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Introdu
tion
In this
hapter, we will
onsider ordinary di erential equations of rst order, that is,
F(t;y;y0)=0:
Unfortunately,onlyvery few spe
ial types of rst order di erential equations admit solutions that
an be expressed in terms of elementary fun
tions. In fa
t, one
ould say that almost all rst order
di erential equations
an not be expressed. In the following, we will dis
uss several spe
ial
lasses
of rst order ODE that are solvable, illustrated by \textbook" examples, as well as some general
existen
e and uniqueness results.
1 Linear First Order ODE
In this se
tion we will fo
us on linear rst order ODE of following type:
dy
dt +p(t)y = g(t); sometime y_ +p(t)y = g(t):
This
lass of di erential equations are easy to solve. Indeed, here we have our rst existen
e and
uniqueness result:
Theorem: Provided p(t) and g(t) are
ontinuous fun
tions on an open interval I =( ; ), there
exists a unique solution to the initial value problem
dy +p(t)y = g(t); y(t )=y
dt 0 0
on whole interval I,forevery t 2 I and y 2 IR. A
tually, the solution is given by
0 0
Rt (s)g(s)ds+y0 4 Rt p(s)ds
t0 t
y(t)= ; where (t) = e 0 ; 8 t 2 I:
(t)
Idea of the proof: The tri
k of solving this di erential equation is the following: Suppose we
an nd a fun
tion (t) su
h that
_ (t)=(t)p(t):
It follows that, bymultiplying (t) on both sides of the equation,
(t)_y + (t)p(t)y = (t)g(t) ) (t)_y +_(t)y = d [(t)y℄=(t)g(t)
dt
1
whi
h yields
(t)y = Z (t)g(t)dt +
) y = R (t)g(t)dt +
:
(t)
However, su
h (t) alwasy exists as
_ (t) d R p(t)dt+
(t) = dt ln(t)=p(t) ) (t)=e :
All su
h fun
tions di ers up to a multipli
ative fa
tor. A possible
hoi
e is
Rt p(s)ds Rt (s)g(s)ds +y0
t t0
(t)=e 0 ) y= ;
(t)
whi
h is a solution to the di erential equation.
Proof: Existen
e: It is straight-forward to verify that fun
tion y de ned above is a solution
to the di erential equation.
Uniqueness: Suppose both y (t) and y (t) are solutions to the di erential equations. That is
1 2
y0 +p(t)y = g(t); y (t )=y ; 8 i =1;2
i i i 0 0
4
De ne z(t) = y (t) y (t). It follows that
1 2
z0 + p(t)z =0; z(t0)=0;
Multiplying both side by (t)we obtain
0 d
(t)z +(t)p(t)z = dt[(t)z℄=0;
whi
h implies that (t)z(t)
onst. In parti
ular, (t)z(t) (t0)z(t0)=0.It follows that
z(t) 0 sin
e (t) > 0. Therefore y (t) y (t). 2.
1 2
Corollary: Suppose p(t) r 6=0,g(t) b are both
onstant. The di erential equation
y0 = ry+b; y(t0)=y0;
has a unique solution
b r(t t0) b
y(t)= y e + :
0 r r
Remark: Here the existen
e and unique result is a global result, sin
e the solution we nd is
de ned on the whole interval. Later we will see that a lo
al existen
e and uniqueness result
for non-linear rst order di erential equation.
Remark: The multipli
ator (t) is
alled integrating fa
tor.
2
Example (Comparison Prin
iple): Suppose x is a solution to the initial value problem
dx
+p(t)x = g (t); x(t )=x
dt 1 0 0
while y is a solution to the initial value problem
dy +p(t)y = g (t); y(t )=y :
dt 2 0 0
If g (t) g (t); 8t t and x y , then x(t) y(t) for all t t .
1 2 0 0 0 0
Proof: Without loss of generality, assume t =0.De ne z(t)=x(t) y(t). It is easy to see
0
that z satis es di erential equation
z0 +p(t)z = g1(t) g2(t):
R p(t)dt
Multiplying both sides byintegrating fa
tor = e ,wehave
d
(t)z +p(t)(t)z = dt (t)z = (t)(g1 g2) 0
This implies that (t)z(t) is a non-de
reasing fun
tion. In parti
ular,
(t)z(t) (0)z(0) 0 ) z(t)0
for all t. This
ompletes the proof. 2
Exer
ise: (Comparsion Prin
iple) Following the pre
eding example, but assume x(t) is instead a
solution to a general the initial value problem
dx
dt = f(t;x); x(t0)=x0
where f(t;x) is a
ontinuous fun
tion su
h that f(t;x) p(t)x + g2(t) for all x and all
t t0. Show the
omparison inequality still holds, i.e. x(t) y(t) for all t t0.
Example: 1. Solve the following di erential equation
y0 + 1y =1; 8 x>0:
x
2. If wehave initial
ondition y(1) = 1, determine y.
Solution:
1. The integrating fa
tor (x) is determined by
0 (x) d 1
(x)= x ) dxln= x ) ln(x)=lnx+
) (x)=
x:
Any
hoi
e of
will yield a integrating fa
tor, and any one will do. We just pi
k
=1
and wehave
0 0 1 2 x
xy +y =x ) (xy) =x ) xy= 2x +
) y= 2 + x:
3
1 1 1
2. y(1) = 1 yields
= 2. Hen
e y(x)=2 x+ x . 2
Exer
ise: Solve the following di erential equation
xy0 +2y = x4 +1; x>0
Solution: Multiply both sides by (x)=x,we obtain that
6 2
2 0 5 2 0 5 2 x x
x y +2xy =x +x ) (x y) =x +x ) x y = 6 + 2 +
Therefore the general solution of the equation is
4
y = x + 1 +
:
6 2 x2
Example (A geometri
problem) Find the family of
urves with the property that the segmentof
a tangent line drawn between a point of tangen
y and Y-axis is bise
ted bytheX-axis.
Solution: Suppose P =(x;y)isapoint on the
urve, and the tangent line drawn from point
P will interse
t Y -axis at point Q whose
oordinates are, say (0;z). Sin
e the mid-pointof
P and Q has
oordinates
x+0 y+z
;
2 2
wemust have z = y,whi
h implies that
dy =slope of tangentline= y y = 2y:
dx 0 x x
Multiplying integrating fa
tor (x)= 1 ,wehave
x2
d y =0 ) y=
x2 for some
onstant
:
2
dx x
The family of
urves we are looking for is y =
x2. 2
Exer
ise: Find the family of
urves su
h that the tangent line drawn between X-axis and Y-axis
is bise
ted by the point of
onta
t. That is, point P is the mid-pointofQ and R.
4
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