Differential Equations: Types, Classification, and Solutions
Differential Equations
Definition
•
A differential equation is a relationship between
an independent variable,
x
, a dependent
variable, y and one or more differential
coefficients of y with respect to x.
•
For e.g.
The Classification of DE
•
O
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•
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.
•
P
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v
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b
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s
.
The Classification of DE (cont)
•
T
h
e
O
r
d
e
r
a
n
d
D
e
g
r
e
e
o
f
a
D
E
•
The order of a DE is the order of the
highest derivative.
•
The degree of a DE is the degree of the
highest ordered derivative in the equation.
For e.g. y
(4)
+ 6y
2
= 1 + x
2
is the 4
th
order
and 2
nd
degree DE
The Classification of DE (cont)
•
L
i
n
e
a
r
a
n
d
N
o
n
L
i
n
e
a
r
D
E
•
Linear DE is an equation of order n is said
to be linear if
A.
the dependent variable and its derivatives
occur to the first degree only,
B.
no products of dependent variable or any
of its derivatives are present and
C.
no transcendental functions
of dependent variable and or its
derivatives occur.
•
A
n
O
r
d
i
n
a
r
y
D
i
f
f
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[
A
]
–
[
C
]
a
r
e
v
i
o
l
a
t
e
d
.
The Classification of DE (cont)
•
H
o
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s
•
It can only apply to linear DE
•
In a standard way, a DE is arranged so
that all terms containing the dependent
variable occur on the left-hand side (LHS),
and those terms that involve only the
independent variable and constant terms
occur on the right-hand side (RHS).
•
If the RHS of a linear DE is equal to zero,
then the linear DE is an homogeneous DE.
•
If the RHS of a linear DE is not equal to
zero, then the linear DE is a
nonhomogeneous DE
Solution of DE
•
A solution of a DE is
is the relation
involving only the variables and arbitrary
constants, free of any derivatives or
differentials that satisfies the given
differential equation.
•
For e.g. Show that y = sin2x satisfies the
DE y”+3y’+4y=6cos 2x.
First Order DE
•
M
e
t
h
o
d
1
:
D
i
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c
t
I
n
t
e
g
r
a
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n
/
E
l
e
m
e
n
t
a
r
y
D
E
•
The simplest form of a first order DE i.e.
•
This can be solved in principle by
integrating f(x) with respect to x.
•
M
e
t
h
o
d
2
:
B
y
s
e
p
a
r
a
t
i
n
g
t
h
e
v
a
r
i
a
b
l
e
s
•
Consider equations of the form
and of the form
or
i.e. equations in which the right hand side
can be expressed as products or quotients
of functions of
x
or of
y
.
•
Example: Find the general solution of the
DE
3
y
2
y
’ = 2
x
– 1.
•
Example: x(x – 1)y’ = y(y + 1), with the
initial condition y(2)=2.
•
M
e
t
h
o
d
3
:
L
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q
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–
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f
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r
•
1st order linear DE in the form of
•
To solve any such equation, we multiply
both sides by an integrating factor (I(x))
which is always .
•
•
•
Thus,
where I(x) =
•
Example: Solve y’ + 2xy = 4x.
•
Example: Solve y’ + tan
x
y = 2 sec
x
•
M
e
t
h
o
d
4
:
E
x
a
c
t
D
i
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q
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•
D
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f
:
T
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x
p
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s
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M
(
x
,
y
)
d
x
+
N
(
x
,
y
)
d
y
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F
(
x
,
y
)
,
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F
(
x
,
y
)
=
M
(
x
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y
)
d
x
+
N
(
x
,
y
)
d
y
f
o
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a
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l
(
x
,
y
)
i
n
D
.
•
Note that dF(x, y) is the total derivative,
can be expressed as
•
D
e
f
:
I
f
M
(
x
,
y
)
d
x
+
N
(
x
,
y
)
d
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M
(
x
,
y
)
d
x
+
N
(
x
,
y
)
d
y
=
0
i
s
c
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x
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•
T
h
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m
:
L
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t
M
(
x
,
y
)
,
N
(
x
,
y
)
,
,
c
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.
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M(x, y)dx + N(x, y)dy = 0
is an exact differential equation if and
only if
for all (x, y) in D.
