# 05 Second Order Constant Coefficients

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```					Second Order Constant Coeffici
ent Homogeneous Linear Differe
ntial equations
In this lecture
• We give methods for finding the general sol
ution of a second order homogeneous linear
differential equations with constant coefficie
nts.
A second order homogeneous linear DE wit
h constant coefficients is an equation of the
form

y¢¢ + py¢ + qy = 0              … (1)

where the coefficients p and q are consta
nts (independent of x and y).
We want to find the general solution of such e
quations. For this we shall first look at a first
order equation.
Look at the first order equation
y¢ - 2 y = 0
We note that    y=e   2x
is a solution.

Hence we “guess” a solution of the second
order equation (1) as

y=e     mx
m a constant.

Substituting in(1) we get
(m + pm + q )e
2                    mx
=0
Since   e   mx
¹ 0,   we find that m satisfies

the so called “Auxiliary equation”

( m + pm + q ) = 0
2

The above is a quadratic equation in m. And
so the general solution of (1) depends on the
nature of the roots of the auxiliary equation.
Case (1) The Auxiliary equation has two rea
l and distinct roots, m1, m2 say.

Hence y = e
m1 x
and y = e             m2 x

are two solutions of (1).
They are LI as their Wronskian is
( m1 + m2 ) x
( m2 - m1 )e                         ¹ 0 as m1 ¹ m2
Hence the general solution of (1) is
y = c1e   m1 x
+ c2e   m2 x
c1, c2 are arbitrary
constants.
Case (2) The Auxiliary equation has two rea
l and equal roots, m1, m1 say.
The roots m1 and m2 are equal real numbers iff
p2 – 4q = 0.
Hence    y = y1 = e                m1 x
is one solution of (1)
with          p
m1 = -
2
A second LI solution is                             1 - ò pdx
y = y2 = v y1 where                           v=ò    2
e
y1
1        ò - pdx dx = x
Hence    v=ò            - px
e
e
m1 x
Thus a second LI solution is y2 = v y1=   xe
Hence the general solution of (1) is
y = c1e + c2 xe
m1 x           m1 x
c1, c2 are arbitrary
constants.
Case (3) The Auxiliary equation has two co
mplex roots, say, a + ib , a - ib
Hence
y=e   (a +ib ) x    ax
=e    (cos b x + i sin b x )
is a “complex” solution of (1).
ax
Hence its real part y1 = e cos b x
ax
and its imaginary part y2 = e sin b x
are two real solutions of the d.e. (1)
The Wronskian of     y1 , y 2 is b e 2a x ¹ 0
Thus {y1, y2} is LI.
Hence the general solution of (1) is

y=e    ax
(c1 cos b x + c2 sin b x )
c1, c2 are arbitrary constants.
Summary
D.E. (a0 D + a1D + a2 ) y = 0
2

Auxiliary equation: a m 2 + a m + a = 0
0    1      2
m1,m2 are the roots of the Auxiliary equation
m1,m2             General Solution
Real and unequ
al
y = c1e        m1 x
+ c2 e   m2 x

Real and equal     y = c1e       m1 x
+ c2 x e   m1 x

y =
Complex: a ± i b e a   x
( c1 c o s b   x + c 2 s in b x )
( c1, c2 are arbitrary constants)
Problems Find the general Solution of

1.   y¢¢ + y¢ - 2 y = 0
i.e. ( D + D - 2) y = 0
2

Auxiliary equation: m 2 + m - 2 = 0
Roots: 1, -2 (real and distinct)
Thus the general solution is
-2 x
y = c1e + c2e
x

( c1, c2 are arbitrary constants)
2. Find the general solution of
3 y¢¢ - 5 y¢ + 2 y = 0
i.e. (3D 2 - 5 D + 2) y = 0
Auxiliary equation: 3m 2 - 5m + 2 = 0
Roots: 1, 2/3 (real and distinct)
Thus the general solution is
2
x
y = c1e + c2e
x       3

