Lecture Notes in Differential Equations - Bruce E. Shapiro

147
Example 17.5. Solve yy ′′ + (y ′ ) 2 = 0.
Making the substitution z = y ′ and y ′′ = z ′ gives
We can factor out a z,
hence either z = 0 or z ′ = −z. The first choice gives
as a possible solution. The second choice gives
zz ′ + z 2 = 0 (17.39)
z(z ′ + z) = 0 (17.40)
dy
dt = 0 =⇒ y 1 = C (17.41)
dz
z = −dt =⇒ ln z = −t + k =⇒ z = Ke−t (17.42)
where K = e −k is a constant. Hence
dy
dt = Ke−t =⇒ dy = Ke −t dt (17.43)
y = −Ke −t + K 1 (17.44)
where K 1 is a second constant of integration. If we let K 0 = −K then this
solution becomes
y 2 = K 0 e −t + K 1 (17.45)
Since we cannot distinguish between the two arbitrary constants K 1 in the
second solution and C in the first, we see that the first solution is actually
found as part of the second solution. Hence (17.45) gives the most general
solution.
Factoring a Linear ODE
The D operator introduced in example (15.10) will be quite useful in studying
higher order linear differential equations. We will usually write it as a
symbol to the left of a function, as in
Dy = dy
dt
(17.46)
where D is interpreted as an operator, i.e., D does something to whatever
is written to the right of it. The proper analogy is like a matrix: think of
D as an n × n matrix and y as a column vector of length n (or an n × 1