Lecture Notes in Differential Equations - Bruce E. Shapiro

18 LESSON 3. SEPARABLE EQUATIONS
This equation is separable with
a(t) = 1 − t 2
(3.5)
and
and it can be rewritten as
Integrating,
∫
b(y) = y 2 (3.6)
dy
(1
y 2 = − t )
dt (3.7)
2
y −2 dy =
∫ (
1 − t )
dt (3.8)
2
− 1 y = t − 1 4 t2 + C (3.9)
The initial condition gives
− 1 = 0 − 0 + C (3.10)
hence
Solving for y,
1
y = 1 4 t2 − t + 1 = 1 4 (t2 − 4t + 4) = 1 4 (t − 2)2 (3.11)
y =
4
(t − 2) 2 (3.12)
Often with separable equations we will not be able to find an explicit expression
for y as a function of t; instead, we will have to be happy with an
equation that relates the two variables.
Example 3.2. Find a general solution of
This can be rearranged as
Integrating,
dy
dt = t
e y − 2y
(3.13)
(e y − 2y)dy = tdt (3.14)
∫
∫
(e y − 2y)dy = tdt (3.15)
e y − y 2 = 1 2 t2 + C (3.16)