FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS
G(x, y, y′) = 0
♦in normal form:
y′=F(x, y)
♦in differential form:
M(x, y)dx +N(x, y)dy = 0
•Last time we discussed first-order linear ODE: y′+q(x)y=h(x).
We next consider first-order nonlinear equations.
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First-Order Ordinary Differential Equations: Nonlinear Cases and Solutions, Study Guides, Projects, Research of Differential Equations
An introduction to first-order nonlinear ordinary differential equations (ODEs), discussing various techniques for solving such equations, including bringing them to separated-variables form, reducing them to linear equations, and bringing them to exact-differential form. The document also includes examples and references for further study.
What you will learn
- How can first-order nonlinear ODEs be brought to separated-variables form?
- What techniques can be used to solve higher-order ODEs that can be reduced to first-order problems?
- What are some examples of first-order nonlinear ODEs and their solutions?
- How can first-order nonlinear ODEs be brought to exact-differential form?
- What is the method for reducing first-order nonlinear ODEs to linear equations?
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FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS
G
x, y, y
′ ) = 0
in normal form
y ′ =
F
x, y
in differential form
M
x, y
dx
N
x, y
dy
Last time we discussed first-order
linear
ODE:
y ′ + q ( x ) y = h ( x ).
We next consider first-order
nonlinear
equations.
NONLINEAR FIRST-ORDER ODEs
No general method of solution for 1st-order ODEs beyond linear case;
rather, a variety of techniques that work on a case-by-case basis.
Examples:
i) Bring equation to separated-variables form, that is,
y ′ = α ( x )
/β
( y ) ;
then equation can be integrated.
Cases covered by this include
y ′ =
ϕ ( ax
(^) +
by
);
y ′ =
ϕ ( y/x
) .
ii) Reduce to linear equation by transformation of variables.
Examples of this include Bernoulli’s equation.
iii) Bring equation to exact-differential form, that is
M
(^) ( x, y
) dx
(^) +
N
(^) ( x, y
) dy
= 0
such that
M
=
∂φ/∂x
, N
=
∂φ/∂y
.
Then solution determined from
φ
( x, y
) =
const.
physically relevant nonlinear equations which can be solved analytically : Although no general method for solution is available, there are several cases ofFirst order nonlinear equations Separable equations
d
( )
d
(
)
y
f
x
x
g y
=
(
)
( )
g y dy
f
x dx
=
Solution :
2
d
e
d
x
y
y
x
=
Ex 1
2
e d x
dy
x
y
=
1
e x
c
y
!
=
i.e
1
(e
)
x
y
c
!
=
or
Almost separable equations
d
(
)
d y
f ax
by
x
=
z
ax
by
=
d
d d
d y
x z
x
a
b
=
Change variables :
1
d
(
( ))
x
z
a
b f
z
=
.
d
( )
d z
a
bf
z
x
!
2
d
( 4
)
d y
x
y
x
!
4
z
y
x
=
!
2
d
d
4
4
d
d
z
y
z
x
x
!
= (^) "
=
"
1
2
4
2
ln(
)
C
z z
x
+!
=
4 4
(
e
)
(
e
)
4
2
x x
k k
y
x
! +
"
=
Ex 2
k a constant
z
ax
by
=
d
( )
d z
a
bf
z
x
!
(^2)
2
d
(
) e
d
y x
y
xy
y
x
y
x
! /
!
=
Ex 3
2
v
v
2
(
v)
e vdv
(v
v)
v
e
ln
v
(
v)
x
x
!
" +! = # =.
v
u
!
1
1
2
1
1
e
e
(
)e d
e
[
]
u
u
u
u
u
u
!
!
!
=
.
"
v
v
v
y
x
y
x
!
!
=
"
=
.
Change variables
To evaluate integral change variables
..
ln
1
x y x y
e
i e
x
Homogeneous
Homogeneous but for constants
2
1
2
dy
x
y
dx
x
y
=
'
,
'
x x a y y b
= + = + ' ' ' '
.
'
'
dy
dy
dy
dx
dy
dx
dx
dx
dx
dx
!
=
=
=
' ' 2 ' 1 2 '
'
'
2
dy
x
y
a
b
dx
x
y
a
b
= + + + + 1 2 0
a
b
=
2
0
a
b
=
3,
1
a
b
= (^)!
=
'
'
2
'
'
'
'
dy
x
y
dx
x
y
=
Homogeneous
Exercise:
Solve the equation
y
′
y/x
x
2 /y
with initial condition
y (1) = 2
This equation is Bernoulli with
n
Set
z
y 2
. Then
z ′ −
z/x
x 2 .
Integrating factor
I
x
) = 1
/x
z ( x
) =
x
[
dx x
2 /x
- const
] =
x
3 / 2 + const
. x
Thus
y = z 1 / 2 = ± √ x 3 /
2 + const
. x
Initial condition
y (1) = 2
y ( x
) =
x^ 3
- 7
x
Homework
1 .
