, D(t) and X (t) denote, respectively, the time (hour),
the depth of snow at t (unknown unit), and the distance (miles) cleared by the
snowplow. We have the following information
X (0) = 0, X (3) = 2, X (5) = 3. We assume that the snow started to come t0 hours before 8:00 am. The
depth D(t) is proportional to the time (t − t0 ).
On the other hand, since the snow was coming at a constant rate k , and since
the snowplow was clearing the road at a rate which was inversely proportional
to the depth of snow, we have
dX
k
k
=
=
,
dt
D(t)
t − t0
where k , k are some constants. By solving this simple diﬀerential equation we
ﬁnd that
X (t) = k ln t − t0  + C,
4 for some constant C which is to be determined. Using information above, we
arrive at
0 = k ln t0  + C, 2 = k ln 3 − t0  + C, 3 = k ln 5 − t0  + C. By subtraction, we obtain
3 − t0
2 = k ln
t0 ,
5 − t0
1 = k ln
3 − t0 . Now by division, we get the following equation in which only t0 is involved:
(3 − t0 )3 = t0 (5 − t0 )2 or (t0 − 3)3 = t0 (5 − t0 )2 . (Note that t0 < 0.) The ﬁrst equation leads to nothing, whereas the second one
gets us to the situation where
t2 + 2t0 − 27 = 0.
0
Since t0 < 0, we have to dispose the positive solution of this equation and
therefore, the only acceptable value for t0 is
√
t0 = −1 − 28 ≈ −6.29
which means that the snow started ﬁve six and seventeen minutes before eight
o’clock in the morning, that is, it started at 1:43 am.
Solution of 3:
(A). From θ = θ1 − θ2 , we derive
tan θ sin θ1 cos θ2 − cos θ1 sin θ2
cos θ1 cos θ2 + sin θ1 sin θ2 = tan(θ1 − tan θ2 ) = = tan θ1 − tan θ2
m1 − m2
=
1 + tan θ1 tan θ2
1 + m1 m2 (B). The slope of the given curve at the point P : (x, y ) is derived as
2x + 2yy = 0,
and at the point P , x
m1 = y (x) = − ,
y c2 = (x2 + y 2 ). Hence the orthogonal trajectory must satisfy the geometric conditions:
x
x
tan(π /4)(1 + m1 m2 ) = (1 − m2 ) = m2 +
y
y
Therefore, we derive the governing equation:
x
x
1 − y = y +
.
y
y
5 or (x + y ) y + x − y = 0. This is a homogeneous equation. To solve this equation, let u(x) = y (x)/x.
Hence, we have
u−1
xu + u =
,
u+1
or
u+1
dx
du = −
2+1
u
x
u+1
1
dx
du = tan−1 u + ln(1 + u2 ) = −
= − ln x + C
2+1
u
2
x
It follows that tan−1
or y
x + 1
1
ln(x2 + y 2 ) − ln x2 = − ln x + C
2
2 tan−1
In polar coordinates, y
x + 1
ln(x2 + y 2 ) = C.
2 θ + ln r = C, 6 r = C1 e−θ . ...
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 Winter '09
 Differential Equations, Equations, Following, Christopher Nolan

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