•
Example: Given the differential equation
3x(xy -2) +(x
3
+ 2y)dy = 0,
determine whether it is an exact
equation. If so, find the general solution.
•
Example: Determine whether cos
x
y
’ –
y
sin
x
+ 1 = 0 is an exact equation. If
so, find the general solution.
Second Order Linear ODE
•
Most practical problems in engineering
give rise to second order differential
equations of the form
•
where a, b and c are constant coefficients
and f(x) is a given function of x.
Homogeneous Case i.e. f(x) = 0
•
Consider the case where f(x)=0, that is
equation of the form
•
If y
1
and y
2
satisfy this equation, then any
linear combination y = Ay
1
+ By
2
also
satisfies this equation.
•
If
a
= 0, we get the first order equation of
the form where
k
=
c
/
b
.
•
Thus, it has the general solution
where C is an arbitrary constant.
•
I
f
w
e
p
u
t
t
h
e
s
y
m
b
o
l
m
f
o
r
–
k
,
t
h
e
s
o
l
u
t
i
o
n
i
s
y
=
C
e
m
x
.
I
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2
n
d
o
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d
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q
u
a
t
i
o
n
a
y
”
+
b
y
’
+
c
y
=
0
i
.
e
.
a
C
m
2
e
m
x
+
b
A
m
e
m
x
+
c
A
e
m
x
=
0
•
Divide both sides by e
mx
am
2
+bm+c=0
•
T
h
i
s
i
s
c
a
l
l
e
d
a
u
x
i
l
i
a
r
y
e
q
u
a
t
i
o
n
(
A
E
)
o
r
c
h
a
r
a
c
t
e
r
i
s
t
i
c
e
q
u
a
t
i
o
n
.
•
Since the AE is always a quadratic
equation, we have three types of solution
which depends on the roots.
a)
R
e
a
l
a
n
d
d
i
f
f
e
r
e
n
t
r
o
o
t
s
Let m = m
1
and m = m
2
be the two roots
of the AE
.
Then , the general solution is
b)
Real and equal roots
Let m = m
1
= m
2
is the solution of the AE.
Then, the general solution is
c)
Complex roots
Let m =
j
, i.e. m
1
=
+ j
and m
2
=
- j
be the solution of the AE. Then, the
general solution is
•
Example: Solve y”-12y’+36y=0
•
Example: Solve y”+2y’-3y=0
•
Example: Solve y”-9y=0
•
Example: Solve y”+7y=0
Non-homogeneous Case
•
The solution of DE of
are found by the following steps.
1.
Solve and the solution is
called the complementary function, y
C
.
2.
Find a particular solution y
p
for
(Particular integral of DE)
3.
The general solution of the non-homogeneous DE
is
y = y
C
+ y
p
Finding Particular Integral – the Method of
Undetermined Coefficients
•
If f(x) = k, assume as P.I. is y = C.
•
If f(x) = kx, assume as P.I. is y = Cx + D.
•
If f(x) = kx
2
, assume as P.I. is y = Cx
2
+
Dx + E.
•
If f(x) = k sin x or k cos x, assume as P.I.
is y = C cos x + D sin x.
•
If f(x) = e
mx
, assume as P.I. is y = Ce
mx
.
Common term found in P.I. and C. F.
•
Since the C.F. makes the L.H.S of
ay”+by’+cy to be equal to 0 , it is pointless
to use a P.I., a term already contained in
the C.F. If this occurs, multiply the assume
P.I. by
x
and proceed as before. If this too
is already included the C.F., multiply by a
further
x
and proceed as usual.
Some Application Problems
A.
A particle moving a straight line has
displacement
x
, velocity
v
and acceleration
a
at time
t
, where
a
= d
v
/d
t
=
v
d
v
/d
x
. These
formula can be used to find
v
given
a
. The
former is used when
v
is required in terms of
t
,
the latter for
v
in terms of
x
.
i.
A particle moves with retardation proportional
to its velocity i.e.
a
= k
v
and has initial velocity
v
0
and initial displacement 0. Find the velocity
in term of time; and in term of displacement.
B.