( c1, c2 are arbitrary constants)
3. Find the general solution of
y¢¢ - 4 y¢ + 4 y = 0
i.e. ( D 2 - 4 D + 4) y = 0
Auxiliary equation: m 2 - 4 m + 4 = 0
Roots: 2, 2 (real and equal)
Thus the general solution is

y = c1e + c2 xe
2x           2x

( c1, c2 are arbitrary constants)
4. Find the general solution of
2 y¢¢ - 2 2 y¢ + y = 0
i.e.(2 D - 2 2 D + 1) y = 0
2

Auxiliary equation: 2m 2 - 2 2m + 1 = 0
1 1
Roots: 2 , 2 (real and equal)
Thus the general solution is
1                1
x                x
y = c1e    2
+ c2 xe    2

( c1, c2 are arbitrary constants)
5. Find the general solution of
y¢¢ + 4 y = 0
i.e.         ( D + 4) y = 0
2

Auxiliary equation:             m +4=0
2

Roots: 2 i , - 2 i (complex)
Thus the general solution is

y = c1 cos 2 x + c2 sin 2 x
( c1, c2 are arbitrary constants)
6. Find the general solution of
y¢¢ + 4 y¢ + 8 y = 0
i.e. ( D + 4 D + 8) y = 0
2

Auxiliary equation: m + 4 m + 8 = 0
2

Roots: -2 + 2i, -2 - 2i (complex)
Thus the general solution is
y=e    -2 x
(c1 cos 2 x + c2 sin 2 x )
( c1, c2 are arbitrary constants)
7. Find the solution of the initial-value problem
1
2 y¢¢ - y¢ - 3 y = 0; y (0) = 2, y¢(0) =
2
Given d.e. is ( 2 D 2 - D - 3) y = 0
Auxiliary equation: 2 m - m - 3 = 0
2

3     (Real and distinct)
Roots: -1,
2
Thus the general solution is
3
x
-x
y = c1e + c2e     2

( c1, c2 are arbitrary constants)
x = 0 gives y = c1 + c2 =2
Differentiating the general solution, we get
3
3           x
y¢ = -c1e + c2e
-x         2
2
3   1
x = 0 gives y - c1 + c2 =
=                        2   2
Solving we get c1 = 1, c2 = 1
Thus the solution to the given problem is
3
x
-x
y =e +e     2
8. Find the solution of the initial-value problem
y¢¢ + 4 y¢ + 13 y = 0; y (0) = 0, y¢(0) = -3
Given d.e. is ( D + 4 D + 13) y = 0
2

Auxiliary equation: m + 4 m + 13 = 0
2

Roots: -2 ± 3i (Complex)
Thus the general solution is
y=e   -2 x
(c1 cos 3 x + c2 sin 3 x )
( c1, c2 are arbitrary constants)
x = 0 gives y = c1 = 0
Differentiating the general solution, we get
y¢ = e (-3c1 sin 3x + 3c2 cos3x )
-2 x

- 2e   -2 x
(c1 cos3x + c2 sin 3x )
x = 0 gives y 3c2 - 2c1 = -3
=
Solving we get c1 = 0, c2 = -1
Thus the solution to the given problem is
-2 x
y = -e          sin 3 x
Euler’s Equidimensional equation
The second order homogeneous LDE
x y¢¢ + px y¢ + qy = 0
2

(p and q are constants)
is called an Euler’s Equidimensional equation.
z.
We put x = e Note that    dx
=e =x
z

dz
dy dy dz dy 1
\ y¢ =   = ´ = ´
dx dz dx dz x
2
d      d y 1 1 dy   1
y¢¢ =    y¢ = 2 ´ ´ + ´ - 2
dx     dz  x x dz  x
dy              2
\ xy¢ =    , x 2 y¢¢ = d y - dy
dz             dz 2
dz
Hence the given Euler’s equation becomes
2
d y            dy
2
+ ( p - 1) + qy = 0
dz             dz
which is a constant coefficient homogeneous l
.d.e.(with z as independent variable) and henc
e can be solved.
9. Find the general solution of
x y¢¢ + xy¢ - 4 y = 0
2