Solve the differential equation
2
dxdy
=
y ( x (^) +
(^) y
)
x 2
“homogeneous”
with
y (1) =
−
1 .
[Answ.:
y
=
(^) x/
(
(^) −
(^2) √
x ) ]
2 .
Solve the differential equation
dx dy
(^) xy
=
xy
2
“Bernoulli”
with
y (0) = 1
/ 2 .
[Answ.:
y
= 1
/ (1 +
(^) e
x 2 / 2 ) ]
Example: Solve the equation
xy
′
2 tan
(^) y
.
This equation can be rewritten as 2 x
(^) sin
(^) y dx
x
2
cos
(^) y dy
i . e .,
M
x, y
x
(^) sin
y ,
N
x, y
x
2
cos
(^) y ,
which is exact because
∂x ∂φ
x
(^) sin
(^) y
φ
( x, y
x
2 sin
(^) y
α
( y )
∂y^ ∂φ
x
2
cos
(^) y
x
2
cos
(^) y
α
′ ( y ) =
x 2
cos
(^) y
α
= constant
Therefore
φ
( x, y
x
2
sin
(^) y
- c
and the general solution is determined by
x
2
sin
(^) y
= const
y
( x
) = arcsin
const
./x
2 )
• DIFFERENTIAL EQUATIONS AND FAMILIES OF CURVES
General solution of a first-order ODE
y ′ =
f (^) ( x, y
contains an arbitrary constant:
y
x, c
one curve in
x, y
plane for each value of
c
general solution can be thought of as one-parameter family of curves
Example:
y ′ =
x/y
separable equation
y dy
x dx
y 2 / 2 =
x 2 / 2 +
c
i . e .,
x
2
y 2
= constant
family of circles centered at origin
Fig.
x
y
Homework
a) Find the family of curves corresponding to solutions of the ODE
y ′ = (
y 2 − x 2 ) /
(
xy
) .
b) Find the orthogonal trajectories to the above family of curves.
homogeneous equation
y ′ =
f (^) ( y/x
)
with
f (^) ( y/x
) = (
y/x
(^) −
x/y
) / 2
solvable by
y →
v
=
y/x
and separation of variables
⇒
x 2
(^) y
2
=
cx
:
family of circles tangent to
y
−
(^) axis at 0
y
x
Fig.
orthogonal trajectories found by solving
y ′ =
−
2 xy/
( y 2 −
(^) x
2 )
⇒ x 2 + y 2 =
ky
:
family of circles tangent to
x
(^) −
(^) axis at 0
EXPLOITING FIRST-ORDER METHODS TO TREAT EQUATIONS OF
HIGHER ORDER IN SPECIAL CASES
♣
y
not present in 2nd-order equation
F
(^) ( x, y, y
′ , y
′′ ) = 0
⇒
setting
y ′ =
q
yields 1st-order equation for
q ( x ) .
♣
x
not present in 2nd-order equation
F (^) ( x, y, y
′ , y
′′ ) = 0
⇒
setting
y ′ =
q, y
′′
=
dq/dx
=
q ( dq/dy
)
yields
G
( y, q, dq/dy
) = 0
.
2
x
y
T
θ T
s
Example: homogeneous, flexible chain
hanging under its own weight
ρ =
(^) linear mass
density
Using Newton’s law, the shape y(x) of the chain obeys
the 2nd−order nonlinear differential equation
y
= a
1 + (y )
2
,
a
ρ
g / T
Setting y
= q
q
= a
1 + q
Homework
1 .
Find the function
y ( x )
obeying the differential equation
y ′ 2
=
x 2 y ′′
and the conditions
y (0) = 2
, y ′ (1) = 2
.
[ Hint: set
y ′ =
q
and apply separation of variables.
]
[Answ.:
y ( x ) = 2(
(^) −
(^) x
) (^) −
(^) 4 ln(
(^) −
(^) x/
]
2 .
Find the function
y ( x )
obeying the differential equation
y ′′
=
y ′ e y
and the conditions
y (0) = 0
, y ′ (0) = 1
.
[ Hint:
y ′ = q ; y
′′
=
dq/dx
=
q ( dq/dy
) ; solve equation for
q ( y ). ]
[Answ.:
y ( x ) =
−
(^) ln(
(^) −
(^) x
) ]
Summary
No general method of solution for 1st-order ODEs beyond linear case; rather, a variety of techniques that work on a case-by-case basis.
Main guiding criteria:
methods to bring equation to separated-variables form
methods to bring equation to exact differential form
transformations that linearize the equation
1st-order ODEs correspond to families of curves in
x ,
y
plane
geometric interpretation of solutions
Equations of higher order may be reduceable to first-order problems in
special cases — e.g. when
y
or
x
variables are missing from 2nd order
equations