First-order Linear Equations-Examples

1. Homogeneous: y' + p(t)y = 0, y = y(t). We assume p(t) continuous
for t > 0. We always have:
i) solutions of the IVP with > 0 are defined for all t > 0,
ii) solutions are either positive for all t > 0, negative for all t > 0, or
zero for all t (depending on the sign of ),
iii) if p(t) doesn't change sign , y(t) is either monotone increasing or
monotone decreasing.

Ex.1 . The solution is :



Solutions are increasing for < 0, decreasing for > 0, in either case, the
asymptotic value is zero .

Ex.2 . The solution is:



Thus if > 0 solutions are decreasing, with asymptotic values (as
t → +∞), (as t → -∞).

Remark. In general, the asymptotic value as t → +∞ will be non-zero
( positive if > 0) when the following improper integral converges:



Ex.3 . We expect solutions with > 0 will be
positive, decreasing and have positive asymptotic value (using the criterion
just mentioned.)
Indeed, the solution is: , with the limit as
t → ∞.

So far the solutions have been monotone. To get oscillations, we consider
examples where p(t) is oscillatory.

Ex.4a. y' + (sin2t)y = 0, y(0) = . The solution is:

Since cos 2t oscillates between -1 and 1, the solution y(t) oscillates between
the constant positive values and .

Ex. 4b. y' + (3 + sin t)y = 0, y(0) = . Here p(t) is everywhere positive
(although it oscillates with mean value 3), so we expect solutions with > 0
will be positive, monotone decreasing, with asymptotic value zero . Indeed

we find:



which alternately touches the curves ( decreasing exponentials ):



while remaining strictly decreasing (as t → +∞). It doesn' t really 'oscillate'.

2. Non- homogeneous equations . y'(t)+ p(t)y = f(t). Assuming p(t) and
f(t) continuous for t > 0, the solutions will be defined for t > 0. Beyond
that, there is little one can say in general. The general solution is:



which has the form ( is the solution of the associated homoge-
neous equation .) The integral defining is often impossible to compute
explicitly.

Ex.5 . The solution is:



With > 0, y(t) tends to infinity as t → 0 and t → +∞, and has a unique
minimum at > 0, with .

Ex.6 y' + 3y = e^7t, y(0) = . Solution: .

Ex.7 y' + 2ty = t, y(0) = . The solution is:



All solutions are even functions, defined for all t ∈ R, and tending to the
constant solution 1/2 as t →±∞: Incidentally, the constant solution can be
found directly, simply by setting y' = 0 in the equation:

2ty = t => y ≡ 1/2.