Introductory Differential Equations using Sage - William Stein

1.4. FIRST ORDER ODES - SEPARABLE AND LINEAR CASES 23
at least in principle - just integrate both sides 5 . For a more general type of ODE, such as
y ′ = f(t,y),
this fails. For instance, if y ′ = t + y then integrating both sides gives y(t) = ∫ dy
∫ dt dt =
y ′ dt = ∫ t+y dt = ∫ t dt+ ∫ y(t)dt = t2 2 +∫ y(t)dt. So, we have only succeeded in writing
y(t) in terms of its integral. Not very helpful.
However, there is a class of ODEs where this idea works, with some modification. If the
ODE has the form
then it is called separable 6 .
To solve a separable ODE:
y ′ = g(t)
h(y) , (1.3)
(1) write the ODE (1.3) as dy
dt = g(t)
h(y) ,
(2) “separate” the t’s and the y’s:
(3) integrate both sides:
h(y)dy = g(t)dt,
∫
∫
h(y)dy =
g(t)dt + C (1.4)
I’ve added a “+C” to emphasize that a constant of integration must be included in
your answer (but only on one side of the equation).
The answer obtained in this manner is called an “implicit solution” of (1.3) since it
expresses y implicitly as a function of t.
Why does this work? It is easiest to understand by working backwards from the formula
(1.4). Recall that one form of the fundamental theorem of calculus is d ∫
dy h(y)dy = h(y).
If we think of y as a being a function of t, and take the t-derivative, we can use the chain
rule to get
g(t) = d dt
∫
g(t)dt = d dt
∫
h(y)dy = ( d ∫
dy
h(y)dy) dy
dt = h(y)dy dt .
So if we differentiate both sides of equation (1.4) with respect to t, we recover the original
differential equation.
5 Recall y ′ really denotes dy
dt , so by the fundamental theorem of calculus, y = R dy
dt dt = R y ′ dt = R f(t)dt =
F(t) + c, where F is the “anti-derivative” of f and c is a constant of integration.
6 It particular, any separable DE must be first order, ordinary differential equation.