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CH2 First-Order Differential Equations
2.1 Solution Curves Without a Solution
Slope:
Because a solution
y= y ( x )
of a first-order DE
dy
dx
=f ( x , y )
is necessarily a
differentiable function on its interval I of definition, it must also be continuous on I.
The function
f
in the normal form
dy
dx
=f ( x , y )
is called slope function or rate
function.
The value f ( x , y )
that the function f
assigns to the point represents the slope of a
line or, as we shall envision it, a line segment called a lineal element.
e.g.
Direction Field:
The collection of all the lineal elements is called a direction field or a slope field of
the differential equation
dy
dx
=f ( x , y )
Autonomous first-order DEs:
An ordinary differential equation in which the independent variable does not appear
explicitly is said to autonomous.
Normal form:
dy
dx
=f ( y )
Critical point/equilibrium point/stationary point:
A real number c is a critical point of the autonomous differential equation if it is a
zero of f
If
c
is a critical point of
dy
dx
=f ( y )
, then
y ( x )=c
is a constant solution of the
autonomous differential equation.
A constant solution y ( x )=c
is called an equilibrium solution.
e.g.
The differential equation
dP
dt
=P ( a−bP)
The equilibrium solutions are
P ( t) = 0
and
P ( t) =
a
b
The arrows on the line shown in Figure 2.1.5 indicate the algebraic sign of
f ( P) =P ( a−bP )
on these intervals and whether a nonconstant solution P ( t)
is
increasing or decreasing on an interval.
Figure 2.1.5 is called a one-dimensional phase portrait , or simply phase portrait , of
the differential equation
dP
dt
=P ( a−bP)
. The vertical line is called a phase line.
The following table explains the figure.
Attracters and repellers:
Suppose that y ( x )
is a nonconstant solution of the autonomous differential equation
dy
dx
=f ( y )
and that
c
is a critical point of the DE.
2.2 Separate Variables
Definition 2.2.1 Separate Equation
A first-order DE of the form
dy
dx
=g ( x ) h ( y )
is said to be separate.
Solution:
dy
dx
=g ( x ) h ( y )
dy
h ( y )
=g ( x ) dx
Let p ( y )=
h ( y )
and
y=ϕ ( x )
p ( y ) dy=g ( x ) dx
∫
p ( y ) dy=
∫
g ( x ) dx
Let H ( y )
and G ( x )
are antiderivatives of p ( y )
and g ( x )
, respectively.
H ( y )=G ( x )+ c
2.3 Linear Equations
Definition 2.3.1 Linear Equation
A first-order ODE of the form
a
1
( x )
d y
d x
+a
0
( x ) y=g ( x )
is a linear equation in the dependent variable y.
If g ( x )= 0
, the linear equation is homogeneous.
If g ( x ) ≠ 0
, the linear equation is nonhomogeneous.
Standard form of a linear equation:
d y
d x
The Property:
Its solution is the sum of the two solutions: y= y
c
p
y
c
is the solution of the homogeneous equation.
y
p
is the solution of the nonhomogeneous equation.
d
d x
[
y
c
p
]
y
c
p
]
[
d y
c
d x
c
]
[
d y
p
d x
p
]
= 0 + f ( x )=f ( x )
The homogeneous equation is separable.
d y
y
y
c
=c e
−
∫
P ( x) dx
Let
y
1
( x ) =e
−
∫
P (x ) dx
y
c
=c y
1
( x )
The procedure:
We can find a particular solution of equation by variation of parameters.
Let
u
is a function such that
y
p
=u ( x ) y
1
( x )=u ( x ) e
−
∫
P ( x )dx
d y
p
d x
p
=f ( x )
d u
d x
y
1
+u
d y
1
d x
+uP ( x ) y
1
=f ( x )
d u
d x
y
1
+u
[
d y
1
d x
1
]
=f ( x )
d u
d x
y
1
=f ( x )
d
d x
[ e
− 3 x
y ]= 6 e
− 3 x
e
− 3 x
y=− 2 e
− 3 x
y=− 2 + c e
3 x
,−∞< x< ∞
Ex3.
Solve x
d y
d x
− 4 y=x
6
e
x
d y
d x
x
y=x
5
e
x
P ( x )=
x
P ( x )
and f ( x )
are continuous on ( 0, ∞ )
e
∫
P ( x) dx
=x
− 4
x
− 4
d y
d x
− 4 x
− 5
y =x e
x
d
d x
[ x
− 4
y ]=x e
x
x
− 4
y=
x − 1
e
x
y=x
5
e
x
−x
4
e
x
+c x
4
, 0 < x< ∞.
Values of x
for which a
1
( x )= 0
are called singular points of the equation.
In ex3. x= 0 is a singular point.
Ex.
Solve
d y
d x
P ( x )= 1
e
∫
P ( x) dx
=e
x
e
x
d y
d x
x
y=x e
x
d
d x
[ e
x
y ]=x e
x
e
x
y=
x− 1
e
x
+c
y=x − 1 + c e
− x
y ( 0 )= 0 − 1 +c = 4
c= 5
y=x − 1 + 5 e
−x
,−∞ < x <∞
The contribute of y
c
=c e
− x
to the values of y becomes negligible for increasing
values of x. We say y
c
=c e
− x
is a transient term.
Ex6.
Solve
d y
d x
where
f
x
1, 0 ≤ x ≤ 1
0, x>1.