A charged oil drop of mass
m
, falling with
velocity
v
, experiences a weight force
mg
downwards, a drag force k
v
upwards
and an electric force
E
sin
t
upwards,
where
g
, k,
E
and
are constant and t is
the time.
i.
Using Newton’s 2
nd
law (i.e
F
=
m
d
v
/d
t
where
F
is the resultant force) derive a
differential equation for
v
.
ii.
Solve the DE to find v in terms of
g
, k,
E,
and
t
.
C.
The equation of motion of a mass on the
end of a vertical spring is y” + 4y =0,
where y is the displacement of the mass
from the static equilibrium position. Given
that the initially (i.e. at time t = 0), the
displacement , y is 3 and the velocity, y’
is 8. Find the solution to the equation in
the form y = C cos(
t +
).
D.
Given an electric circuit with a resistance (
R
) of
2 ohms, an inductance (
L
) of 1 henry, and a
capacitance (
C
) of 0.2 farad, and electromotive
force (
V
) of 17 sin2
t
volts (where t is the time in
seconds), find the current (
i
, in ampere), the
transient current (i.e. that part of the solution
which approaches zero as
t
) and the
steady current (i.e. that part of the solution
which does not approaches zero as
t
),
given that
i
satisfies the differential equation
and that
i
= 0 and di/dt = 0 when t =0.
Differential equations are mathematical equations that relate independent and dependent variables through differential coefficients. They can be classified as ordinary or partial, based on the types of derivatives involved. The order and degree of a differential equation, as well as its linearity and homogeneity, play crucial roles in solving them. Solutions to these equations involve finding relations that satisfy the given conditions. Various methods like direct integration are used to solve first-order differential equations.
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Presentation Transcript
Definition A differential equation is a relationship between an independent variable, x, a dependent variable, y and one or more differential coefficients of y with respect to x. For e.g.
The Classification of DE Ordinary and Partial Differential Equations ODE is an equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable. PDE is an equation involving partial derivatives of one or more dependent variables with respect to more than one independent variables.
The Classification of DE (cont) The Order and Degree of a DE The order of a DE is the order of the highest derivative. The degree of a DE is the degree of the highest ordered derivative in the equation. For e.g. y(4)+ 6y2= 1 + x2 is the 4thorder and 2nddegree DE
The Classification of DE (cont) Linear and Non Linear DE Linear DE is an equation of order n is said to be linear if A. the dependent variable and its derivatives occur to the first degree only, B. no products of dependent variable or any of its derivatives are present and C. no transcendental functions of dependent variable and or its derivatives occur.
The Classification of DE (cont) Homogeneous and nonhomogeneous equations It can only apply to linear DE In a standard way, a DE is arranged so that all terms containing the dependent variable occur on the left-hand side (LHS), and those terms that involve only the independent variable and constant terms occur on the right-hand side (RHS). If the RHS of a linear DE is equal to zero,
Solution of DE A solution of a DE is is the relation involving only the variables and arbitrary constants, free of any derivatives or differentials that satisfies the given differential equation. For e.g. Show that y = sin2x satisfies the DE y +3y +4y=6cos 2x.
First Order DE Method 1: Direct Integration/Elementary DE The simplest form of a first order DE i.e. This can be solved in principle by integrating f(x) with respect to x.
Method 2: By separating the variables Consider equations of the form and of the form or i.e. equations in which the right hand side can be expressed as products or quotients of functions of x or of y. Example: Find the general solution of the DE 3y2y = 2x 1. Example: x(x 1)y = y(y + 1), with the
Method 3: Linear Equations use integrating factor 1st order linear DE in the form of To solve any such equation, we multiply both sides by an integrating factor (I(x)) which is always .
Thus, where I(x) = Example: Solve y + 2xy = 4x. Example: Solve y + tanx y = 2 secx
Method 4: Exact Differential Equation Def: The expression M(x, y)dx + N(x, y)dy is called an exact differential form in a domain D in the xy plane if there exists a function F(x, y), called the potential function, such that dF(x, y) = M(x, y)dx + N(x, y)dy for all (x, y) in D. Note that dF(x, y) is the total derivative, can be expressed as Def: If M(x, y)dx + N(x, y)dy is an exact
Theorem: Let M(x, y), N(x, y), , continuous functions in a domain D. The first order differential equation M(x, y)dx + N(x, y)dy = 0 is an exact differential equation if and only if for all (x, y) in D.
