# 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.

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