Putting x = ez, the given equation becomes
2
d y
2
- 4y = 0
dz
Auxiliary equation: m - 4 = 0
2

Roots: 2 , -2 (Real and distinct)
Thus the general solution is
-2 z                1
y = c1e + c2e
2z
= c1 x + c2 2
2

x
( c1, c2 are arbitrary constants)
10. Find the general solution of
x y¢¢ + xy¢ + 9 y = 0
2

Putting x = ez, the given equation becomes
2
d y
2
+ 9y = 0
dz
Auxiliary equation:          m +9 = 0
2

Roots: ± 3i (Complex)
Thus the general solution is
y = c1 cos3 z + c2 sin 3 z ( c1, c2 are arbitrary constants)
= c1 cos(3ln x ) + c2 sin(3ln x )
Problem 6 (page 98):
Consider the general homogeneous equ
Consider the general homogeneous equ
ation
ation     y¢¢ + P ( x ) y¢ + Q ( x ) y = 0 (1)
& change the independent variable from
& change the independent variable from
x to z= z(x), where z(x) is an unspecified
x to z= z(x), where z(x) is an unspecified
function of x. Show that (1) can be trans
function of x. Show that (1) can be trans
formed in this way into an equation with c
formed in this way into an equation with c
onstant coefficient if and only if
onstant coefficient if and only if
(Q¢ + 2 PQ ) is constant, in which case
3/ 2
Q
will effect the desired result.
z = ò Q( x)dx
Let y " + P( x) y '+ Q( x) y = 0        (1)
where P(x) and Q(x) are function of x and l
et independent variable be changed from x
to z = z(x).
2                     2
dy dy dz       d y d æ dy ö dz dy d z
= .      and 2 = ç ÷ . + . 2
dx dz dx       dx    dx è dz ø dx dz dx
d2y    d æ dy ö dz dz dy d 2 z
i.e.    2
= ç ÷. . + . 2
dx    dz è dz ø dx dx dz dx

Substituting in (1) we get
æ d 2 y ö æ dz ö2 dy d 2 z dy dz
ç 2 ÷ ç ÷ + . 2 + P( x) . + Q( x) y = 0
è dz ø è dx ø dz dx        dz dx
2
d y    dy
i.e.    2
+ p1 + q1 y = 0 ... (2)
dz       dz
2
d z   dz            Q
2
+P     & q1 =           2
where p1 =
dx       dx
2
(dz / dx )
(dz / dx )
Q
Let q1 =             2
= k (Const.)
(dz / dx )
2      dz
Q( x) = k (dz / dx )         or    = ± Q( x) / k
dx
2
d z     1    1
2
=            Q¢( x )
dx      k 2 Q( x)

d 2 z / dx + P ( x)dz / dx     Q¢ / 2 kQ + P Q / k
p1 =                    2           =
(dz / dx )                      Q/k
Q¢ + 2 PQ k é Q¢ + 2 PQ ù k
p1 =          . =ê     3/ 2 ú 2
.
2 k Q 2 ë Q             û
Thus, Eq (2) can be solved if and only if
é Q¢ + 2 PQ ù
p1 = ê     3/ 2  ú = const.
ë Q         û
Now             2
æ dz ö = Q   dz
ç ÷        Þ    = Q/k
è dx ø   k   dx
1
z = ò Q( x)dx
k
Problem 7, (Page 98): Solve

Solution: Compare the given eqn with
2
We choose z such that    æ dz ö = Q
ç    ÷
è dx ø   k
Now changing the independent variable fro
m x to z, by using the above given relation t
he given eqn is transformed into
Where

Solve the above eqn we get
Solve the DE

Solution is not possible by the above met
hod, because p1 is not constant.

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 views: 20 posted: 12/8/2011 language: English pages: 32