2.2 Separable Equations
dy
= g(t) · f (y)
dt
1
h(y) dy = g(t) dt, where h(y) =
f (y)
y 0 = g(t) · f (y) ⇔
The equation f (y) = 0 gives possible lost solutions.
Z
Z
h(y) dy = g(t) dt
Integrate the last equation and find y(t).
Examples:
1. y 0 = y 2
(1)
The equation (1) is in a normal form. We have:
dy
= y2 ⇒
dt
Z
y
−2
Z
dy =
dt
1
1
1
=C −t ⇒ y =
⇒ − =t−C ⇔
y
y
C −t
When divided by y 2 the equation could lose the solution y = 0. To check if it is a solution or
not we plug it into the equation (1).
y = 0 is a solution because (1) becomes a true equality 0 = 0.
2. y 0 = 2ty 2
dy
= 2t · y 2 ⇔
dt
(2)
Z
y
−2
Z
dy =
1
1
2t dt ⇔ − = t2 + c ⇔ y = − 2
y
t +c
A possible lost solution is y = 0 (division by y 2 had occurred). To check we plug it into the
equation (2) which holds if y = 0.
Hence, y = 0 is also a solution.
Explicit vs. Implicit Solutions
3. t2 y 2 y 0 + 1 = y
(3)
y−1
dy
1 y−1
Normal form: y = 2 2 ⇔
= 2· 2
⇔
ty
dt
t
y
0
1
Z
y2
dy =
y−1
Z
dt
t2
y2 − 1 + 1
(y − 1)(y + 1) + 1
1
=
=y+1+
y−1
y−1
y−1
y2
1
+ y + ln |y − 1| = − + c
2
t
is an implicit solution.
When divided by t2 y 2 the equation could lose the solution y = 0. Also y − 1 = 0 ⇔ y = 1
could be a solution. To check if they are solutions or not we plug them into the equation (1).
y = 0 is not a solution. y = 1 is a solution.
4. y 0 =
1
2t2 y
(4)
(It is the equation (2) where y and t are switched.)
Z
Z
1
1
y2
1
1
dy
−2
= 2
⇒
y dy =
t dt ⇔
=
−
+ c1
dt
2t y
2
2
2
t
1
⇔ y 2 = − + c is an implicit solution.
t
Interval of existence
Initial conditions
5. (#13, page 35)
y0 =
y
,
x
y(1) = −2
dy
dx
=
⇔ ln |y| = ln |x| + c ⇔ |y| = ec |x| = A1 |x|
y
x
where A1 = ec > 0.
Let A = ±A1 , i.e. the constant A can be positive or negative.
Then the general solution is y = Ax.
A possibly lost solution y = 0 (not of IVP) is a solution which is easy to check. Assuming that
A can be equal 0 we write the general solution as
y = Ax with A ∈ (−∞, ∞)
IC: y(1) = A = −2
Therefore, the particular solution of the IVP is y = −2x.
The function y = −2x is defined for all x but the DE is not defined at x = 0 ⇒ the interval
of existence of the solution is (0, ∞) since from the IC we have t = 1 and 1 ∈ (0, ∞).
(Note: the lost solution y = 0 is a solution of the equation not of the IVP).
2
6.
y0 =
cos x
y
Consider two initial conditions:
IC(1): y(0) = 1
IC(2): y(0) = −2
Solution:
y2
c
y dy = cos x dx,
= sin x +
2
2
√
y 2 = 2 sin x + c, y = ± 2 sin x + c is the general solution.
R
R
IC(1): y(0) = 1 > 0,
y(0) =
√
c = 1,
Hence, the particular solution is y =
Interval of existence: 2 sin x + 1 > 0
c=1
√
2 sin x + 1
⇐⇒
sin x > − 12
⇐⇒
− π6 + 2πk 6 x 6
7π
6
+ 2πk,
where k is an integer number.
The interval of existence has to contain the point x = 0. Therefore it is the one when k = 0,
i.e. − π6 , 7π
.
6
IC(2): y(0) = −2 < 0,
√
y(0) = − c = −2,
c=4
√
And the particular solution is y = − 2 sin x + 4.
Interval of existence: (−∞, ∞).
Half-life problem:
The DE
N 0 = −λN,
λ>0
is used to model the number of remaining nuclei N in a radioactive substance. This equation
is separable.
Its solution is N (t) = Ae−λt .
Indeed,
dN
dN
= −λN ⇒
= −λ dt ⇒
dt
N
ln N = −λt + C ⇒ N = e−λt+C
⇒ N (t) = Ae−λt , where A = eC .
3
If an initial number of nuclei N (0) = N0 then N (t) = N0 e−λt
Let T1/2 be the half-life. That means N (t + T1/2 ) = 21 N (t) for any t. Then
N (t + T1/2 ) = Ae
−λT1/2
⇒ e
=
1
2
−λ(t+T1/2 )
= Ae−λt e
⇒ T1/2 =
ln 2
λ
−λT1/2
= N (t) · e−λT1/2 = N (t) ·
because
ln
1
2
1
2
= − ln 2 .
Example (#25 p. 35) Suppose that 100mg of tritium 3 H decays to 80mg in 4 hours.
Find the half-life of tritium.
Solution:
Let N (t) define the amount of 3 H at time t.
We have N (0) = 100,
N (4) = 80 (t is given in hours)
N (t) = N (0)e−λt = 100e−λt
Let T be the half-life. Then N (T ) = 21 N (0) = 50
N (4) = 80 ⇒ 100e−4λ = 80,
(e−λ )4 = .8,
e−λ = (.8)1/4 .
Then N (t) = 100e−λt = 100(e−λ )t = 100((.8)1/4 )t = 100(.8)t/4
N (T ) = 100(.8)T /4 = 50 ⇒ (.8)T /4 = 12 ,
Hence T = −
(.8)T =
ln 16
ln(.8)
4
1
,
16
T ln(.8) = − ln 16