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Linear Differential Equations with Constant Coefficients

An equation of the form
dn y dn−1 y dn−2 y
a0 + a 1 + a2 + . . . + an y = F (x),
dxn dxn−1 dxn−2
where a0 , a1 , a2 , . . . , an are constants, is called a Linear differential equation of degree n with
constant coefficients.
d d2
Let = D, 2 = D2 , etc. Then the above equation can be written as
dx dx
(a0 Dn + a1 Dn−1 + a2 Dn−2 + . . . + an )y = F (x)

i.e. ϕ(D)y = F (x) (1)

The general or complete solution of (1) consists of two parts namely (i) Complementary Func-
tion (C.F.) and the (ii) Particular Integral (P.I).
i.e. y =C.F.+P.I.
To find the complementary function
Putting D = m and F (x) = 0 in (1).
Therefore the auxiliary equation of (1) is ϕ(m) = 0
i.e. a0 mn + a1 mn−1 + . . . + an = 0.
By solving this equation, we get n roots say m1 , m2 , m3 , . . . , mn .
Case (i): If all the roots are real and unequal, i.e. if m1 ̸= m2 ̸= m3 ̸= . . . ̸= mn , then
C.F.=c1 em1 x + c2 em2 x + c3 em3 x + . . . + cn emn x
Case (ii): If two roots are equal (i.e. m1 = m2 = m) and the remaining be real and unequal,
then
C.F.=(c1 + c2 x)emx + c3 em3 x + . . . + cn emn x
Case (iii): If three roots are equal (i.e. m1 = m2 = m3 = m) and the remaining be real and
unequal, then
C.F.=(c1 + c2 x + c3 x2 )emx + c4 em4 x + . . . + cn emn x
Case (iv): If all the roots are equal (i.e. m1 = m2 = m3 = . . . = mn = m) then
C.F.=(c1 + c2 x + c3 x2 + . . . + cn xn−1 )emx
Case (v): If roots are imaginary i.e. if m = α ± iβ, then
C.F.=eαx (c1 cos βx + c2 sin βx)
1

,2

To find the Particular integral (P.I.)
If the RHS is zero, i.e. F (x) = 0, then there is no particular integral. In this case the comple-
mentary function is the general solution of the given differential equation. On the other hand
if F (x) ̸= 0, then we have P.I. also.
1
The P.I. is given by P.I.= F (x)
ϕ(D)
where F (x) is any one of the following form
(1) F (x) = eax
(2) F (x) = sin ax or cos ax
(3) F (x) = xn , where n is a constant (+ve integer)
(4) F (x) = eax f (x), where f (x) = xn or sin ax or cos ax
Type 1: If F (x) = eax
1 ax
P.I. = e
ϕ(D)
1 ax
= e provided ϕ(a) ̸= 0
ϕ(a)
1 1 ′
If ϕ(a) = 0 then P.I.=x. ′ eax = x. ′ eax , provided ϕ (a) ̸= 0
ϕ (D) ϕ (a)
′ 1 1 ′′
If ϕ (a) = 0 then P.I.=x2 . ′′ eax = x2 . ′′ eax , provided ϕ (a) ̸= 0.
ϕ (D) ϕ (a)
This process may be repeated till the denominator becoming non zero when replacing D by a.
d2 y dy
Example 1: Solve 2
−7 + 12y = 0.
dx dx
Solution: (D2 − 7D + 12)y = 0

The auxiliary equation is m2 − 7m + 12 = 0
⇒ (m − 3)(m − 4) = 0 ⇒ m = 3, 4
The general solution is y = Ae3x + Be4x .
d2 y dy
Example 2: Solve 2
−6 + 13y = 0.
dx dx
Solution: (D2 − 6D + 13)y = 0
The auxiliary 2
√equation is m √ − 6m + 13 = 0
6 ± 36 − 52 6 ± −16
⇒m= =
2 2
⇒ m = 3 ± 2i
The general solution is y = e3x (A cos 2x + B sin 2x).
Example 3: Solve (D2 − 4D + 4)y = 0

, 3

Solution: The auxiliary equation is m2 − 4m + 4 = 0
⇒ (m − 2)2 = 0 ⇒ m = 2, 2
The general solution is y = (A + Bx)e2x .
Example 4: Solve (D2 + 3D + 2)y = e5x
Solution:
The auxiliary equation is m2 + 3m + 2 = 0
⇒ (m + 1)(m + 2) = 0 ⇒ m = −1, −2
The complementary function (C.F.) is Ae−x + Be−2x
To find Particular integral (P.I.):

1
P.I. = e5x
D2
+ 3D + 2
1
= 2 e5x
5 + 3(5) + 2
e5x
=
42
The general solution is
e5x
y = C.F. + P.I. = Ae−x + Be−2x + .
42
d2 y dy
Example 5: Solve 2
+4 + 4y = e−2x .
dx dx
Solution: (D2 + 4D + 4)y = 0
The auxiliary equation is m2 + 4m + 4 = 0
⇒ (m + 2)2 = 0 ⇒ m = −2, −2
The complementary function (C.F.) is (A + Bx)e−2x
To find Particular integral (P.I.):

1
P.I. = e−2x
D2 + 4D + 4
1
= 2
e−2x
(−2) + 4(−2) + 4
e−2x 1
= = x. e−2x
0 2D + 4
1
= e−2x
2(−2) + 4
e−2x x2
= = e−2x
0 2

, 4

The general solution is
x2 e−2x
y = C.F. + P.I. = (A + Bx)e−2x + .
2
Example 6: Solve (D2 + 2D + 1)y = e−x + 3
Solution:
The auxiliary equation is m2 + 2m + 1 = 0
⇒ (m + 1)2 = 0 ⇒ m = −1, −1
The complementary function (C.F.) is (A + Bx)e−x
To find Particular integral (P.I.):


1
P.I. = (e−x + 3e0x )
D2
+ 2D + 1
1 1
= 2 e−x + 3. 2 e0x
D + 2D + 1 D + 2D + 1
1 −x 1
= e + 3. e0x
(−1)2 + 2(−1) + 1 0+0+1
e−x 1
= + 3 = x. e−x + 3
0 2D + 2
e−x x2
= + 3 = e−x + 3
0 2
The general solution is
x2 −x
y = C.F. + P.I. = (A + Bx)e−x + e + 3.
2
Example 7: Solve (D2 + 9)y = e−2x
Solution:
The auxiliary equation is m2 + 9 = 0
⇒ m = − ± 3i
The complementary function (C.F.) is A cos 3x + B sin 3x
To find Particular integral (P.I.):


1
P.I. = 2
(e−2x
D +9
1
= e−2x
(−2)2 + 9
1
= e−2x
13