Example: Given the differential equation 3x(xy -2) +(x3+ 2y)dy = 0, determine whether it is an exact equation. If so, find the general solution. Example: Determine whether cos x y y sin x + 1 = 0 is an exact equation. If so, find the general solution.
Second Order Linear ODE Most practical problems in engineering give rise to second order differential equations of the form where a, b and c are constant coefficients and f(x) is a given function of x.
Homogeneous Case i.e. f(x) = 0 Consider the case where f(x)=0, that is equation of the form If y1and y2satisfy this equation, then any linear combination y = Ay1+ By2also satisfies this equation. If a = 0, we get the first order equation of the form where k=c/b.
Thus, it has the general solution where C is an arbitrary constant. If we put the symbol m for k, the solution is y = Cemx. It will also be a solution of the 2nd order equation ay +by +cy=0 i.e. aCm2emx+ bAmemx+ cAemx= 0 Divide both sides by emx am2+bm+c=0 This is called auxiliary equation (AE) or characteristic equation.
Since the AE is always a quadratic equation, we have three types of solution which depends on the roots. a) Real and different roots Let m = m1and m = m2be the two roots of the AE. Then , the general solution is b) Real and equal roots Let m = m1= m2is the solution of the AE. Then, the general solution is
c) Complex roots Let m = j , i.e. m1= + j and m2= - j be the solution of the AE. Then, the general solution is Example: Solve y -12y +36y=0 Example: Solve y +2y -3y=0 Example: Solve y -9y=0 Example: Solve y +7y=0
Non-homogeneous Case The solution of DE of are found by the following steps. 1. Solve and the solution is called the complementary function, yC. 2. Find a particular solution ypfor (Particular integral of DE) 3. The general solution of the non-homogeneous DE is y = yC + yp
Finding Particular Integral the Method of Undetermined Coefficients If f(x) = k, assume as P.I. is y = C. If f(x) = kx, assume as P.I. is y = Cx + D. If f(x) = kx2, assume as P.I. is y = Cx2+ Dx + E. If f(x) = k sin x or k cos x, assume as P.I. is y = C cos x + D sin x. If f(x) = emx, assume as P.I. is y = Cemx.
Common term found in P.I. and C. F. Since the C.F. makes the L.H.S of ay +by +cy to be equal to 0 , it is pointless to use a P.I., a term already contained in the C.F. If this occurs, multiply the assume P.I. by x and proceed as before. If this too is already included the C.F., multiply by a further x and proceed as usual.
Some Application Problems A. A particle moving a straight line has displacement x, velocity v and acceleration a at time t, where a = dv/dt = v dv/dx. These formula can be used to find v given a. The former is used when v is required in terms of t, the latter for v in terms of x. A particle moves with retardation proportional to its velocity i.e. a = kv and has initial velocity v0and initial displacement 0. Find the velocity in term of time; and in term of displacement. i.
B. A charged oil drop of mass m, falling with velocity v, experiences a weight force mg downwards, a drag force kv upwards and an electric force Esin t upwards, where g, k, E and are constant and t is the time. i. Using Newton s 2ndlaw (i.e F = m dv/dt where F is the resultant force) derive a differential equation for v. ii. Solve the DE to find v in terms of g, k, E, and t.
C. The equation of motion of a mass on the end of a vertical spring is y + 4y =0, where y is the displacement of the mass from the static equilibrium position. Given that the initially (i.e. at time t = 0), the displacement , y is 3 and the velocity, y is 8. Find the solution to the equation in the form y = C cos( t + ).
D. Given an electric circuit with a resistance (R) of 2 ohms, an inductance (L) of 1 henry, and a capacitance (C) of 0.2 farad, and electromotive force (V) of 17 sin2t volts (where t is the time in seconds), find the current (i, in ampere), the transient current (i.e. that part of the solution which approaches zero as t ) and the steady current (i.e. that part of the solution which does not approaches zero as t ), given that i satisfies the differential equation and that i = 0 and di/dt = 0 when t =0.