For x ∈ [ 0,1]
P ( x )= 1
e
∫
P ( x) dx
=e
x
e
x
d y
d x
x
y=e
x
d
d x
[ e
x
y ]=e
x
e
x
y=e
x
y= 1 + c e
− x
y ( 0 )= 1 +c= 0
c=− 1
y= 1 −e
−x
For x ∈ (1, ∞ )
d
d x
[ e
x
y ]= 0
e
x
y=c
2
y=c
2
e
− x
y=
1 −e
−x
, 0 ≤ x ≤ 1
c
2
e
− x
, x> 1
Functions defined by integrals:
Error function:
erf ( x )=
√
π
0
x
e
−t
2
dt
erfc ( x )=
√
π
x
∞
e
−t
2
dt
Ex7.
Solve the IVP
d y
d x
− 2 xy= 2 , y ( 0 )=1.
M ( x , y ) dx+ N ( x , y ) =
∂ f
∂ x
dx +
∂ f
∂ y
dy
∂ M
∂ y
2
f
∂ x ∂ y
∂ N
∂ x
2
f
∂ x ∂ y
∂ M
∂ y
Method of Solution
∂ f
∂ x
=M ( x , y )
f ( x , y )=
∫
M ( x , y ) dx +g ( y )
∂ f
∂ y
∂ y
∫
M ( x , y ) dx+ g
'
( y )=N ( x , y )
g
'
( y )=N ( x , y )−
∂ y
∫
M ( x , y ) dx
Integrate it, then we can get the f ( x , y )
Integrating factors
If M ( x , y ) dx+ N ( x , y ) dy= 0
is not an exact equation. We can multiply an integrating
factor μ ( x , y )
so that μ ( x , y ) M ( x , y ) dx + μ ( x , y ) N ( x , y ) dy = 0
is an exact equation.
To find μ
( μM )
y
=( μN )
x
μ
y
M +μ M
y
=μ
x
N + μ N
x
μ
x
N−μ
y
M=
M
y
−N
x
μ
Suppose μ depends on x
dμ
dx
M
y
−N
x
N
μ
μ ( x )=e
∫
M y
− N x
N
dx
dμ
dy
N
x
−M
y
M
μ
μ ( y )=e
∫
N x
− M y
M
dy
e.g.
The nonlinear first-order differential equation
xydx +( 2 x
2
2
− 20 ) dy= 0
is not exact.
M
y
−N
x
N
x− 4 x
2 x
2
2
− 3 x
2 x
2
2
N
x
−M
y
M
4 x −x
xy
y
dμ
dy
y
μ
μ
y
= y
3
x y
4
dx+( 2 x
2
y
3
5
− 20 y
3
) dy = 0
∂ f
∂ x
=x y
4
f =
x
2
y
4
∂ f
∂ y
= 2 x
2
y
3
'
( y )
g ( y )=
y
6
− 5 y
4
x
2
y
4
y
6
− 5 y
4
=c
2.5 Solutions by Substitutions
Substitutions
Often the first step in solving differential equation consists of transforming it into
another differential equation by means of substitution.
Ex.
Suppose we wish to transform the first-order ODE
dy
dx
=f ( x , y )
by the substitution
y=g ( x ,u )
, where u
is regarded as a function of the variable x.
dy
dx
=g
x
y
du
dx
g
x
y
du
dx
=f ( x , u)
du
dx
=F ( x ,u )
If we can determine u=ϕ ( x )
of this last equation, then a solution of the original
differential equation is y=g
x , ϕ ( x )
Homogeneous Equations
If a function f
possesses the property
f
tx , ty
=t
α
f
x , y
for α ∈ R
, then f
is said to
When n ≠ 0
and n ≠ 1
, the substitution
u= y
1 −n
reduces the equation to a linear
equation.
Ex.
Solve x
dy
dx
2
y
2
dy
dx
x
y=x y
2
let u= y
− 1
y=u
− 1
dy
dx
=−u
− 2
du
dx
−u
− 2
du
dx
u
− 1
x
=x u
− 2
du
dx
x
u=−x
g ( x )=e
− ∫
1
x
dx
x
x
du
dx
x
2
u
x
=−x +c
u=−x
2
+cx
y=
−x
2
+cx
Reduction to Separation of Variables
A differential equation has the form
dy
dx
=f ( Ax+ By+C )
can always be reduced to an equation with separable variables by means of the
substitution u=Ax +By +C, B≠ 0.
Ex.
Solve
dy
dx
=(− 2 x+ y )
2
− 7 , y ( 0 )=0.
Let u=− 2 x + y
du
dx
dy
dx
du
dx
2
du
u
2
=dx
(
u− 3
u+ 3
)
du=dx
ln∨
u− 3
u+ 3
∨¿ x +c
ln
|
u− 3
u+ 3
|
= 6 x+ 6 c
u− 3
u+ 3
=c
1
e
6 x
u=
(
1 + c
1
e
6 x
)
1 −c
1
e
6 x
y=u+ 2 x=
(
1 +c
1
e
6 x
)
1 −c
1
e
6 x
y ( 0 )= 0
3 + 3 c
1
c
1
y=
1 −e
6 x
1 +e
6 x
Riccati’s equation:
The differential equation has the form
dy
dx
=P ( x ) +Q ( x ) y + R ( x ) y
2
is called Riccati’s equation.
(unsolved)
2.6 A Numerical Method
Use the Tangent Line:
Assume that the first-order IVP
y
'
=f
x , y
, y
x
0
= y
0
possesses a solution. One way of approximating this solution is to use tangent lines.
Relative error ¿
absolute error
|actual value|
Percentage relative error:
Percentage relative error ¿
absolute error
|actual value|
× 100
A caveat:
Euler’s method is seldom used in serious calculations. Instead, we will use fourth
order Runge-Kutta method (RK4 method).
